Class 8 Maths - Squares And Square Roots

Find the square root of the following numbers correct to two decimal places:
$5.462$

$\surd 5.462=2.337=2.34$
It can be written as
1703888_1806500_ans_6151d0076d75405689cc37ff90992381.png

Find the square root of the following decimal numbers by division method:
$18.4041$

$\sqrt {18.4041} =4.29$

1704004_1806488_ans_4753a9950c8e42489c58a719030c7332.png

Find the square root of the following numbers by prime factorization method:
$1849$

We know that
square root of

$1849$ can be written as

$\sqrt{1849}=\sqrt{43 \times 43 }$

$=43$

Find the square root of the following numbers correct to two decimal places:
$645.8$

$\surd 645.8=25.41$
It can be written as
1703904_1806497_ans_a3959811af0442bda5e1b806c445c1b1.png

Find the square root of the following numbers correct to two decimal places:
$107.45$

$\sqrt{107.45}=10.36$
It can be written as
1703894_1806498_ans_1e9eede7086e4354b18199942ae82e1c.png

Find the square root of the following numbers by prime factorization method:
$6241$

We know that
square root of

$6241$
It can be written as

$\sqrt{6241}=\sqrt{79 \times 79 } = 79$
1703476_1806501_ans_b15b3a5e418e454a864d8137a354843e.png

Find the square root of the following decimal numbers by division method:
$5.774409$

$\surd 5.774409=2.403$
By division method
1703962_1806491_ans_6357b19bf315467abe7d806116936caf.png

Find the square root of the following numbers by prime factorization method:
$441$

We know that
square root of

$441$
It can be written as

$\sqrt{441}=\sqrt{3 \times 3 \times 7 \times 7}$
So we get

$=3 \times 7$

$=21$

1703983_1806489_ans_43757576ac874724b26e2f2683902fe5.png

Find the square root of the following numbers by prime factorization method:
$784$

We know that
square root of

$784$
It can be written as

$\sqrt{784}=\sqrt{2 \times 2 \times 2 \times 2 \times 7\times 7}$
So we get

$=2 \times 2 \times7$

$=28$

Find the square root of the following numbers by prime factorization method:
$4356$

We know that
square root of

$4356$
It can be written as

$\sqrt{4356}=\sqrt{2 \times 2 \times 3 \times 3 \times 11 \times 11 }$
So we get

$=2 \times 3\times 11$

$66$

1703908_1806495_ans_13f935fbc46e4dacb369b7f8fb6a30de.png

Find the square root of the following numbers correct to two decimal places:
$3$

$\surd 3=1.73$
It can be written as
1703451_1806506_ans_e0c84f64e36f4c6b951edb5ac85f64e0.png

Find the square root of the following numbers by prime factorization method:
$9025$

We know that
square root of

$9025$
It can be written as

$\sqrt{9025}=\sqrt{5 \times 5 \times 19 \times 19 }$
So we get

$=5 \times 19$

$=95$

1704277_1806550_ans_92f43a0e06da4af99a5b1f8d979afe58.png

Find the square root of the following number by prime factorization method:
$8281$

Square root of

$8281$ can be written as;

$\sqrt{8281}=\sqrt{7 \times 7 \times 13 \times 13 }$

$=7 \times 13$

$=91$

1704261_1806548_ans_3be65bb5b910477dbc40046778c09895.png

Which of the following natural numbers are perfect squares? Give reasons in support of your answer.
$5488$

We know that
It can be written as

$5488=2 \times 2 \times 2 \times 2 \times 7 \times 7 \times 7$
Here
After pairing the same prime factors, one factor

$7$ is left unpaired.
Therefore,

$5488$ is not a perfect square.
1705741_1806672_ans_d0e276ad5d2346dab18a8d58e6911427.PNG

Which of the following natural numbers are perfect squares? Give reasons in support of your answer.
$243$

We know that
It can be written as

$243=3 \times 3 \times 3 \times 3 \times 3$
Here
After pairing the same prime factors, factor

$3$ is left unpaired.
Therefore,

$243$ is not a perfect square.
1705755_1806675_ans_367125a497fa4f518f6b0d4c24ea6c56.PNG

Find the square root of the following numbers by prime factorization method:
$0.0064$

$0.0064 = 64/10000$
By squaring we get

$\sqrt{\dfrac{64}{10000}} = \dfrac{\sqrt{64}}{\sqrt{10000}}$
We know that
So we get

$=(2 \times 2 \times 2 )/(2 \times 2 \times 5 \times 5)$

$=8/100$

$=0.08$

1705605_1806562_ans_3652bec540cb448989bdb14856dcb3f7.PNG

Show that each of the following numbers is a perfect square. Also, find the number whose square is the given number.
$7056$

We know that
It can be written as

$7056=2 \times 2 \times 2 \times 2\times 3\times 3 \times 7 \times 7$
Here
After pairing the same prime factors, no factor is left
Therefore,

$7056$ is a perfect square of

$2\times 2 \times 3 \times 7 =84$

Find numbers from $100$ to $400$ that end with $0, 1, 4, 5, 6$ or $9$ , which are perfect squares.

Following are the perfect squares between

$100$ to

$400$

$10^2 = 100, 11^2 = 121$

$12^2 = 144, 13^2 = 169$

$14^2 = 196, 15^2 = 225$

$16^2 = 256, 17^2 = 289$

$18^2 = 324, 19^2 = 361, 20^2 = 400$

All the above numbers ends with either

$0,1,4,5,6$ or

$9$

Thus, the numbers from

$100$ to

$400$ that end with

$0,1,4,5,6$ or

$9$ , which are perfect squares are given above.

Find the square root of the following numbers by factorization $256$

Prime Factors of

$256$ are

$256 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2$

Taking one prime number put from each pair we get:

$\sqrt{256} = 2 \times 2 \times 2 \times 2$

Hence,

$\sqrt{256} = 16$

Find the smallest positive integer with which one has to multiply $847$ to get a perfect square.

The given number is

$847$

But,

$847 = 11 \times 11 \times 7=11^2 \times 7$
Here,

$7$ occur only once. So, to get a perfect square number we need to have one more

$7$ .

Thus, to get perfect square we have to multiply the given number

$847$ by

$7$ .

$1156$

Prime factors of 1156 will be

$1156 = 2\times 2\times 17\times 17$

Taking one number out of the pair of prime numbers we get:

$\sqrt{1156} = 2 \times 17$

Square root of

$1156 = 34$

Find the square root of $10404$ by prime factorization method.

Prime factors of

$10404$ are,

$10404 = 2 \times 2 \times 3 \times 3 \times 17 \times 17$

Now,

$\sqrt{10404} = \sqrt{2 \times 2\times 3\times 3 \times 17\times 17}$

$\sqrt{10404} = 2 \times 3\times 17$

Therefore, the Square root of

$10404$ will be

$2\times 3\times 17=102$

Simplify: $\sqrt{169} \times \sqrt{361}$

Given:

$\sqrt{169} \times \sqrt {361}$

Prime factors of

$169=13\times 13=13^2$

$\implies \sqrt{169}=\sqrt{13^2}=13$

Also, prime factors of

$361=19\times 19=19^2$

$\implies \sqrt{361}=\sqrt{19^2}=19$

Thus,

$\sqrt{169}\times \sqrt{361}=13 \times 19=247$

Simplify: $\sqrt{1764} - \sqrt{1444}$

Prime factors of

$1764$ are

$1764=2\times 2\times 21\times 21$

so,

$\sqrt { 1764 } =2\times 21=42$

Prime factors of

$1444$ are ,

$1444=2\times 2\times 19\times 19$

so,

$\sqrt { 1444 } =2\times 19=38$

as per problem,

$\sqrt { 1764 } -\sqrt { 1444 }$

$=42-38$

$=4$

Simplify: $\sqrt{1360 + 9}$

$\sqrt { 1360+9 }$

$=\sqrt { 1369 }$

$=\sqrt{37\times 37}$

$=\sqrt{37^2}$ .......

$37$ is a prime and it can't be expanded

$=+37$ or -37.

Find the square root by prime factorization method $13225$

Prime factors of

$13225$ will be

$13225 = 5\times 5\times 23\times 23$

Taking one number from each pair of the prime numbers we get:

$\sqrt{13225} = 5 \times 23$

Hence

$\sqrt{13225} = 115$

Simplify $\sqrt{2704} + \sqrt{144} + \sqrt{289}$

Prime factors of 2704 are

$2704=$

$2\times 2\times 2\times 2\times 13\times 13$

$\sqrt { 2704 } =2\times 2\times 13=52$

Prime factors of

$144$ are

$144=12\times 12$

$\sqrt { 144 } =12$

Prime factors of

$289$ are

$17\times 17$

so,

$\sqrt { 289 } =17$

$\sqrt { 2704 } +\sqrt { 144 } +\sqrt { 289 }$

$=52+12+17$

$=81$

Simplify: $\sqrt{225} - \sqrt{25}$

It is known that

$\sqrt{225}=15$ and

$\sqrt{25}=5$

According to the given condition,

$\sqrt{225}-\sqrt{25}=15-5$

$=10$

Simplify $\sqrt{100} + \sqrt{36}$

$\sqrt{100}=\sqrt{(10)^2}$

$=10$

Also,

$\sqrt{36}=\sqrt{(6)^2}$

$=6$

Thus,

$\sqrt{100}+\sqrt{36}=10+6$

$=16$

Find the largest perfect square factor of $48$

the prime factors of

$48$ are

$2\times 2\times 2\times 2\times 3=48$

Hence after making the group of all

$2's$ we get the largest perfect square factorof

$48$ will be,

$2\times 2\times 2\times 2=16$

Find the largest perfect square factor in the number $1352$ .

$1352 = 2 \times 2 \times 2 \times 13 \times 13$

$=(2 \times 13) \times (2 \times 13) \times 2$

$=26 \times 26 \times 2$
Now,

$26 \times 26=676$
Therefore,

$676$ is the largest perfect square factor of

$1352$ .

Find the smallest positive integer with which one has to multiply $450$ to get a perfect square.

The given number is

$450$

But,

$450 = 5 \times 5 \times 3 \times 3 \times 2=5^2\times 3^2 \times 2$
Here,

$2$ occur only once.

So, to get a perfect square we need to multiply by

$2$
Thus, we have to multiply by

$450$ by

$2$ to get a perfect square.

Find the smallest positive integer with which one has to multiply to $1445$ to get a perfect square.

The given number is

$1445$

But,

$1445 = 17 \times 17 \times 5=17^2 \times 5$ .
Here,

$5$ occur only once.

Thus, we have to multiply

$1445$ by

$5$ to get a perfect square.

Find the smallest positive integer with which $1352$ be multiplied to get a perfect square.

The given number is

$1352$ . Breaking the number as product of prime factors.

$1352 = 2 \times 2 \times 13 \times 13 \times 2=2^2 \times 13^2 \times 2$

To get a perfect square, the prime numbers should occur even number of times.
Here,

$2$ occurs three times (odd number of times).
Thus, we have to multiply

$1352$ by

$2$ to get a perfect square.

Find the nearest integer to the square root of $232$ .

We know,

$15^2 = 225$ and

$16^2 = 256$
Also we have,

$225 < 232 < 256$

$\implies 15^2 < 232 < 16^2$

Difference of

$225$ and

$232$ is

$7$ whereas difference of

$256$ and

$232$ is

$24$ .
So, the square root of

$232$ is nearest to

$15$ .

Hence, the answer is

$15$

Find the nearest integer to the square root of $824.$

We know,

$28^2 = 784$ and

$29^2 = 841$
Also, we have,

$784 < 824 < 841$

$\implies 28^2 < 824 < 29^2$

Difference of

$784$ and

$824$ is

$40$ whereas difference of

$824$ and

$841$ is

$17$ .
So, the square root of

$824$ is nearest to

$29$ .

Hence, the answer is

$29$ .

Find the largest perfect square factor in $729$

Largest perfect square factor in 729 is 729

as prime factors of 729 are

$3\times 3\times 3\times 3\times 3\times 3=729$

If the area of a square is $90$ cm $^2$ , what is its side-length, rounded to the nearest integer?

Since

$A$

$= l^2$ , we have

$l^2 =90$ . But

$81 < 90 < 100$ and

$81$ is nearer to

$90$ than

$100$ . Hence the nearest integer to

$\sqrt{90}$ is

$\sqrt{81}$ = 9.

A square piece of land has area 112 m $^2$ . What is the closest integer which approximates the perimeter of the land?

$l$ is the side-length of a square, its perimeter is

$4l$ . We know that

$l^2$ =

$112$ . Hence,

$(4l)^2 = 16l^2 = (16) \times (112) = 1792.$
But

$42^2 = 1764 < 1792 < 1849 = 43^2$ and

$1764$ is nearer to

$1792$ than

$1849$ . Hence the integer approximation for

$\sqrt{1792}$ is

$42$ . The approximate value of the perimeter is

$42$ cm.

Find the square root of 0.017424.

Write

$0.017424 = \frac{17424}{1000000}$ . Observe

$17424=18\times 968 = 18\times 8\times 121= 3^2\times 4^2\times 11^2$
and

$1000000 = 1000^2$ . Hence

$\sqrt{17424} = 3\times 4\times 11=132$ and

$\sqrt{1000000}=1000$ . We obtain

$\sqrt{0.017424} =\frac{\sqrt{17424}}{\sqrt{1000000}} = \frac{132}{1000} = 0.132$

Find the square root of the following numbers by the factorisation numbers : 82944

Resolving

$82944$ as the product of primes, we get

$82944=2^{10} \times 3^4$

$\Rightarrow \sqrt{82944}= \sqrt{2^{10} \times 3^4}= 2^5 \times 3^2$

$\therefore \sqrt{ 82944} =288.$

Explain with reason whether the given number is perfect square or not: $2433$

Given a number 6098 is not perfect square. Need to find out the reason.

Now, we have the perfect square m = n× n = n² , where m and n are integers

⇒ 6098 is not a perfect square as it cannot be expressed as n × n form the product of two equal integer
$\text{and the number that have 2,3,7 or 8 in the units place are not perfect squares.}$

Hence, the given number 6098 is not a perfect square.

Find whether the square of the given number is even or odd: 8204

Consider units digit of a number

$8204$ and square the units digit number.
⇒ Here, Units digit number is

$4$
⇒

$4^2 = 16$
Thus units digit of square of

$8204$ is

$6$
∴ square of 8204 will be again even number

Explain with reason whether the given number is perfect square or not: $6098$

Given a number 6098 is not perfect square. Need to find out the reason.

Now, we have the perfect square m = n × n = n² , where m and n are integers

⇒ 6098 is not a perfect square as it cannot be expressed as n × n form the product of two equal integer
$\text{and the number that have 2,3,7 or 8 in the units place are not perfect squares.}$

Hence, the given number 6098 is not a perfect square.

What will be the units digit of the square of the given number:
$8742$

Unit place number in

$8742$ is

$2.$
So,

$2^2 = 4$

Hence units digit in

$8742^2$ will be

$4.$

Find the square root of $1296$ by Prime Factorization

Resolving

$1296$ into prime factors, we get

$$1296=(2\times 2)\times(2\times2)\times(3\times3)\times(3\times3)$$

So, square root of $$1296=2\times2\times 3\times 3=36$$So, square root of $$12961296=(2×2)×(2×2)×(3×3)×(3×3)

How many numbers lie between the square of the given numbers: $25$ and $26$

We know that, 25² = 625
And, 26² = 676
Now, 676 - 625 = 51
So, there are 51 - 1 = 50 numbers lying between 25² and 26²

Find the square roots of the given number by Prime factorization method: 441

441 = 3 × 3 × 7 × 7

$\sqrt{441} = \sqrt{3 × 3 × 7 × 7}$
= 3 × 7
= 21

Explain with reason whether the given number is perfect square or not: 4592

Given a number 4592 is not perfect square. Need to find out the reason.

Now, we have the perfect square m = n × n = n² , where m and n are integers ⇒ 4592 is not a perfect square as it cannot be expressed as n × n form the product of two equal integer
$\text{and the number that have 2,3,7 or 8 in the units place are not perfect squares}$ .

Hence, the given number 4592 is not a perfect square.

Find the square root of
$24+4\sqrt { 15 } -4\sqrt { 21 } -2\sqrt { 35 }$

$24+4\sqrt{15}-4\sqrt{21}-2\sqrt{35}$

$=24+2\times\sqrt{5}\times2\sqrt3+2\times(-\sqrt{7})\times2\sqrt3+2\sqrt{5}\times(-\sqrt7)$

$=5+7+12+2\times\sqrt{5}\times2\sqrt3+2\times(-\sqrt{7})\times2\sqrt3+2\sqrt{5}\times(-\sqrt7)$

$=(\sqrt5+2\sqrt3-\sqrt7)^2$

Hence square root of

$24+4\sqrt{15}-4\sqrt{21}-2\sqrt{35}$ is

$\pm(\sqrt5+2\sqrt3-\sqrt7)$

Express the following square numbers as the sum of two odd numbers.
i) $81$ ii) $144$

Sum of two odd numbers will always be an even number

$1)81$ is odd number so it cannot be expressed as sum of two odd numbers

$2)144=1+143=3+141=5+139$ (similarly, it can be written in many ways)

A society collected $\text{Rs. }9,216$ wherein, each member contributed as many rupees as there were members in the society. Find the number of members.

If there were

$a$ members in the society, then each contributes

$\text{Rs}$

$a$ as per the given statement.

So, total amount collected

$a \times a = 9216$

$\implies a = \sqrt {9216} = \sqrt {{2} ^{10} \times { 3 } ^ {2}} = {2} ^{5} \times 3 = 32 \times 3 = 96$

$\therefore$ There are

$96$ members in the society

Find the smallest whole number by which $1458$ should be multiplied, so as to get a perfect square number.

By prime factorisation of

$1458$ , we get

$\displaystyle 1458 = 2\times 3\times 3\times 3\times 3\times 3\times 3$

Here

$3$ are in pair, but

$2$ needs a pair to make

$1458$ a perfect square.

So,

$1458$ needs to be multiplied by

$2$ to become a perfect square.

For the following number, find the smallest whole number by which it should be multiplied so as to get a perfect square number.
252

By prime factorisation we get

$\displaystyle 252=2\times 2\times 3\times 3\times 7$

Here

$2$ and

$3$ are in pairs but

$7$ needs a pair.

Thus 7 can become pair after multiplying

$252$ with

$7$ .

$252$ will become a perfect square when multiplied by

$7$ .

Thus Answer

$= 7$

Find the least number that must be added to $1300$ so as to get a perfect square. Also find the square root of the perfect square.

We can find

$\displaystyle \sqrt{1300}$ by long division method and the remainder is

$4$

This shows that

$\displaystyle 36^{2} <1300$

Next perfect square number is

$\displaystyle 37^{2} =1369$

Hence the number to be added is

$\displaystyle 37^{2} -1300=1369-1300=69$

Find the square root of each of the following by Long division method:
(i) $2304$ (ii) $4489$
(iii) $3481$ (iv) $529$
(v) $3249$ (vi) $1369$
(vii) $5776$ (viii) $7921$
(ix) $576$ (x) $3136$

1391397_616358_ans_28207e27f66841a8aaa3442405cee545.JPG

Find the square root of the following decimal numbers:
(i) $2.56$ (ii) $7.29$
(iii) $51.84$ (iv) $42.25$
(v) $31.36$ (vi) $0.2916$
(vii) $11.56$ (viii) $0.001849$

1391402_616379_ans_baabb0649ccd46bd8e7108d65545a440.JPG

Find the square root of the following numbers by long division method:
531441
4401604

$531441$

Ref. image

$\sqrt{531441} = 729$

$4401604$

Ref. image

$\sqrt{4401604} = 2098$ .

1331722_1258798_ans_af3d164512f849b88b12ff5c9840f96a.png

Find the square root of each of the following numbers by using the method of prime factorization:
$729$

Given

$729$
A square can always be expressed as a product of pairs of equal factors
Resolve the given number into prime factors
We get

$729=3\times 3\times 3\times 3\times 3\times 3$

$\surd 729=3\times 3\times 3=27$

Find the square root of each of the following numbers by using the method of prime factorization:
$4096$

Given

$4096$
A square can always be expressed as a product of pairs of equal factors
Resolve the given number into prime factors
We get

$4096=2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2$

$\surd 4096=2\times 2\times 2\times 2\times 2\times 2 =64$

Find the square root of each of the following numbers by using the method of prime factorization:
$7056$

Given

$7056$

A square can always be expressed as a product of pairs of equal factors

Resolve the given number into prime factors

We get

$7056=2\times 2\times 2\times 2\times 3\times 3\times 7\times 7$

$\surd 7056=2\times 2\times 3\times 7=84$

Find the square root of each of the following numbers by using the method of prime factorization:
$2025$

Given

$2025$
A square can always be expressed as a product of pairs of equal factors
Resolve the given number into prime factors
We get

$2025=3\times 3\times 3\times 3\times 5\times 5$

$\surd 2025=3\times 3\times 5 =45$

Find the square root of each of the following numbers by using the method of prime factorization:
$8100$

Given

$8100$
A square can always be expressed as a product of pairs of equal factors
Resolve the given number into prime factors
We get

$8100=2\times 2\times 3\times 3\times 3\times 3\times 5\times 5$

$\surd 8100=2\times 3\times 3\times 5=90$

Write a Pythagorean triplet whose smallest number is
$20$

Given

$20$ is the smallest number in Pythagorean triplet

But we have for every number

$m > 1$ , we have

$(2m, m^2-1, m^2+1)$ as a Pythagorean triplet

$2m=20$

$m=20/2=10$

Then

$m^2=100$

$m^2-1=100-1=99$

$m^2+1=100+1=101$

Therefore the Pythagorean triple is

$=(2m,m^2-1, m^2+1)=(20,99,101)$

Write a Pythagorean triplet whose smallest number is
$14$

Given

$14$ is the smallest number in Pythagorean triplet

But we have for every number

$m > 1$ , we have

$(2m, m^2-1, m^2+1)$ as a Pythagorean triplet

$2m=14$

$m=14/2=7$

Then

$m^2=49$

$m^2-1=49-1=48$

$m^2+1=49+1=50$

Therefore the Pythagorean triple is

$=(2m,m^2-1, m^2+1)=(14,48,50)$

Find the least square number which exactly divisible by each of the numbers $8,12,15$ and $20$

Given numbers

$=8,12,15$ and

$20$

The smallest number divisible by

$8,12,15$ and

$20$ is their

$L.C.M$ i.e.

$120$

Resolving the

$L.C.M.$ as prime factors we get

$120=2\times 2\times 2\times 3\times 5$

To make it perfect square multiply by

$2\times 3\times 5=30$

Which implies

$120\times 30=3600$

Least square number which is exactly divisible by

$6,9,15$ and

$20$ is

$3600$

Find the largest number of $3$ digits which is perfect square.

We know that

$3$ digit largest number is

$999$ .
We also know that square root of

$999$ is

$13.61$
So the square of any number greater than

$31.61$ is

$4$ digit
Therefore,

$31^2=961$ will be the largest number of

$3$ digits which is Perfect square

Evaluate:
$\surd 576$

Given

$\surd 576$ , using long division method we get
Therefore

$\surd 576=24$
1620701_1771869_ans_fbf2f6be4602466c9ce0221cbfcb3db3.PNG

Write a Pythagorean triplet whose number is
$63$

$63$
Take

$n^2-1=63$
By further calculation

$n^2=63+1=64=8^2$
So

$n=8$
Here

$2n=2\times 8=16$

$n^2-1=63$

$n^2+1=8^2+1=64+1=65$
Therefore, the triplets are

$16, 63$ and

$65$ .

Find the square root of the following decimal numbers by division method:
$42.25$

$\surd 42.25=6.5$
By division method
1704093_1806485_ans_ca3deee19a6d4f1faf43658fc7951de9.png

Find the square root of each of the following by division method:
$106929$

$\surd{106929}=327$
By division method
1702491_1806451_ans_f9199d4a877d455eac5e1c6874b070c6.png

Write a Pythagorean triplet whose number is
$8$

$8$
Take

$n=8$
So the triplet will be

$2n, n^2-1, n^2+1$
Here
If

$2n=8$ , then

$n=8/2=4$
Substituting the values

$n^2-1=4^2 -1=16-1=15$

$n^2+1=4^2+1=16+1=17$
Therefore, the triplets are

$8, 15$ and

$17$ .

Find the square root of 1764 by the prime factorisation method.

$\sqrt {1764} =\sqrt { 2\times 2\times 3\times 3\times 7\times 7 }$

$\therefore \sqrt { 1764 } =2\times 3\times 7$

$=42$

Evaluate :
$\sqrt { 92416 }$

Thus,

$92416=2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 19\times 19$

$\sqrt{92416}=2\times 2\times 2\times 2\times 19$

$=16\times 19=304$

Find the unit digit in the square of the numbers $81$ and $79$ .

The unit digit of the number

$81^2$ is the unit digit of

$1^2$ which is

$1$ .

Similarly the unit digit of the number

$79^2$ is the unit digit in

$9^2$ i.e. unit digit in

$81$ i.e. is

$1$ .

$22453$

1873038_1671376_ans_4f054af17bc54a6baa481259a7fecb77.jpg

The square of which of the following numbers would be an odd number?
$731$

1873636_1671388_ans_bc690a35b34c4885840f05d8f9d710b9.jpg

Give reason.
$8948$

1873024_1671370_ans_f3f6a41556934dffa3b4fb3c32463e3f.jpg

Show that the following number is not perfect square:
$4058$

1873037_1671375_ans_dc6b3435dc1d40b48eb61eac057884a8.jpg

. Give reason.
$333333$

1873025_1671372_ans_ae39922406f44491bee08b5f382110e1.jpg

What will be the units digit of the squares of the following numbers?
$977$

1873936_1671399_ans_1454cf83a7fc4ca291e1cbec94bbcf30.png

What will be the units digit of the squares of the following number?
$53924$

1873961_1671413_ans_581c7f08d375423187c4bc8bbb9db324.png

What will be the units digit of the squares of the following numbers?
$4583$

1873937_1671401_ans_3697ae62f3304a338b5e84c381a968bc.png

What will be the units digit of the squares of the following numbers?
$52698$

1873943_1671408_ans_d46a507fa90a4cc1a128f159acb0dc56.jpg

What will be the units digit of the squares of the following numbers?
$12796$

1873954_1671411_ans_8c1d8a54895b4a619d330ca8992156f2.png

What will be the units digit of the squares of the following numbers?
$99880$

1873948_1671410_ans_e71abab20c7e4a8981f0eba7e6f00825.png

Check whether the following triplet is Pythagorean?
$(18, 80, 82)$

1874073_1671431_ans_572dcf5e8ede4796850a0a05964abfb2.jpg

What will be the unit's digit of the square of the following number?
$78367$

1873941_1671405_ans_44a822db118944cab9b96a3448f54eac.png

What will be the units digit of the squares of the following numbers?
$52$

1873934_1671397_ans_b69071e684b94f62b7e641bb23caf5b0.png

What will be the units digit of the squares of the following numbers?
$55555$

1873960_1671412_ans_93e253e404f444a6a85b4d175ee93034.png

The number $1023$ is a perfect square.
Type 1 if above statement is true, otherwise 0.

All the perfect square numbers end with any one from

$00,1,4,5,6,9$

Since

$1023$ ends with

$3$

Hence, it is not a perfect square.

Write five numbers which you cannot decide whether they are square just by looking at the unit’s digit.

1898855_1671481_ans_25251c4011f94a4496c24c1bbcde5e02.jpg

The number $1027$ is a perfect square.
Type 1 if above statement is true, otherwise 0.

All the perfect square numbers end with any one from

$00,1,4,5,6,9$

Since

$1028$ ends with

$8$

Hence, it is not a perfect square.

Find the square root of the following number by division method.
$2304$ .

Thus, square roots of

$2304$ is

$48$

Hence, option A is correct.

1161938_1067531_ans_bc14629dd4054ea88173179fb8cf865b.png

Find the square root of the following number by factorisation.
$1156$

1171040_1068281_ans_9dda495e4f2048edaad240d8e6d74a50.jpg

Check whether the following triplet is Pythagorean?
$(12, 35, 38)$

1874160_1671439_ans_02470c02d923457aa4dcc8c15251e3eb.jpg

The number $1027$ is a perfect square.
Type 1 if above statement is true, otherwise 0.

All the perfect square numbers end with any one from

$0,1,4,5,6,9$

Since

$1022$ ends with

$2$

Hence, it is not a perfect square.

Find the square root of each of the following by long division method:
$1471369$

Write the prime factorization of the following number and hence find its square root:
$5929$

$5929 = 7 \times 7 \times 11 \times 11$

So, the square root of

$5929$ is

$77$ .

Express the following surds as continued fraction, and find the sixth convergent to each
$\sqrt 11$

1574688_1744901_ans_8de837275ba4430c994d756841078829.jpg

Estimate the square root : $\dfrac {1}{\sqrt {33}}$

1993364_1744910_ans_2ed46adadc4746068f127bb79ec21a53.jpeg

Estimate the value of square root : $\sqrt {14}$

1578938_1744903_ans_6abd8e63f8574bfe9366011ee3f2d794.jpg

Give reason to show that none of the numbers given below is a perfect square:
$360$

Given

$360$

We know that property

$2$ of perfect square i.e., number ending in an add number of zeros is never a perfect square.

Therefore the given number

$360$ is not a perfect square.

Find the square root of 5929 by the prime factorisation method.

$5929=7\times 7\times 11\times 11$

$\Rightarrow \sqrt { 5929 } =\sqrt { 7\times 7\times 11\times 11}$

$=7\times 11$

$=77$

Find the square root of the number 729 by the prime factorisation method.

$729 = 3\times3\times3\times3\times3\times3$

$729 = 3^{6}$

$\sqrt{729} = 3^{3}$

$\sqrt{729} = 27$

The difference of two perfect squares is a perfect square.
If true then enter $1$ and if false then enter $0$

False.

Let the two numbers be

$6$ and

$5$ .

${6}^{2} - {5}^{2} = 36 - 25 \\\ \ \ \ \ \ \ \ \ \ \ = 11$

$11$ is not a square number.

The product of two perfect squares is a perfect square.
If true then enter $1$ and if false then enter $0$

True.

${2}^{2} \times {3}^{2} = 4 \times 9 = 36$ , which is a square number

The number of digits in a perfect square is even.
If true then enter $1$ and if false then enter $0$

False.

Square of

${11}^{2} = 121$ , which has odd number of digits.

Find the square root of the $196$ by factorisation.

Prime factors of 196 are

$196 = 2 \times 2 \times 7 \times 7$

taking one prime number out from the pair of prime numbers

Hence

$\sqrt{196} = 2 \times 7$

square root of

$196$ will be

$2\times 7=14$

Find the square root of the following in the form of a binomial surd.
$8+2\sqrt{7}$ .

Let

$x = \sqrt{8+2\sqrt{7}} = \sqrt n + \sqrt m$

$8+2\sqrt{7} = n + m + 2\sqrt{nm}$

Hence,

$n+m = 8 , nm = 7$

$n^2-8n+7 =0$

$n = 7$ or

$n = 1$

Similarily,

$m = 1$ or

$m=7$

Find the greatest number of three digit which is a perfect square.

Now,

$961 = 31 \times 31 = (31)^2$

Thus,

$961$ is the greatest three digit number which is a perfect square.

1877745_1671862_ans_63f71218a45a4a0cb8931ce154a07707.jpg

Estimate the square root : ${\dfrac {7}{\sqrt {11}}}$

1578950_1744912_ans_f297be282b4b46d8b3b1673b4ba5a332.jpg

Estimate the square root : $\dfrac {1}{\sqrt {21}}$

1993363_1744909_ans_98324b3a6fb54cc5b4e8bcbe2478ed48.jpeg

Express the following surds as continued fraction, and find the sixth convergent to each:
$\sqrt 5$

1574687_1744872_ans_e4e823c09f914752a81d464763d0199b.jpg

Find limits of the error when $\dfrac {208}{65}$ is taken for $\sqrt {17}$

1993372_1744913_ans_08d408f54b1d47c5b8d0f1fd8626187e.jpeg

Estimate the value of square root : $\sqrt 22$

1578939_1744904_ans_830976b5b3d54b35ad54532176527c9d.jpg

Find the square root of the following in the form of a binomial surd.
$14+6\sqrt{5}$ .

Let

$x = \sqrt{14+6\sqrt{5}} = \sqrt n + \sqrt m$

$14+6\sqrt{5} = n + m + 2\sqrt{nm}$

Hence,

$n+m = 14 , nm = 45$

$n^2-14n+45 =0$

$n = 9$ or

$n = 5$

Similarily,

$m = 5$ or

$m=9$

Give reason to show that none of the numbers given below is a perfect square:
$5963$

Given

$5963$

By observing the property

$1$ of perfect square, it always ends on (1,4,5,6,9,0)i.e., number ending with

$3$ is never a perfect square.

Therefore the given number

$5963$ is not a perfect square.

Give reason to show that none of the numbers given below is a perfect square:
$64000$

Given

$64000$
We know that property

$2$ of perfect square i.e., number ending in an add number of zeros is never a perfect square.

Therefore the given number

$64000$ is not a perfect square.

Give reason to show that none of the numbers given below is a perfect square:
$8457$

Given

$8457$

By observing the property

$1$ of perfect square, it always ends on (1,4,5,6,9,0)i.e., number ending with

$7$ is never a perfect square.

Therefore the given number

$8457$ is not a perfect square.

Find limits of the error when $\dfrac {916}{191}$ is taken for $\sqrt {23}$ .

1993376_1744914_ans_ed73a97ea988467292ebe96fb9cb380f.jpeg

Which of the following squares are squares of even number?
$196$

Given

$196$
We know that from the property of perfect square, the square of an even number is always even.
So the given number

$196$ is even, therefore it is square of even number.

Give reason to show that none of the numbers given below is a perfect square:
$9468$

Given

$9468$

By observing the property

$1$ of perfect square, it always ends on (1,4,5,6,9,0)i.e., number ending with

$8$ is never a perfect square.

Therefore the given number

$9468$ is not a perfect square.

Give reason to show that none of the numbers given below is a perfect square:
$5372$

Given

$5372$

By observing the property

$1$ of perfect square, it always ends on (1,4,5,6,9,0)i.e., number ending with

$2$ is never a perfect square.

Therefore the given number

$5372$ is not a perfect square.

Fill in the blanks to make the statements true.
The square of 0.7 is _________.

The square of 0.7 is 0.49 .

Fill in the blanks to make the statements true.
The digit at the ones place of $57^2$ is _________.

The digit at the ones place of

$57^2$ is

$\underline{9}$ .
Unit digit of

$(57)^2 = 7$

$7 \times 7 = 49$

Fill in the blanks to make the statements true.
The square root of $24025$ will have _______ digits.

The square root of 24025 will have

$\underline{3}$ digits.

Given,

$24025$

Number of digits

$(n)= 5$ (odd)

Number of digits in the square root

$=\dfrac{n+1}{2} = \dfrac{5+1}{2} = 3$

Write a Pythagorean triplet whose number is
$80$

The number given is

$80$
Take

$2n=80$

$n=\dfrac{80}{2}=40$
Here

$n^2-1=40^2-1=1600-1=1599$

$n^2+1=40^2+1=1600+1=1601$
Therefore, the triplets are

$80, 1599$ and

$1601$ .

Find the square of the following number containing $5$ in unit's place:
$305$

$305^2=(30\times 31)$ hundred

$+25$
By further calculation

$=93000+25$

$=93025$

Find the square root of each of the following by division method:
$167281$

$\surd 167281=409$
By division method
1702497_1806454_ans_3c84ac53364f4edeabc38fed36668d0d.png

Find the square of the following number containing $5$ in unit's place:
$525$

$525^2=(52\times 53)$ hundred

$+25$
By further calculation

$=275600+25$

$=275625$

$13.69$ ?

Square root of

$13.69$

$\Rightarrow \sqrt{13.69}=\sqrt{\left( 3.7\right)^2}$

$=3.7$

Hence, the answer is

$3.7.$

Write two Pythagorean triplets each having one of the numbers as $5$ .

The sum of two smaller sides is always equal to the square of longer side.
Two Pythagorean triplets each were having one of the numbers as 5 are:

$(5)^2 = (3)^2 + (4)^2$ and

$(13)^2 = (12)^2 + (5)^2$
Hence the pythagorean triplets are

$(3,4,5)$ and

$(12,5,13)$

Find the square root of the following by long division method.
a) $27.04$
b) $1.44$

1650542_1792434_ans_6a1b2539de984743bcaf1428699c4cb4.jpg

Write the first five square numbers.

The first five square numbers.

$(1)^2 = 1$

$(2)^2) = 4$

$(3)^2 = 9$

$(4)^2 = 16$

$(5)^2 = 25$

Write five number which you can decide by looking at their one's digit that they are not square numbers.

We know that
A number which ends with the digits

$2, 3, 7$ or

$8$ at its unit places is not a perfect square.
Example -

$111,372,563,978,1282$ are not square numbers.

The following numbers are obviously not perfect. Give reason.
$5298$

In the given number
If the square of a number does not have

$2, 3, 7, 8$ or

$0$ as its unit digit, the square

$567, 2453, 5208, 46292$ and

$74000$ cannot be the perfect squares as they have

$7, 2, 8, 2$ digits at the unit place.

Find the square root of the following decimal numbers by division method:
$51.84$

$\surd 51.84=7.2$
By division method

Find the square root of 2304 by long division method.

We get a remainder

$0$ and quotient as

$48$ using division method.

This shows that

$\sqrt{2304}=48$

Hence, the answer is

$48$ .

Find the square root of 529 by long division method.

We get a remainder

$0$ and quotient as

$23$ using division method.

This shows that

$\sqrt{529}=23$

Hence, the answer is

$23$ .

Find the square root of 3481 by long division method.

We get a remainder

$0$ and quotient as

$59$ using division method.

This shows that

$\sqrt{3481}=59$

Hence, the answer is

$59$ .

Find the square root of 400 by prime factorisation method.

$400={ 2 }^{ 2 }\times { 2 }^{ 2 }\times { 5 }^{ 2 }$

$=>\sqrt { 400 } ={ 2 }\times { 2 }\times { 5 }$

$=20$

Find the smallest positive integer with which one has to divide 336 to get a perfect square.

We observe that

$336 = 2 \times 2 \times 2 \times 2 \times 3 \times 7$ . Here both 3 and 7 occur only once. Hence we have to remove them to get a perfect square.
We divide 336 by 3

$\times$ 7 = 21 and get

$\dfrac{336}{21} = 16 = 4^2$
The least number required is 21.

What could be the possible 'one's' digits of the square root of the following number?
$657666025$

Since,

$5^2=25$

Therefore,

$5$ is the possible 'one's' digits of the square root of

$657666025$ .

Solve : $3\sqrt{25 \times 6 \times 30 \times 6}$

$3\sqrt{25\times 6\times 30\times 6}$

$=3\sqrt{(5)^{2}\times (6)^{2}\times 2\times 3\times 5}$

$=3\times 5\times 6\sqrt{2\times 3\times 5}$

$=90\sqrt{30}$

Find:
$\sqrt {4096}$

$\sqrt { 4096 } =\sqrt { 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2 } =2\times 2\times 2\times 2\times 2\times 2=64$

Find the square root of $9216$

$9216$ can be written as

$2^{10}.3^2$

root of

$9216$ is

$2^5.3=96$

If $\sqrt { 2 } =1.414$ , then find the value of $\cfrac { 1 }{ 1-\sqrt { 2 } }$

$\dfrac{1}{1-\sqrt{2}}=\dfrac{\sqrt{2}+1}{1-2}=-(1+\sqrt{2})=-2.414$

Find the least number which must be subtracted to get a perfect square from $1000$ .

Thus,

$39$ must be subtracted from

$1000$ to get a perfect square.

1035243_1098839_ans_178f2ea3948b425e91fe5034ed96ebb2.jpg

Find the square root of a number $529$ by prime factorisation method.

By using prime factorisation method:

$\displaystyle 529=23\times 23$

Taking square root on both the sides, we get

$\displaystyle \sqrt{529}=\sqrt{23\times 23}=23$

For each of the following numbers, find the smallest whole number by which it should be divided so as to get a perfect square number. Also, find the square root of the square number obtained.

(i) $252$ (ii) $2925$ (iii) $396$ (iv) $2645$ (v) $2800$ (vi) $1620$

$\textbf{Step 1: Solving for (i)}$

$252 = 2 \times 2 \times 3 \times 3 \times 7$

$\text{Here, prime factor 7 has no pair, Hence 252 must be divided by 7 to get a perfect square.}$

$\dfrac{252}{7} = 36$

$\text{Square root of 36 =} \sqrt{36} = 2 \times 3 = 6$

$\textbf{Step 2: Solving for (ii)}$

$2925 = 3 \times 3 \times 5 \times 5 \times 13$

$\text{Here, prime factor 13 has no pair, Hence 2925 must be divided by 13 to get a perfect square.}$

$\dfrac{2925}{13} = 225$

$\text{Square root of 225 =} \sqrt{225} = 3 \times 5 = 15$

$\textbf{Step 3: Solving for (iii)}$

$396 = 2 \times 2 \times 3 \times 3 \times 11$

$\text{Here, prime factor 11 has no pair, Hence 396 must be divided by 11 to get a perfect square.}$

$\dfrac{396}{11} = 36$

$\text{Square root of 36 =} \sqrt{36} = 2 \times 3 = 6$

$\textbf{Step 4: Solving for (iv)}$

$2645 = 5 \times 23 \times 23$

$\text{Here, prime factor 5 has no pair, Hence 2645 must be divided by 5 to get a perfect square.}$

$\dfrac{2645}{5} = 529$

$\text{Square root of 529 =} \sqrt{529} = 23$

$\textbf{Step 5: Solving for (v)}$

$2800 = 2 \times 2 \times 2 \times 2 \times 5 \times 5 \times 7$

$\text{Here, prime factor 7 has no pair, Hence 2800 must be divided by 7 to get a perfect square.}$

$\dfrac{2800}{7} = 400$

$\text{Square root of 400 =} \sqrt{400} = 2 \times 2 \times 5 = 20$

$\textbf{Step 6: Solving for (vi)}$

$1620 = 2 \times 2 \times 3 \times 3 \times 3 \times 3 \times 5$

$\text{Here, prime factor 5 has no pair, Hence 1620 must be divided by 5 to get a perfect square.}$

$\dfrac{1620}{5} = 324$

$\text{Square root of 324 =} \sqrt{324} = 2 \times 3 \times 3 = 18$

$\textbf{Therefore , Square root of each number given is obtained .}$

Find $\sqrt{96.04}$ .

Therefore,

$\sqrt{96.04} = 9.8$
1391392_565836_ans_09c109e8aeb1471cad6e85ddc278a947.jfif

Find the square root of $2025$ .

Resolving

$2025$ into Prime factors, we get

$2025 = (3 \times 3)\times (3\times 3) \times (5 \times 5)$

$\sqrt{2025} = 3 \times 3 \times 5$

$\therefore \sqrt{2025} = 45$

Find the square root of 42.25 using division method.

1391391_565834_ans_4cc8bed9e5704bb5a7355c66f8957f53.JPG

Find the square root of $8281$

Prime factorisation of $8281 = 7 × 7 × 13 × 13$

So, $\sqrt{8281}$ $= 7 × 13$

$\sqrt{8281} = 91$

Find the square root of $7744$ by the prime factorisation method.

$7744=2\times 2\times 2\times 2\times 2\times 2\times 11\times 11$

$=>\sqrt { 7744 }=\sqrt { 2\times 2\times 2\times 2\times 2\times 2\times 11\times 11 }$

$=2\times 2\times 2\times 11$

$=88$

What will be the unit digit of the square of $3853$ ?

The unit's digit of

$3^2$ is

$9$

Hence the units digit of

$(3853)^2$ is also

$9$ .

What will be the unit digit of the square $12796$ ?

The unit's digit of

$6^2$ is

$6$

Hence the units digit of

$(12796)^2$ is also

$6$ .

What will be the unit digit of the square of $799$ ?

The unit's digit of

$9^2$ is

$1$

Hence the units digit of

$(799)^2$ is also

$1$ .

What will be the unit digit of the square of $52698$ ?

Units digit of square of

$52698$ will be equal to units digit of square of

$8$ .

The unit's digit of

$8^2$ is

$4$

Hence the units digit of

$(52698)^2$ is also

$4$ .

What will be the unit digit of the square of $81$ ?

The unit's digit of

$1^2$ is

$1$ . Also units digit of

$(81^2)$ will also be equal to

$1^2$ . Hence the units digit of

$(81)^2$ is also

$1$ .

What will be the unit digit of the square of $26387$ ?

The unit's digit of

$7^2$ is 9

Hence the units digit of

$(26387)^2$ is also

$9$ .

What will be the unit digit of the square of $272$ ?

The unit's digit of

$2^2$ is

$4$

Hence the units digit of

$(272)^2$ is also

$4$ .

What will be the unit digit of the square of $99880$ ?

The unit's digit of

$0^2$ is

$0$

Hence the units digit of

$(99880)^2$ is also

$0$ .

Find the square root of 1369 by long division method.

We get a remainder

$0$ and quotient as

$37$ using division method.

This shows that

$\sqrt{1369}=37$

Hence, the answer is

$37$ .

Find the square root of 4489 by long division method.

We get a remainder

$0$ and quotient as

$67$ using division method.

This shows that

$\sqrt{4489}=67$

Hence, the answer is

$67$ .

Find the square root of 5776 by long division method.

We get a remainder

$0$ and quotient as

$76$ using division method.

This shows that

$\sqrt{5776}=76$

Hence, the answer is

$76$ .

Find the square root by long-division method:
$3249$

For the following number, find the smallest whole number by which it should be divided so as to get a perfect square. Also, find the square root of the square number so obtained.
$252$
$2925$

By prime factorisation, we get

$252=\underline{2\times2}\times\underline{3\times3}\times7$
Since the prime factor

$7$ cannot be paired.

$\therefore$ The given number should be divided by

$\therefore\;\displaystyle\frac{252}{7}=\displaystyle\frac{2\times2\times3\times3\times7}{7}$

${ \underline { 2\mid 252 } }\\ { \underline { 2\mid 126 } }\\ { \underline { 3\mid 63 } }\\ { \underline { 3\mid 21 } }\\ { \underline { 7\mid 7 } }\\ \; \; \mid 1$

$=\underline{2\times2}\times\underline{3\times3}$

$=36$ is a perfect square.
and,

$\sqrt{36}=\sqrt{\underline{2\times2}\times\underline{3\times3}}$

$=2\times3=6$
By prime factorisation, we get

$2925=\underline{3\times3}\times\underline{5\times5}\times13$
Since prime factor

$13$ cannot be paired.

$\therefore$ The given number should be divided by

$13$ .

$\therefore\;\displaystyle\frac{2925}{13}=\displaystyle\frac{3\times3\times5\times5\times13}{13}$

${ \underline { 3\mid 2925 } }\\ { \underline { 3\mid 975 } }\\ { \underline { 5\mid 325 } }\\ { \underline { 5\mid 65 } }\\ { \underline { 13\mid 13 } }\\ \; \; \; \; \mid 1$

$=\underline{3\times3}\times\underline{5\times5}=225$ is a perfect square
and,

$\sqrt{225}=\sqrt{\underline{3\times3}\times\underline{5\times5}}$

$=3\times5=15$ .

One of the following number is not a perfect square, state the number.
$2453, 2500$

We know that the number ends with

$00, 1, 4, 5, 6$ or

$9$ are perfect square

$2453$ is not a perfect square because it ends with

$3$ and

$2500$ is a square of

$50.$

What will be the unit digit of the square of the number $26387$ ?

Since

$\displaystyle 7^{2}$

$=49$

Therefore

$\displaystyle (26387)^{2}$ will have

$9$ at unit's place.

What will be the unit digit of the square of the number $12796$ ?

Since

$\displaystyle 6^{2}$

$=36$

So,

$\displaystyle (12796)^{2}$ will have

$6$ at unit's place.

What will be the unit digit of the square of the number $52698$ ?

Since,

$\displaystyle 8^{2}$

$=64$

So,

$\displaystyle( 52698)^{2}$ will have

$4$ at unit's place.

What will be the unit digit of the square of the number $272$ ?

Since, the unit digit of

$272$ is

$2$ whose square is

$4$ .

Therefore, unit digit of squares of

$272$ is

$4$ .

State whether $1335$ is a perfect square or not.

We need to find square root of

$1335$ .

Lets use prime factorization method:

$\sqrt {1335}=\sqrt {5\times 3\times 89}$

We does not get pair of any of these numbers.

Hence,

$1335$ is not a perfect square.

Find the square root of the following decimal numbers
$(i) 2.56 (ii) 7.29 (iii) 51.84 (iv) 42.25 (v) 31.36$

$\textbf{Step 1: Converting all the numbers to exponential form}$

$\text{i) }2.56=256\times10^{-2}$

$\text{i) }7.29=729\times10^{-2}$

$\text{i) }51.84=5184\times10^{-2}$

$\text{i) }42.25=4225\times10^{-2}$

$\text{i) }31.36=3136\times10^{-2}$

$\textbf{Step 2: Finding square root of each number}$

$\sqrt{2.56}=\sqrt{256\times10^{-2}}=\sqrt{256}\times\sqrt{10^{-2}}=16\times10^{-1}=1.6$

$\sqrt{7.29}=\sqrt{729\times10^{-2}}=\sqrt{729}\times\sqrt{10^{-2}}=27\times10^{-1}=2.7$

$\sqrt{51.84}=\sqrt{5184\times10^{-2}}=\sqrt{5184}\times\sqrt{10^{-2}}=72\times10^{-1}=7.2$

$\sqrt{42.25}=\sqrt{4225\times10^{-2}}=\sqrt{4225}\times\sqrt{10^{-2}}=65\times10^{-1}=6.5$

$\sqrt{31.36}=\sqrt{3136\times10^{-2}}=\sqrt{3136}\times\sqrt{10^{-2}}=56\times10^{-1}=5.6$

$\textbf{Hence, The square root of each of the numbers are i)1.6 ii)2.7 iii)7.2 iv)6.5 v)5.6}$

Find the square root of $1057$ .

The given number is

$1057$ .

Let us find its square root by using the long division method, as shown above.

We will first group the digits into two from the right. In this case, it would be group

$57$ first and then

$10$ .

Now, find a number whose square is less than or equal to

$10$ . That would be

$3$ .

So, the remainder we get after division is

$1$ .

Bring the next group as it is.

So, now the dividend becomes

$157$ .

Next, we will multiply the existing quotient by

$2$ to get the first digit of the divisor in the second step.

Here, we need to fill in the blank of the divisor in the second step and the second digit of the quotient with the same digit such that the product is

$\leq$

$157$ .

So, we have

$62 \times 2 = 124$ .

The remainder would then be

$33$ . Now, we shall place a decimal point in the quotient and add

$2$ zeroes, which will make the dividend in third step as

$3300$ .

Repeating the above steps will result in a remainder of

$999$ and the quotient as

$32.51$ .

Hence,

$\sqrt{1057}\approx32.51$ .

Is $676$ a perfect square? If yes, explain why?

Consider the given number.

We know that $25^2=625$

$676=26\times 26={{26}^{2}}$

Also we can check that perfect squares end in $0,1,4,9,6,5.$

Hence, this is the perfect square.

The following numbers are obviously not perfect squares. Give reason.
(i) 1057
(ii) 23453
(iii) 7928
(iv) 222222
(v) 64000
(vi) 89722
(vii) 222000
(viii) 505050

We know that, numbers having

$2,\ 3,\ 7\ or\ 8$ at units place are not perfect squares.

That is because:

$1^2=1$

$2^2=4$

$3^2=9$

$4^2=16$

$5^2=25$

$6^2=36$

$7^2=49$

$8^2=64$

$9^2=81$

Also,

$10^2=100$

$100^2=10000$

Hence, any perfect square will end with

$1,\ 4,\ 5,\ 6,\ 9$ or even number of zeros.

All the given numbers end with either

$2,\ 3,\ 7,\ 8$ or with odd number of zeros.

Hence, the given numbers are not perfect squares.

Make a list of all perfect squares from $1$ to $500$ .

Perfect squares from

$1$ to

$500$ are

$1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196, 225, 256, 289, 324, 361, 400, 441, 484.$

Find the smallest integer larger than 1 which is a perfect square as well as a perfect cube.

Let us start with any number

$n$ and multiply it with itself

$6$ times to get a number

$N$ . We observe that

$N= n \times n \times n \times n \times n \times n = (n \times n) \times (n \times n) \times (n \times n)$

$= (n^2) \times (n^2) \times (n^2) = (n^2)^3$ .
Thus

$N$ is the cube of

$n^2$ . On the other hand you may also observe that

$N = n \times n \times n \times n \times n \times n = (n \times n \times n) \times (n \times n \times n)$

$=(n^3) \times (n^3) = (n^3)^2.$
Hence

$N$ is also the square of

$n^3$ . Thus

$N$ is both a perfect cube and a perfect square. Taking

$n =2$ , we get the least number:

$N = 2 \times 2 \times 2 \times 2 \times 2 \times 2 = 64$ . We may verify that

$64 = 4^3$ and

$64 = 8^2$ .

Find the least number to be added to 9300 to get a perfect square.

We use the steps of finding the square root till all the groups are exhausted.
The calculation shows that

$9300 > 96^2$ and

$9300 = 96^2+84$ . Hence we have to look for the next square,

$97^2$ . We observe

$97^2=9409=9300+109$ . Hence we have to add 109 to get the nest perfect square.

Find the square root of the following numbers by the factorisation numbers : 19881

Resolving

$19881$ as the product of primes, we get

$19881=3^2 \times 47^2$

$\Rightarrow \sqrt{19881} = \sqrt{3^2 \times 47^2} = 3 \times 47$

$\therefore \sqrt{19881} =141$

Find the square root of 70.56.

We write 70.56 in the form

$70.56=\frac{7056}{100}$ ,
and find the square root of 7056 and 100 separately using factorisation. Observe

$7056 = 16\times 441 = 16\times 9\times 49=4^2\times 3^2\times 7^2$
Hence

$\sqrt{7056}=4\times 3\times 7=84$ . We also have

$\sqrt{100}=10$ . Thus

$\displaystyle \sqrt{70.56}=\sqrt{\frac{7056}{100}}=\frac{\sqrt{7056}}{\sqrt{100}}=\frac{84}{10}=8.4$

Find the least number to be subtracted from 2011 to get a perfect square.

We follow the steps of finding the square root using division method.
Observe

$84\times 4 = 336 < 411$ and

$85\times 5=425 > 411$ . Thus

$2011=44^2+75$ . Hence the least number to be subtracted from 2011 to get a perfect square is 75. The perfect square we get is

$44^2$

Find the square root of the following numbers by the factorisation numbers : 34596

Resolving

$34596$ as the product of primes, we get

$\Rightarrow 34596=2^2 \times 3^2 \times 31^2$

$\Rightarrow \sqrt{34596}= \sqrt{2^2 \times 3^2 \times 31^2}= 2 \times 3 \times 31$

$\therefore \sqrt{34596}=186$

Find the square root of the following numbers by the factorisation numbers : 76176

Resolving

$76176$ as the product of primes, we get

$76176=2^4 \times 3^2 \times 23^2$

$\Rightarrow \sqrt{76176} = \sqrt{2^4 \times 3^2 \times 23^2} = 2^2 \times 3^1 \times 23^1$

$\therefore \sqrt{76176} = 276$

Find the approximation of $\sqrt{3.11}$ both from below and above to 3 decimal places.

We find the square root of 3.11 up to 3 decimal places.
Thus the square root of

$\sqrt{3.11}$ to 3 decimal places is 1.763. We observe

$(1.763)^2=3.108169 < 3.11 < 3.111696 = (1.764)^2$
Thus, 1.763 is the approximation from below of 3.11 to 3 decimal places; and 1.764 is the approximation from above of 3.11 to 3 decimal places.

Find the square root of the following number correct to 2 decimal places: 12

$\therefore \sqrt{12}=3.46\text{(upto two decimal places)}$
1391343_559775_ans_618ded0e6e0747fd9ab841f953a9343b.JPG

Round off following number to 3 decimal places : 14.56789

14.56789 = 14.568
The digit in the third decimal place is 7 (greater than 5). Therefore, we add 1 to the previous digit 7.

Round off following number to $3$ decimal places : $1.5678$

$1.5678 = 1.568$
The digit in the third decimal place is

$7$ (greater than

$5$ ). Therefore, we add

$1$ to the previous digit

$7$ .

Find the square root of the following number correct to 2 decimal places: 234.234

1391365_559782_ans_82ef84e8ec204d0b89e7254d945456be.JPG

Find the square root of the following number correct to 2 decimal places: 12.34

1391361_559779_ans_1e518752e6844c2cb0d5bd0e50f11c2a.JPG

Round off following number to 3 decimal places : 2.84671

2.84671 = 2.847
The digit in the third decimal place is 6 (greater than 5). Therefore, we add 1 to the previous digit 6.

Find the square root of the following number correct to 2 decimal places: 1.8

Hence,

$\sqrt {1.8}=1.34$
1391347_559777_ans_1c546004fa414a73b3595e4097142bf7.JPG

Round off following numbers to 3 decimal places : 3.3333567

3.3333567= 3.333
The digit in the third decimal place is 3 (less than 5). Therefore, we leave the digit in the third place 3 unchanged.

Find the square roots of the given number by Prime factorization method: 784

Prime factors of 784: 2 × 2 × 2 × 2 × 7 × 7

$\sqrt{784} = \sqrt{2 × 2 × 2 × 2 × 7 × 7} =2\times2\times7= 28$

Which of the following numbers are perfect squares?
(i) 121 (ii) 1136 (iii) 256 (iv) 321 (v) 600

121 = 11²

256 = 16²

^{So, 121 and 256 these two numbers are perfect squares.}

Find the square root of the following numbers by the factorization method:
(i) $729$ (ii) $400$
(iii) $1764$ (iv) $4096$
(v) $7744$ (vi) $9604$
(vii) $5929$ (viii) $9216$
(ix) $529$ (x) $8100$

(i) $\sqrt{729} = \sqrt{(3 * 3 * 3 * 3 * 3 * 3)}$

= 3 * 3 * 3

= 27
(ii) $\sqrt{400} = \sqrt{(2 * 2 * 2 * 2 * 5 * 5)}$

= 2 * 2 * 5

= 20
(iii) $\sqrt{1764} = \sqrt{(2 * 2 * 3 * 3 * 7 * 7)}$

= 2 * 3 * 7

= 42
(iv) $\sqrt{4096} = \sqrt{(2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2)}$

= 2 * 2 * 2 * 2 * 2 * 2

= 64
(v) $\sqrt{7744} = \sqrt{(2 * 2 * 2 * 2 * 2 * 2 * 11 * 11)}$

= 2 * 2 * 2 * 11

= 88
(vi) $\sqrt{9604} = \sqrt{(2 * 2 * 7 * 7 * 7 * 7)}$

= 2 * 7 * 7

= 98
(vii) $\sqrt{5929} = \sqrt{(7 * 7 * 11 * 11)}$

= 7 * 11

= 77

(viii) $\sqrt{9216} = \sqrt{(2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 3 * 3)}$

= 2 * 2 * 2 * 2 * 2 * 3

= 96

(ix) $\sqrt{529} = \sqrt{(23 *23)}$

= 23
(x) $\sqrt{8100} = \sqrt{(2 * 2 * 3 * 3 * 3 * 3 * 5 * 5)}$

= 2 * 3 * 3 * 5

= 90

$84.64$

1389502_565856_ans_5e6ab65384e84b82ba13d74b72c00cae.JPG

Find the square roots of the given number by Prime factorization method: 7056

$\text{7056 = 2²×2²×3²×7²}$

$\sqrt{7056}$

$\sqrt{2²×2²×3²×7²}$

$=2×2×3×7$

$= 84$

Find the square root of the given number by division method: $1089$

Consider given the number $1089$ ,

Hence, $33$ is the answer.

1103864_565841_ans_7b1162a4823a437daa7bbd2818b72389.png

Find the square root of the given number by division method: 2304

1391393_565842_ans_396d8c9ce57d4a6c999b131d38f66f13.JPG

$2.56$

1389498_565851_ans_8559342aa6ab48d79dbc056155ee5e86.JPG

Find the square root of each expression given below:
(i) $3\times 3\times 4\times 4$
(ii) $2\times 2\times 5\times 5$
(iii) $3\times 3\times 3\times 3\times 3\times 3$
(iv) $5\times 5\times 11\times 11\times 7\times 7$

$(i) 3\times 3\times 4\times 4$

$3\times 3\times 4\times 4=3^{2}\times 4^{2}=(3\times 4)^{2}$

So square root of

$3\times 3\times 4\times 4$ is

$3\times 4=12$

(ii)

$2\times 2\times 5\times 5$

$2\times 2\times 5\times 5=2^{2}\times 5^{2}=(2\times 5)^{2}$

So square root of

$2\times 2\times 5\times 5$ is

$2\times 5=10$

(iii)

$3\times 3\times 3\times 3\times \times 3\times 3$

$3\times 3\times 3\times 3\times 3\times 3=3^{2}\times 3^{2}\times 3^{2}=(3\times 3\times 3)^{2}$

So square root of

$3\times 3\times 3\times 3\times 3\times 3\times$ is

$3\times 3\times 3=27$

$(iv) 5\times 5\times 11\times 11\times 7\times 7$

$5\times 5\times 11\times 11\times 7\times 7=5^{2}\times 11^{2}\times 7^{2}=(5\times 11\times 7)^{2}$

So square root of

$5\times 5\times 11\times 11\times 7\times 7$ is

$5\times 11\times 7=385$

Write down the unit digits of the following:
(i) $78^{2}$
(ii) $27^{2}$
(iii) $41^{2}$
(iv) $35^{2}$
(v) $42^{2}$

(i) As we can see that at unit's place of

$78$ is

$8$ and square of

$8$ is

$64$ .

So, the unit of digit

$78^{2}$ will be

$4$ .

(ii) As we can see that at unit's place of

$27$ is

$7$ and square of

$7$ is

$49$ .

So, the unit digit of

$27^{2}$ will be

$9$ .

(iii) As we can see that at unit's place of

$41$ is

$1$ and square of

$1$ is

$1$ .

So, the unit digit og

$41^{2}$ will be

$1$

(iv) As we can see that at unit's place of

$35$ is

$5$ and square of

$5$ is

$25$ .

So, the unit digit of

$35^{2}$ will be

$5$ .

(v) As we can see that at unit's place of

$42$ is

$2$ and square of

$2$ is

$4$ .

So, the unit digit of

$42^{2}$ will be

$4$ .

Find the square roots of the given number by Prime factorization method: 4096

$\text{Prime factors of 4096 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2}$

$\sqrt{4096}$

$\text{= 2 × 2 × 2 × 2 × 2 × 2 = 64}$

Find the square root of the given number by division method: $$9025$$

1391396_565847_ans_70926a9605134bf59c33cfb36b930d8d.JPG

Find the square root of $5929$

$5929 = \underbrace {7\times 7}\times \underbrace {11\times 11} = 7^{2}\times 11^{2}$

$\sqrt {5929} = \sqrt {7^{2}\times 11^{2}} = 7\times 11$

$\therefore \sqrt {5929} = 77$

Find the least number by which $384$ must be divided to make it a perfect square

$384 = 3\times \underbrace {2\times 2}\times \underbrace {2\times 2} \times \underbrace {2\times 2}\times 2$

$'3'$ and

$'2'$ remain without a pair.
Hence,

$384$ must be divided by

$6$ to make it a perfect square.

Write down the unit digits of the squares of the following numbers:
(i) $24$ (ii) $78$ (iii) $35$

(i) The square of

$24 = 24\times 24$ . Here

$4$ is in the unit's place.
Therefore, we have

$4\times 4 = 16. \implies 6$ is the unit's digit of square of

$24$ .
(ii) The square of

$78 = 78\times 78$ . Here,

$8$ is in the unit's place.
Therefore, we have

$8\times 8 = 64. \implies 4$ is the unit's digit of square of

$78$
(iii) The square of

$35 = 35\times 35$ . Here,

$5$ is in the unit's place.
Therefore, we have

$5\times 5 = 25. \implies 5$ is the unit's digit of square of

$35$ .

Find the square root of $529$ using long division method

Step

$1$ : We write

$529$ as

$5 \overline {29}$ by grouping the numbers in pairs, starting from the right end. (i.e. from the units place ).
Step

$2$ : Find the number whose square is less than (or equal to)

$5$ . Here it is

$2$ .
Step

$3$ : Put

$2$ on the top, and also write

$2$ as a divisor as shown.
Step

$4$ : Multiply

$2$ on the top with the divisor

$2$ and write

$4$ under

$5$ and subtract. The remainder is

$1$ .
Step

$5$ : Bring down the pair

$29$ by the side of the remainder

$1$ , yielding

$129$ .
Step

$6$ : Double

$2$ and take the resulting number

$4$ . Find that number

$n$ such that

$4n \times n$ is just less than or equal to

$129$ .
For example:

$42\times 2 = 84$ ; and

$43\times 3 = 129$ and so

$n = 3$ .
Step

$7$ : Write

$43$ as the next divisor and put

$3$ on the top along with

$2$ . Write the product

$43 \times 3 = 129$ under

$129$ and subtract. Since the remainder is

$0$ , the division is complete. Hence

$\sqrt {529} = 23$ .

Find the least number by which $200$ must be multiplied to make it a perfect square

$200 = 2\times \underbrace {2\times 2}\times \underbrace {5\times 5}$

$'2'$ remains without a pair.
Hence,

$200$ must be multiplied by

$2$ to make it a perfect square.

By observing the units digits, which of the numbers $3136, 867$ and $4413$ cannot be perfect squares?

Since

$6$ is in units place of

$3136$ , there is a chance that it is a perfect square.

$867$ and

$4413$ are surely not perfect squares as

$7$ and

$3$ are the unit digit of these numbers.

Find the square root of $169$

$169 = \underbrace {13\times 13} = 13^{2}$

$\sqrt {169} = \sqrt {13^{2}} = 13$

Find the square root of $64$

$64 = \underbrace {2\times 2} \times \underbrace {2\times 2} \times \underbrace {2\times 2} = 2^{2}\times 2^{2} \times 2^{2}$

$\sqrt {64} = \sqrt {2^{2}\times 2^{2} \times 2^{2}} = 2\times 2\times 2 = 8$

$\sqrt {64} = 8$

Find $\sqrt {3969}$ by the long division method

Step

$1$ : We write

$3969$ as

$\overline {39} \overline {69}$ by grouping the digits into pairs, starting from right end.
Step

$2$ : Find the number whose square is less than or equal to

$39$ . It is

$6$ .
Step

$3$ : Put

$6$ on the top and also write

$6$ as a divisor.
Step

$4$ : Multiply

$6$ with

$6$ and write the result

$36$ under

$39$ and subtract. The remainder is

$3$ .
Step

$5$ : Bring down the pair

$69$ by the side of this remainder

$3$ , yielding

$369$ .
Step

$6$ : Double

$6$ , take the result

$12$ and find the number

$'n'$ . Such that

$12n\times n$ is just less than or equal to

$369$ .
Since

$122\times 2 = 244; 123\times 3 = 369, n = 3$
Step

$7$ : Write

$123$ as the next divisor and put

$3$ on the top along with

$6$ . Write the product

$123 \times 3 = 369$ under

$369$ and subtract. Since the remainder is

$0$ , the division is complete.
Hence

$\sqrt {3969} = 63$ .

Find the square root of $12.25$

$\sqrt {12.25} = \sqrt {\dfrac {12.25\times 100}{100}}$

$\dfrac {\sqrt {1225}}{\sqrt {100}} = \dfrac {\sqrt {5^{2}\times 7^{2}}}{\sqrt {10^{2}}} = \dfrac {5\times 7}{10}$

$\sqrt {12.25} = \dfrac {35}{10} = 3.5$

Find the square root of
$6+\sqrt { 12 } -\sqrt { 24 } -\sqrt { 8 }$

$6+4\sqrt{12}-4\sqrt{24}-\sqrt{8}$

$=6+2\times\sqrt{3}\times1+2\times(-\sqrt{2})\times\sqrt3+2\times1\times(-\sqrt2)$

$=1+2+3+2\times\sqrt{3}\times1+2\times(-\sqrt{2})\times\sqrt3+2\times1\times(-\sqrt2)$

$=(1+\sqrt3-\sqrt2)^2$

Hence square root of

$6+4\sqrt{12}-4\sqrt{24}-\sqrt{8}$ is

$=\pm(1+\sqrt3-\sqrt2)$

Solve the following equation.
$16x^2-24x+9=0$

Consider the given equation.

$16{{x}^{2}}-24x+9=0$

$\Rightarrow 16{{x}^{2}}-12x-12x+9=0$

$\Rightarrow 4x\left( 4x-3 \right)-3\left( 4x-3 \right)=0$

$\Rightarrow \left( 4x-3 \right)\left( 4x-3 \right)=0$

So, the value of $x$ is $\dfrac{3}{4}$

Verify the validity of the identity
$(a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca$
for the following given values.
$a = 2$ $b = 4$ $c = 6$
$a = 5$ $b = -2$ $c = 3$
$a = -2$ $b = 3$ $c =-4$

i) For

$a=2,b=4,c=6$ we have

$(a+b+c)^2=(12)^2=144$ and

$a^2+b^2+c^2+2ab+2bc+2ca$

$=4+16+36+16+48+24$

$=144$

ii) For

$a=5,b=-2,c=3$ we have

$(a+b+c)^2=(6)^2=36$ and

$a^2+b^2+c^2+2ab+2bc+2ca$

$=25+4+9-20-12+30$

$=36$

iii) For

$a=-2,b=3,c=-4$ we have

$(a+b+c)^2=(-3)^2=9$ and

$a^2+b^2+c^2+2ab+2bc+2ca$

$=4+9+16-12-24+16$

$=9$

Find the square root of the following in the form of a binomial surd.
$n+\sqrt{n^2-1}$ .

Let

$x = \sqrt{n+\sqrt{n^2-1}} = \sqrt x + \sqrt y$

$n+\sqrt{n^2-1} = x + y + 2\sqrt{xy}$

Hence,

$x+y = n , xy = \cfrac{n^2-1}{4}$

$4x^2-4nx+n^2-1 =0$

$x = \cfrac{(n+1)}{2}$ or

$x = \cfrac{(n-1)}{2}$

Similarily,

$y = \cfrac{(n-1)}{2}$ or

$y=\cfrac{(n+1)}{2}$

Triangle $XYZ$ is right-angled at vertex $Z$ . Calculate the length of $YZ$ , if $XY=13$ and $XZ=12$

$\triangle XYZ$ is a right angled triangle

By Pythagoras Theorem,

${ XY }^{ 2 }={ YZ }^{ 2 }+{ XZ }^{ 2 }$

${ 13 }^{ 2 }={ YZ }^{ 2 }+{ 12 }^{ 2 }$

$169-144=25={ YZ }^{ 2 }$

$YZ=\sqrt { 25 } =5 \ units$

943388_1015900_ans_9232cf16abf7411c9dfd7738a52f22c2.png

Find the least number, which must be subtracted from $3250$ to make it a perfect square

This shows that

$57^{2}$ is less than

$3250$ by

$1$ . If we subtract the remainder from the number, we get a perfect square. So the required least number is

$1$ .

Find the square root of $-7-24\sqrt { -1 }$ .

Assume

$\sqrt { -7-24\sqrt { -1 } } =x+y\sqrt { -1 }$
then

${ -7-24\sqrt { -1 } } ={ x }^{ 2 }-{ y }^{ 2 }+2xy\sqrt { -1 }$

$\therefore { x }^{ 2 }-{ y }^{ 2 }=-7....(1)$
and

$2xy=-24$

$\therefore { \left( { x }^{ 2 }+{ y }^{ 2 } \right) }^{ 2 }={ \left( { x }^{ 2 }-{ y }^{ 2 } \right) }^{ 2 }+{ \left( 2xy \right) }^{ 2 }$

$=49+576$

$=625$

$\therefore { x }^{ 2 }+{ y }^{ 2 }=25.....(2)$
From (1) and (2),

${x}^{2}=9;{y}^{2}=16$

$\therefore x=\pm 3,y=\pm 4$
Since the product

$xy$ is negative, we must take

$x=3,y=-4$ ; or

$x=-3,y=4$
Thus the roots are

$3-4\sqrt { -1 }$ and

$-3+4\sqrt { -1 }$ ;
that is

$\sqrt { -7-24\sqrt { -1 } } =\pm (3-4\sqrt { -1 } )$

Find a square of a natural number between 75 and 90.

We know that

$8^2 = 64$ and

$10^2=100$

Since

$64$ and

$100$ does not lie in the given range therefore square of

$9$ may lie in this range.

We know that the square of a number ending with

$9$ or

$1$ ends with

$1$ at its unit's place. Only

$81$ is such a number that exists between the given range and which is also the square of number

$9$ .

Therefore, square between

$75$ and

$90$ is

$81$ .

Check if the following number is perfect square.
1243

The given number,

$1243$ is not a perfect square for the fact that a perfect square never ends in

$2, 3, 7 \ or \ 8$

The factors of

$1243$ are

$11\times113$

Evaluate: $\dfrac{\sqrt{32} + \sqrt{48}}{\sqrt{8} + \sqrt{12}}$

$\dfrac{\sqrt{32} + \sqrt{48}}{\sqrt{8} + \sqrt{12}}$

$=\dfrac{\sqrt{16\times2} + \sqrt{16\times 3}}{\sqrt{4\times 2} + \sqrt{4\times3}}$

$=\dfrac{4\sqrt{2} + 4\sqrt{3}}{2\sqrt{2} + 2\sqrt{3}}$

$=\dfrac{4(\sqrt{2} + \sqrt{3})}{2(\sqrt{2} + \sqrt{3})}$

$=2$ .

Check if the following number is perfect square.
3300

The factors of

$3300$ are

$11\times3\times2\times2\times5\times5$

$=11\times3\times2^2\times5^2$

The factors are not of the form,

$p^2$ .

Hence,

$3300$ isn't a perfect square

Find the square root of a given number by prime factorization method.
3136

Dividing by prime numbers

2	3136
2	1568
2	784
2	392
2	196
2	98
7	49
7	7
	1

Dividing by prime numbers

$3136=2 \times2 \times2 \times2 \times2 \times2 \times7 \times7$

Hence,

$\sqrt{3136}=2 \times2 \times2 \times7=56$

Find the value of $\dfrac { 3 }{ \sqrt { 5 } +\sqrt { 2 } } +\dfrac { 7 }{ \sqrt { 5 } -\sqrt { 2 } } \sqrt { 5 } =2.236$ and $\sqrt { 2 } =1.414$ ?

$\displaystyle = 3\times 2.236 + 1. 414 + \frac{7}{(2.236 - 1.414)}$

$\displaystyle = 16.6378.$

1048186_1038115_ans_ef049b43bed340b3a4f6b0f73fb9b52e.jpg

Find the value of $\sqrt {3}$

1078405_1035998_ans_7a8cc3d2ecdb4ec7a76966186f0629f3.png

Number of solutions of $2^{x} + 3^{x} + 4^{x} = 5^{x}$ is

we know that,

$3^2 + 4^2 = 5^2$ (Pythagorean triplet)

So, there is no such value of

$x$ such that,

$2^x + 3^x + 4^x = 5^x$

Number of solutions

$=0$

A General wishes to draw up his $7500$ soldiers in the form of a largest square. After arranging, he found out that some of them are left out. How many soldiers were left out?

No. of soldiers

$=7500$

Clearly it is between

$85^2$ &

$90^2$

$85^2=7225$ and

$90^2=8100$

$86^2=7396$

$87^2=7569$

So nearest perfect square number to

$7500$ is

$7396$

So, it is

$86^2=7396$

Left out

$=7500-7396$

$=104$ .

Find the square root of $11 + 2\sqrt {30}$

To find :

$\sqrt{11+2 \sqrt{30}}$

We can rewrite

$11+2\sqrt{30}$ as

$5+2\sqrt{30}+6$

$= \dfrac{25+10\sqrt{30}+(\sqrt{30})^{2}}{5}$

$(\because$ divided and multiplied by

$5)$

$\Rightarrow \sqrt{11+2\sqrt{30}} = \sqrt{\dfrac{(\sqrt{30}+5)^{2}}{5}}$

$= \dfrac{\sqrt{30}+5}{\sqrt{5}}$

$= \sqrt{6} + \sqrt{5}$

$\therefore \sqrt{11+2\sqrt{30}} = \sqrt{6} + \sqrt{5}$

Solve the problem:-
$\sqrt 5 = ?$

$\sqrt 5 = 2.23$

Find the square root of $\left(x+\dfrac{1}{x}\right)^2-4\left(x-\dfrac{1}{x}\right)$ .

Now,

$\left(x+\dfrac{1}{x}\right)^2-4\left(x-\dfrac{1}{x}\right)$

$=\left(x-\dfrac{1}{x}\right)^2+4-4\left(x-\dfrac{1}{x}\right)$ [ Since

$(a-b)^2=(a+b)^2-4ab$ ]

$=\left(x-\dfrac{1}{x}\right)^2-2.\left(x-\dfrac{1}{x}\right).2+2^2$

$=\left(x-\dfrac{1}{x}-2\right)^2$

So square root of the given expression is

$\pm\left(x-\dfrac{1}{x}-2\right)$ .

Prove the followings,
i) $(a+b)^2+(a-b)^2=2(a^2+b^2)$
ii) $(a+b)^2-(a-b)^2=4ab$

We have,

$(a+b)^2+(a-b)^2$

$=(a^2+2ab+b^2)+(a^2-2ab+b^2)$

$=2(a^2+b^2)$

ii)

$(a+b)^2-(a-b)^2$

$=(a^2+2ab+b^2)-(a^2-2ab+b^2)$

$=4ab$ .

Find square root of $= 7.1289$ by long division method.

975432_1052154_ans_87eb94e5567b4e0bb417ec3f80d1372c.png

Find the greatest five digit number which is a perfect square. Also, find the square root of the number so obtained.

Division method

We know

$99999$ is largest

$316$

_____________

$3$

$99999$

$+3$

$-9$

_________ _____________

$61$

$99$

$+1$

$-61$

_________ _____________

$626$

$3899$

$-3756$

_____________

$143$

$\therefore 99999-143=99856$ is perfect

$\Rightarrow \sqrt{99856}=316$ .

Find the square root of $361201$ by long division method

1326263_1054947_ans_af5d04964dc34653af06705bf93ea156.png

Find the smallest number by which $2400$ is to be multiplied to get a perfect square and also find the square root of the resulting number.

$2400=$

$=100\times 4\times 69$

$100\times 4\times 2\times 3$

$=(10)^{2}\times (2)^{2}\times (6)$

It should be multiplied with

$6$ to get

$a$ square root

$=(10)^{2}\times (2)^{2}\times (6)^{2}=14400$

Sq. roots

$=10\times 2\times 6=120$

Find whether the square of the following numbers are even or odd?
$2826$ .

$2826$ is an even no. and square of an even no is always even.

Find square root of $625$ by long divison method.

1024581_1060030_ans_1186f5d55fb94abfafda5e0c5c5357f9.png

Find the square of $\dfrac{-12}{13}$

$=\left (\dfrac{-12}{13}\right )^{2}$

$=\dfrac{144}{169}$

Solve: $25{\left( {2x + y} \right)^2} - 16{\left( {x - y} \right)^2}$

$25(2x+y)^2-16(x-y)^2$

$=25(4x^2+y^2+4xy)-16(x^2+y^2-2xy)$

$=100x^2+25y^2+100xy-16x^2-16y^2+32xy$

$=84x^2+9y^2+132xy$

Hence, the value is

$84x^2+9y^2+132xy$ .

Rajini says $1\ sq\ m=100^{2}\ sq\ cm$ . Do you agree? Explain.

Yes, this is correct.

1 m = 100cm

1 sq m = 1m x 1m = 100cm x 100cm =

$100^2 sq cm$

If $x=\dfrac{5}{2}, y=\dfrac{3}{4}$ then find the value of $(x^2+y^1)(x^2-y^1)$ .

1330926_1075404_ans_b0a1e1aacf2746218e4adf3bd65112f7.jpg

If $\sqrt {15625}=125$ , then find the value of $\sqrt {15625}+\sqrt {15625}+\sqrt {0.015625}$

Given

$\sqrt{15625}=125$

$\sqrt{15625}+\sqrt{15625}+\sqrt{0.015625}=125+125+\sqrt{15625\times 10^{-6}}=125+125+125\times 10^{-3}=250.125$

Write a Pythagorean triplet whose one member is
$(i)6$
$(ii)14$
$(iii)16$
$(iv)18$

$(i)\ \textbf{6},8,10$

$(ii)\ \textbf{14},48,50$

$(iii)\ \textbf{16},30,34$

$(iv)\ \textbf{18},24,30$

Find the square root of $529$

2 | 5 29

2 |-4

43 | 1 29

3 | -1 29

$\times$

$\therefore\sqrt { 529 }=23$

Find the square roots of $6084$ by division method.

$6084=2^2\times3^2\times13^2$

$\sqrt{6084}=2\times3\times13$

$=78$

Hence, the square root of

$6084$ is equal to

$78.$

1171708_1090503_ans_3d3388d1d9364ae199349d526566f40d.png

What could be the possible 'one's' digits of the square root of the following number?
$99856$

Since,

$4^2=16$ and

$6^2=36$

Therefore,

$4$ and

$6$ are the possible 'one's' digits of the square root of

$99856$ .

Find the square roots of the following decimal numbers.
i) $2.56$ ii) $18.49$ iii) $68.89$ iv) $84.64$

$2.56=1.6\times 1.6$

$\therefore \sqrt{2.56} = 1.6$ .

$18.49=4.3\times 4.3$

$\therefore \sqrt{18.49}=4.3$

$68.89=8.3\times 8.3$

$\therefore \sqrt{68.89} =8.3$

$84.64=9.2\times 9.2$

$\therefore \sqrt{84.64}= 9.2$

What could be the possible 'one's' digits of the square root of the following numbers?
$9801$

The possible one's digit of the square of the given number will be

$1$ or

$9$ .

As square of the numbers having

$1$ &

$9$ at one's place has always the unit digit is

$1$

Simplify: $(2)^{\dfrac{1}{4}}.(4)^{\dfrac{1}{8}}.(8)^{\dfrac{1}{16}}.(16)^{\dfrac{1}{32}}$ .

$(2)^{1/4} (4)^{1/8} (8)^{1/16} (16)^{1/32}$

$= (2)^{1/4} (2^2)^{1/8} (2^3)^{1/16} (2^4)^{1/32}$

$= (2)^{1/4} (2)^{2/8} (2)^{3/16} 2^{4/32}$

$= (2)^{1/4 + 2/8 + 3/16 + 4/32}$

$= (2)^{\dfrac{8+8+6+4}{32}}$

$= (2)^{26/32}$

$= 2^{13/16}$

$\boxed{Ans\,\, 2^{13/16}}$

1143791_1105502_ans_24e5773bc04a4c04929915d54c26470b.jpg

Evaluate : ${\left( {{{17}^2} - {8^2}} \right)^{\frac{1}{2}}}$

Using this formula

${ a }^{ 2 }-{ b }^{ 2 }=({ a }-{ b })({ a }+{ b })\\ { 17 }^{ 2 }-{ 8 }^{ 2 }=({ 17 }-{ 8 })({ 17 }+{ 8 })\\ \quad ={ 25 }\times { 9 }\\ ={ 5 }^{ 2 }\times { 3 }^{ 2 }\\ ({ { 17 }^{ 2 }-{ 8 }^{ 2 }) }^{ \frac { 1 }{ 2 } }={ 15 }$

Simplify $\sqrt { 10 \frac { 151 } { 225 } }$ .

$\sqrt{10\dfrac{151}{225}}\\$

$=\sqrt{\dfrac{2250+151}{225}}\\$

$=\sqrt{\dfrac{2401}{225}}\\$

$=\sqrt{\dfrac{49\times 49}{15\times 15}}\\$

$=\dfrac{49}{15}$

Which of $77^2,82^2,161^2,109^2$ will end with the digit $1$ ?

Here, the squares of those numbers end with

$1$ which end with

$1$ or

$9$ .

Here,

${ 77 }^{ 2 }=5929,\\ { 82 }^{ 2 }=6724,\\ { 161 }^{ 2 }=25921,\\ { 109 }^{ 2 }=11881.$

Hence, the squares of the numbers ending with the digit

$1$ are

$161$ and

$109$ .

Solve $\sqrt{444}$

$\sqrt {444}$

$= \sqrt {2 \times 2 \times 111}$

$= 2\sqrt {111}$

$= 21.07$

simplify $\sqrt {98}$

$\sqrt { 98 } =\sqrt { 2\times 7\times 7 } \quad =\quad \sqrt { 2\times { 7 }^{ 2 } } =\quad 7\sqrt { 2 }$

A rational number whose square is $\dfrac { 81 }{ 64 }$ is ______ .

Let the rational number be

$\dfrac{p}{q}$ whose square is

$\dfrac{81}{64}.$

$\Rightarrow \bigg( \dfrac{p}{q} \bigg)^2=\dfrac{81}{64}$

$\Rightarrow \dfrac{p}{q}=\sqrt{\dfrac{81}{64}}$

$=\dfrac{9}{8}$

Hence,

$\dfrac{9}{8}$ is the required rational number.

Square of $-\dfrac { 13 }{ 17 }$ is_____

Square of $\dfrac { -13 }{ 17 } = \dfrac { -13 }{ 17 }^2$
$=\dfrac { -13 }{ 17 } \dfrac { -13 }{ 17 } =(-1) \dfrac { 13 }{ 17 } (-1) \dfrac { 13 }{ 17 }$
$=(+1) \dfrac { 169 }{ 289 } =\dfrac { 169 }{ 289 }$

The approximate value of $\sqrt { 50 }$ is

$\sqrt{50}=\sqrt{25\times 2}$

$=5\times \sqrt{2}$

$=5\times 1.414$

$=7.07$

Solve:-
$\dfrac{2\sqrt{6}-\sqrt{5}}{3\sqrt{5}-2\sqrt{6}}$

$\dfrac{{2\sqrt 6 - \sqrt 5 }}{{3\sqrt 5 - 2\sqrt 6 }}$

$= \dfrac{{2 \times 2.44 - 2.23}}{{3 \times 2.23 - 2 \times 2.44}}$

$= \dfrac{{2.65}}{{1.81}} = 1.464$

Find the square roots of the following numbers by prime factorization method.
$441$

1115740_1202375_ans_74b9c9ea1b514759ac8f1e70ae6c56a1.jpg

Find the square roots of the following numbers by prime factorization method.
$784$

1115734_1202376_ans_9524654ac6044998869820e9c6bdebb2.jpg

Find the square roots of the following numbers by prime factorization method.
$4096$

Given number is

$4096$ .

To find out,

The square root of the number using the prime factorization method.

We can represent

$4096$ as a product of its prime factors as:

$4096=2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2$

$\Rightarrow 4096=2^{12}$

$\Rightarrow 4096=(2^6)^2$

$\Rightarrow \sqrt{4096}=2^6$

$\Rightarrow \sqrt{4096}=2\times 2\times 2\times 2\times 2\times 2$

$=64$

Hence, the square root of

$4096$ is

$64$ .

Find the square roots of the following numbers by prime factorization method.
$7056$

1115722_1202380_ans_749184194b5f40f0a58dc1ba8dc95e13.jpg

Find the square of the following :
$7\frac{1}{2}$

$7\dfrac {1}{2} = \dfrac {15}{2}$

$\left(7\dfrac {1}{2}\right)^2$

$=$

$\left(\dfrac {15}{2}\right)^2$

$=$

$\dfrac {225}{4}$

Find $\sqrt{1276}$

$\sqrt { 1276 } \\ 1276=2\times 2\times 11\times 29\\ \therefore \sqrt { 1276 } =\sqrt { 4\times 11\times 29 } =2\sqrt { 319 }$

Find:
$5\sqrt {25} + \sqrt {25} - 5\sqrt {25} = .......$

$5\sqrt{25}+\sqrt{25}-5\sqrt{25}$

$=\sqrt{25}$

$=5$

Find the square root of $(x^{2}+4x+4)$ .

$\sqrt{x^{2}+4x+4}$

$= \sqrt{x^{2}+2^{2}+(2x)}$

$= \sqrt{(x+2^{2})}$

$= \pm (x+2)$

1113751_1200684_ans_c89615a36bf94c508538005deed80ea4.jpg

Solve : $20\sqrt { 15 }$

We have,

$20\sqrt1{5}$

$\Rightarrow 20\times\sqrt{5}\times\sqrt3$

Hence, this is the answer.

What is the square root of $18$ ?

$3.\sqrt{2}$
Explanation:
Factor it to primes and pull the duplicate

$3$ out or the root:

$\sqrt{18}=\sqrt{2.3.3}=3.\sqrt{2}$

Solve
$5^{2}+6^{2}+7^{2}+...+20^{2}$

$A]5^{2}+6^{2}+7^{2}+...+20^{2}$

$=(1^{2}+2^{2}+...+20^{2})-(1^{2}+2^{2}+3^{2}+4^{2})$

sum of square is

$\frac{n(n+1)(2n+1)}{6}$

for n = 20

$Sum=\frac{20(20+1)(40+1)}{6}=\frac{20\times 21\times 41}{6}=2870$

for n = 4

$sum=\frac{4(4+1)(8+1)}{6}=2\times 5\times 3=30$

$\therefore 5^{2}+6^{2}+7^{2}+...20^{2}$

= 2870 - 30

= 2840

$\therefore$ Required sum is 2840

1182476_1293407_ans_cbff59c8b9b94541a7e1f5ba8362a2a4.jpg

Find the last $4$ digit no which is a perfect square.

$100^{2} \rightarrow 10000$ first five digit . no.

$99^{2} \rightarrow 9810.$

$\therefore$ 9810 is the last 4 digit no. which is perfect square.

1213339_1305408_ans_3b0aa062d7394eeeb5a75c9acac72456.JPG

Expression the following as the product of expression through prime factorization
$144$

We have,

$\begin{matrix} \dfrac { { 144 } }{ 2 } =72 \\ \dfrac { { 36 } }{ 2 } =18 \\ \dfrac { { 18 } }{ 2 } =9 \\ \dfrac { 9 }{ 3 } =3 \\ \end{matrix}$

Then,

We get

$144=2 \times 2 \times 2 \times 2 \times 3 \times 3 = {2^4} \times {3^2}$

Identify with reason whether the following are Pythagoras triplet $24, 70, 74$ .

Pythagoras triplets can be checked as below :

$a^{2} + b^{2} = c^{2}$

As we can see that,

$24^{2}+70^{2} = 74^{2}$

Hence (24, 70 ,74) is a Pythagorean triplet.

Evaluate : $\sqrt{5.38\times 0.47}$

$\sqrt { 5.38\times 0.47 } =\sqrt { 2.5286 } =1.59$

Evaluate: $\sqrt { 616225 }$

For finding square root, first we find the prime factors of the given number.

Hence,

$\sqrt{616225}=\sqrt{5\times 5\times157\times157}$

$=\sqrt{(5\times 157)^2}$

$=5\times 157$

$=785$

Find answer
$\sqrt {5x}\times \sqrt {5x}=$

We have,

$\sqrt{5x}\times \sqrt{5x}$

$\Rightarrow \sqrt{5x\times 5x}$

$\Rightarrow 5x$

Hence, this is the answer.

Find the square roots of $11\dfrac {9}{18}$

Let

$y=11\frac{9}{18}$

$y=\frac{207}{18}$

then

$\sqrt{y}=\sqrt{\frac{207}{18}}$

$\sqrt{y}=\sqrt{\frac{23}{2}}= 3.39$

1214721_1369671_ans_eb6f7b3363c94c54ab0b7db9c81781e8.JPG

Simplify: $\sqrt { \dfrac { 35.87\times 0.0914 }{ 0.0578 } }$

$\sqrt{\dfrac{35.87\times 0.0914}{0.0578}}$

$=\sqrt{\dfrac{3587}{100}\times \dfrac{914}{10000}\times \dfrac{10000}{578}}$

$=\dfrac{1}{10}\sqrt{\dfrac{3587\times 914}{578}}$

$=\dfrac{1}{10}\sqrt{5672.18}=\dfrac{75.314}{10}=7.5314$ . (approx.)

1222982_1319647_ans_8e68265f098c4fd7b41f5f78ea3ad09e.PNG

Evaluate :
$\sqrt{\dfrac{2809}{4096}}$

We have,

$\sqrt{\dfrac{2809}{4096}}$

$\sqrt { \dfrac { 53\times 53 }{ 64\times 64 } } =\boxed { \dfrac { 53 }{ 64 } }$

Hence, this is the answer.

Find the following .
$\sqrt{6146.56}$

$\sqrt{6146.56}$

$= \sqrt{614656\times 10^{-2}}$

$= \sqrt{2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 74\times 10^{-2}}$

$= \sqrt{2^{8},7^{4},10^{-2}}$

$=2^{8/2},7^{4/2},10^{-2/2}$

$= 2^{4},7^{2},10^{-1}$

=78.4

1175676_1348069_ans_9caaf899560b41c38f1c1a8f5ebde8cd.jpg

Evaluate $\left( x-\dfrac { 1 }{ x } \right) ^{ 2 }-\left( x+\dfrac { 1 }{ x } \right) ^{ 2 }$

$\left(x-\dfrac{1}{x}\right)^2-\left(x+\dfrac{1}{x}\right)^2$

$\Rightarrow x^2+\dfrac{1}{x^2}-2-\left(x^2+\dfrac{1}{x^2}+2\right)$

$\Rightarrow x^2+\dfrac{1}{x^2}-2-x^2-\dfrac{1}{x^2}-2$

$\Rightarrow -4$ .

1199902_1445216_ans_29b1d05d2f474141a411afc86c6347aa.jpg

$\sqrt { 784 }$ by long division method

$2| 784$

___________

$2| 392$

___________

$2| 196$

___________

$2| 98$

___________

$7| 49$

___________

$7| 7$

___________

$1$

$=2\times 2\times 2\times 2\times 7\times 7$

$=\sqrt{784}=2\times 2\times 7$

$=28$ .

1205744_1453922_ans_e6b3d6bc454941c28dcd69a6055d21e5.jpg

$\sqrt {129}=z$ . Find the value of $z$ .

$z=\sqrt{129}$

$z=\sqrt{43\times 3}$

$z=11.35$

1214720_1369673_ans_363a2850ab6d48c9b9b41cd28ba983cd.JPG

Write the Pythagorean triplet whose one member is
$\left(i\right)16$
$\left(ii\right)14$

1211587_1376186_ans_b8f13d5d4b024bf7971a3d32eaecc357.jpg

What is the value of $3^{2} +4^{2} ?$

$3^{2}+4^{2} = 9+16=25$

Find the smallest number by which $882$ be multiplied so that the product becomes a perfect square.

Prime factors of

$882=2\times 3\times 3\times 7\times 7$

Here, 2 dosen't make a pair.

$\therefore$ We need to multiply 882 by 2 to make it a perfect square.

1204282_1534542_ans_358794fb8c5142d3a8ee681e4e4c00f8.PNG

$\sqrt{0.6}$

Let

$y = \sqrt{x}, x = 0.64, y = 0.8$

$\Delta x = 0.6 - 0.64 = -0.04$

$\because y = \sqrt{x}$

Diff. W.r.t. x

$\because \frac{dy}{dx} = \frac{1}{2\sqrt{x}}$

$\therefore dy = (\frac{dy}{dx}) \Delta x$

$\therefore dy = \frac{1}{2\sqrt{x}} . \Delta x$

$\Rightarrow dy = \frac{1}{2\sqrt{0.64}} \times (-0.04)$

$\Rightarrow dy = - \frac{0.04}{2 \times 0.8} = - \frac{0.04}{1.6}$

$\Rightarrow dy = - \frac{4}{160} = - \frac{1}{40} = - 0.025$

$= \Delta y$

$(\because \Delta y \cong dy)$

$\therefore \sqrt{0.6} = y + \Delta y$

$= 0.8 + (-0.025)$

$= 0.8 - 0.025 = 0.775$

Hence, approximate value of

$\sqrt{0.6}$ is

$0.775$ .

Estimate the square root : $4\sqrt {10}$

1993362_1744908_ans_c8d1b145bd86480785c69909a595b4bc.jpeg

Estimate the square root : $3\sqrt 5$

1993361_1744907_ans_52d96460325d47d19abf0f3589f6a422.jpeg

Estimate the square root : $2\sqrt 3$

1993369_1744905_ans_7c45ec63d0f44c0ba3d0516e28ca290b.jpeg

Estimate the square root : $4\sqrt 2$

1993358_1744906_ans_c3d870c29a32494db7d76d9a2be5dcb7.jpeg

Write down the cube root of $64$

Ans : 4

cube root of 64 is 4.

$\sqrt{401}$

Let

$y = x^{1/2}, x = 400, y = 20,$

$\Delta x = 401 - 400 = 1$

$\because y = x^{1/2}$

Diff. w.r.t. x

$\frac{dy}{dx} = \frac{1}{2}x^{1/2 - 1}$

$\Rightarrow \frac{dy}{dx} = \frac{1}{2}x ^{-1/2}$

$= \frac{1}{2x^{1/2}} = \frac{}{2\sqrt{x}}$

$\therefore dy = (\frac{dy}{dx}) \Delta x = \frac{1}{2\sqrt{x}} \times \Delta x$

$\Rightarrow dy = \frac{1}{2\sqrt{400}} \times 1$

$= \frac{1}{2 \times 20} = \frac{1}{40}$

$\Rightarrow dy = 0.025 = \Delta y (\because \Delta y \cong dy)$

$\therefore (401)^{1/2} = y + dy$

$= 20 + 0.025 = 20.025$

Hence, approximate value of

$\sqrt{401}$ is

$20.025$ .

Find the smallest whole number by which $1458$ should be multiplied so as to get a perfect square number.

By prime factorisation of

$1458$ , we get

$1458=2\times 3\times 3\times 3\times 3\times 3\times 3$
Here,

$3$ are in pairs but

$2$ needs a pair to make

$1458$ a perfect square.
So,

$1458$ needs to be multiplied by

$2$ to make it a perfect square.
Thus, the number is

$1458\times 2=2916$

$\sqrt { 2916 } =\sqrt { 2\times 2\times 3\times 3\times 3\times 3\times 3\times 3 }$

$=2\times 3\times 3\times 3$

$=54$

Thus

$2$ is the required smallest whole number to make the given number a perfect square.

$2025$ plants are to be planted in a garden in such a way that each row contains as many plants as the number of rows. Find the number of plants in each row.

Let the number of rows be

$x$
Thus, the each row contains

$x$ plants.
The total number of plants in the garden = number of rows

$\times$ number of plants in each rows
Thus,

$2025=x\times x$

$=>x^{ 2 }=2025$

$=>x=\sqrt { 2025 }$

$=\sqrt { 3\times 3\times 3\times 3\times 5\times 5 }$

$=3\times 3\times 5$

$=45$
Thus, there are

$45$ rows and each row contains

$45$ plants.

Determine all pairs of positive integers (m, n) for which $2^{m}+3^{n}$ is a perfect square.

Let

$2^{m}+3^{n}=k^{2}$
Since

$(-1)^{m}\equiv 2^{m}\equiv k^{2}\equiv 1 (mod 3)$

$(since, 3 | k)$ m is even, say 2p.
Now,

$(k-2^{p})(k+2^{p})=3^{n}$

$\Rightarrow k-2^{p}=1 and k+2^{p}=3^{n}$

$\Rightarrow 2^{p+1}+1 =3^{n}$
Since

$(-1)^{n}\equiv 3^{n} (mod 4)=2^{p+1}+1\equiv 1, n$ is even, say 2q.
Now

$(3^{q}-1)(3^{q}+1)=2^{p+1}\Rightarrow 3^{q}-1=2$

$\Rightarrow 3^{q}=3\Rightarrow q=1$ and hence

$p=2$
So, we have only one solution (4, 2)

A man, after a tour, finds that he had spent every day as many rupees as the number of days he had been on tour. How long did his tour last, if he had spent in all $Rs.\ 1,296$ ?

${\textbf{Step-1: Multiply days spent tour and amount of money he spent each day. }}$

${\text{Let the number of days be}}$

$x.$

${\text{Let the amount of money he spent each day be}}$

$Rs.x.$

${\text{Total amount spent in the entire tour}}$

$= Rs. x \times x = Rs. 1296.$

${\textbf{Step-2: Apply square root on both sides.}}$

$\Rightarrow$

${x}^{2} = 1296$

$\Rightarrow$

$x = \sqrt{1296}$

$=36.$

${\textbf{Hence, the tour was for 36 days and amount Rs.36 he spent each day.}}$

Out of 745 students, maximum are to be arranged in the school field for a P.T. display, such that the number of rows is equal to the number of columns. Find the number of rows if 16 students were left out after the arrangement.

$\textbf{Step -1: Find the number of students that are required for the given arrangement.}$

$\text{After making the arrangement such that the no. of rows and no. of columns are same, }$

$16\text{ students were left out.}$

$\text{So, if we subtract }16 \text{ students, we will get the required number of students.}$

$\therefore\text{Number of students should be }745-16=729$

$\textbf{Step -2: Find the number of rows in the arrangement}$

$\text{Let }729\text{ students are arranged in }x\text{ rows and }x\text{ columns in the school field.}$

$\Rightarrow x\times x=729$

$\Rightarrow x^2=27^2$

$\Rightarrow x=\pm27$

$\text{Since, the number of rows cannot be negative, therefore x = 27}$

$\textbf{Hence, the number of rows }\mathbf{27.}$

State true or false.
The square root of 739.461 correct to two decimal places is 27.193.

$\quad \\ \quad \quad \quad \quad \quad 27.1\\ \quad \quad \quad \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \\ \quad \quad 2)739.461\\ \quad \quad \quad -4\quad \downarrow \\ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \\ 47)03\quad 39\\ \quad \quad -0329\quad \quad \downarrow \\ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \\ 541)\quad 010\quad 46\\ \quad \quad -\quad \quad 0541\quad \quad \downarrow \\ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \\ \quad \quad \quad \quad \quad \quad 505\quad$

Find the square root of $4356$ by prime factorization method.

Resolving into prime factors,

$4356=\underline{2\times2}\times\underline{3\times3}\times\underline{11\times11}$ .

Forming pairs of like prime factors and picking out one factor from each pair,

$\sqrt{4356}=2\times3\times11=66.$

The following numbers are not perfect squares. Give reason
$1057$
$23453$
$7928$

A number that ends either with

$2,\,3,\,7,\;8$ or an odd number of zeroes cannot be a perfect square.

Since the given number $1057$ ends with $7$ , so it cannot be a perfect square.
Since the given number $23453$ ends with $3$ , so it cannot be a perfect square.
Since the given number $7928$ ends with $8$ , so it cannot be a perfect square.

What will be the unit digit of the squares of the following numbers?
$81$
$272$
$799$

The unit digit of the squares of the given numbers is shown against the number in the following table.

Number	Unit digit in the square of the number	Reason
$81$	$1$	$1\times1=1$
$272$	$4$	$2\times2=4$
$799$	$1$	$9\times9=81$

Evaluate $\sqrt{7056}$ .

$7056 = \underline { 2\times 2 } \times \underline { 2\times 2 } \times \underline { 3\times 3 } \times \underline { 7\times 7 }$

Hence, $\sqrt {7056} = \sqrt {\underline { 2\times 2 } \times \underline { 2\times 2 } \times \underline { 3\times 3 } \times \underline { 7\times 7 }} = 2 \times 2 \times 3 \times 7 = 84$

Find the greatest $4$ -digit number, which is a perfect square. Find its square root.

The greatest

$4$ -digit number is

$9999$ . First, find the least

number which when substracted from

$9999$ gives a perfect square.

$\;\;\;\;\;\,\mid\;\;\;9\;\;\;9 \\{\overline{9\;\;\;\mid\;\;\underline{99}\underline{99}}}\\\;\;\;\;\;\mid-81\\{\overline{189\mid\;\;1899}}\\\;\;\;\;\;\mid-1701\\{\overline{\;\;\;\;\;\;\mid\;\;\;\;198}}$
The remainder is

$198$ . The required number, which is a perfect square is

$9999-198=9801$
and clearly,

$\sqrt{9801}=99$ .

Estimate the square root of $140$ to the nearest integer.

We know that

$11^2=121\;and\;12^2=144$
So,

$\sqrt{140}$ will lie between

$\sqrt{121}\;and\;\sqrt{144}$ i.e. lie between

$11\;and\;12$ .
Now

$140-121=19$
and

$144-140=4$ .
So the square root will be close to

$\sqrt{144}$ .
The approximate value of

$\sqrt{140}$ is

$12$ .

$5929$ students are sitting in an auditorium in such a manner that there are as many students in a row as there are rows in the auditorium. How many rows are there in the auditorium?

Let there be

$'a'$ rows in the auditorium
Since the number of students in a row is same as the number of rows in the auditorium.

$\therefore$ Number of students in a row

$=a$

$\Rightarrow$ Number of students in

$'a'$ rows

$=a\times a=a^2$
It is given that the total number of students in the auditorium

$=5929$ .

$\therefore\;a^2=5929$

$\Rightarrow\;a=\sqrt{5929}$

$7\;\;\mid 5929 \\{\overline{7\;\;\mid\;\;847}}\\{\overline{11\mid\;\;121}} \\ {\overline{\;\;\;\;\;\mid\;\;\; 11}}$

$\Rightarrow\;a=\sqrt{(7\times7)\times(11\times11)}$

$\;\;\;\;$ [By prime factorisation]

$\Rightarrow\;a=7\times11$ [Taking one factor in each pair]

$\Rightarrow\;a=77$
Hence, there are

$77$ rows in the auditorium.

If $ABC x CBA = 65125$ where $A, B$ and $C$ are single digits the $A + B + C =$ ?

As the unit digit of the product is

$5$ therefore the unit digit of one of

the numbers is

$5$ and the unit digit of the other number is odd

Therefore

$AB5\times 5BA = 65125$ where

$A = 1, 3, 5, 7$ or

$9$
As the product

of two three-digit number is a five-digit number and not a six-digit

number A can only be equal to

$1. IB5\times 5B1 = 65125$
The digit sums of

both numbers

$1B5$ and

$5B1$ will be same Therefore the product would give

digit sum of a perfect square The digit sum on the

$R. H. S$ . is

$1$

Therefore the digit sum of each number can be

$1$ or

$8$ Correspondingly

$B$

will be

$4$ or

$2$ (as digit sum cannot be equal to

$1$ )
Keeping

$B = 2$ we can see that

$125\times 521 = 65125$

Find the square root of a number $400$ by the prime factorisation method.

Using prime factorisation method, we have

$\displaystyle 400=2^{2}\times 2^{2}\times 5^{2}$

Taking root on both sides, we get

$\displaystyle \sqrt{400}=\sqrt {2^2\times 2^2\times 5^2}$

$=2\times 2\times 5=20$

For the following number, find the smallest whole number by which it should be multiplied so as to get a perfect square number.
$2028$

By prime factorization of

$2028$ , we get

$\displaystyle 2028=2\times 2\times 3\times 13\times 13$

Here

$2$ and

$13$ are in pair, but

$3$ needs a pair to make

$2028$ a perfect square.

Thus

$2028$ needs to be multiplied by

$3$ to become a perfect square.

Find the square root of a number $4096$ by prime factorisation method.

By prime factorisation method, we have

$\displaystyle 4096=2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2$

$\displaystyle \Rightarrow 4096=2^{2}\times 2^{2}\times 2^{2}\times 2^{2}\times 2^{2}\times 2^{2}$

Taking square root, we have

$\displaystyle \sqrt{4096}=\sqrt{2^{2}\times 2^{2}\times 2^{2}\times 2^{2}\times 2^{2}\times 2^{2}}$

$\displaystyle=2\times 2\times 2\times 2\times 2\times 2=64$

Find the square root of the following number by the prime factorisation method.
$9216$

We can factorize

$\displaystyle 9216$ as

$2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 3\times 3$

$\displaystyle \Rightarrow 9216=2^{2}\times2^{2}\times 2^{2}\times 2^{2}\times 2^{2}\times 3^{2}$

$\displaystyle \Rightarrow \sqrt{9216}=\sqrt{2^{2}\times2^{2}\times 2^{2}\times 2^{2}\times 2^{2}\times 3^{2}}$

$\displaystyle \Rightarrow \sqrt{9216}=2\times 2\times 2\times 2\times 2\times 3 =32\times 3 =96$
Thus answer is

$96$ .

Find the smallest whole number by which $180$ should be multiplied, so as to get a perfect square number.

By prime factorisation, we get

$\displaystyle 180=3\times 3\times 2\times 2\times 5$

Here

$3$ and

$2$ are in pair, but

$5$ needs a pair to make

$180$ a perfect square.

$180$ needs to be multiplied by

$5$ to become a perfect square.

Find the smallest whole number by which $768$ should be multiplied, so as to get a perfect square number.

By prime factorisation of

$768$ , we get

$\displaystyle 768 = 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 3$

Here

$2$ are in pair, but

$3$ needs a pair to make

$768$ a perfect square.

So,

$768$ needs to be multiplied by

$3$ to become a perfect square.

Find the square root of a number $1764$ by the prime factorisation method.

By using prime factorisation method:

$\displaystyle 1764=2\times 2\times 3\times 3\times 7\times 7=2^{2}\times 3^{2}\times 7^{2}$

Taking square root, we have

$\displaystyle \sqrt{1764}=\sqrt{2^{2}\times 3^{2}\times 7^{2}}$

$=2\times 3\times 7$

$\displaystyle =42$

$1008$

By prime factorization of

$1008$ , we get

$\displaystyle 1008=2\times 2\times 2\times 2\times 3\times 3\times 7$

Here

$2$ and

$3$ are in pair, but

$7$ needs a pair to make

$1008$ a perfect square.

Thus

$1008$ needs to be multiplied by

$7$ to become a perfect square.

Find the smallest whole number by which $2645$ should be multiplied so as to get a perfect square number.

By prime factorisation of

$2645$ , we get

$\displaystyle 2645=5\times 23\times 23$

Here

$23$ is in pair, but

$5$ needs a pair.

Thus

$5$ can be eliminated after dividing

$2645$ by

$5$ .

Hence,

$2645$ needs to be divided by

$5$ to become a perfect square.

Find the smallest whole number by which $252$ should be divided, so as to get a perfect square.

By prime factorisation of

$252$ , we get

$\displaystyle 252=2\times 2\times 3\times 3\times 7$

Here

$2$ and

$3$ are in pair, but

$7$ has no pair which can be eliminated after dividing

$768$ by

$7$ .

Hence,

$252$ needs to be divided by

$7$ to become a perfect square.

Find the smallest whole number by which $396$ should be multiplied, so as to get a perfect square number.

By prime factorisation of

$396$ , we get

$\displaystyle 396=2\times 2\times 3\times 3\times 11$

Here

$2$ ans

$3$ are in pair, but

$11$ needs another

$11$ to make a pair.

Thus

$11$ can be eliminated after dividing

$396$ by

$11$ .

Thus

$396$ needs to be divided by

$11$ to become a perfect square.

Find the square root of the number 729 by the prime factorisation method.

$729$ can be written as:

$729=3\times 243$

$=3\times 3\times 81$

$=3\times 3\times 27$

$=3\times 3\times 3\times 9$

$=3\times 3\times 3\times 3\times 3\times 3$

Taking root on both the sides, we get

$\sqrt {729}=\sqrt {3\times 3\times 3\times 3\times 3\times 3}$

$=3\times 3\times 3$

$=27$

Find the square root of 7744.

Given number is

$7744$

The number can be represented as a product of its prime factors, as shown below:

$7744=2\times 2\times 2\times 2\times 2\times 2\times 11\times 11$

$2^{2}\times 2^{2}\times 2^{2}\times 11^{2}=(2\times 2\times 2\times 11)^{2}$

By pairing the prime factors, the square root of

$7744=2\times 2\times 2\times 11=88$

Hence, the square root of

$7744$ is

$88$

Find the square of the integer just greater than the square root of the following number:
540

We have,

$\sqrt{540}=23.23$

The integer just greater than

$23.23$ is

$24$ .

$\therefore$ we need to find the square of

$24$ and that is

$24^2=576$ .

Hence, the answer is

$576$ .

Find the square of the integer just less than the square root of the following number:
5500

We have,

$\sqrt{5500}=74.16$

The integer just less than

$74.16$ is

$74$ .

$\therefore$ we need to find the square of

$74$ and that is

$74^2=5476$ .

Hence, the answer is

$5476$ .

For each of the following numbers find the smallest whole number by which it should be multiplied so as to get a perfect square number. Also find the square root of the square number so obtained .

(i) 252
(ii) 180
(iii) 1008
(iv) 2028
(v) 1458
(vi) 768

(i)

$252= 2\times 2\times3\times3\times7$
Here,

$2$ and

$3$ are in pairs but

$7$ needs a pair. Thus,

$7$ can become a pair after multiplying

$252$ with

$7$ .
So,

$252$ will become a perfect square when multiplied by

$7$ .
Thus, Answer

$= 7$

Perfect square no.

$= 252 \times 7 = 1764$
Square root of

$1764 = 42$

(ii)

$180 = 3 \times 3 \times 2 \times 2 \times 5$
Here,

$3$ and

$2$ are in pair but

$5$ needs a pair to make

$180$ a perfect square.

$180$ needs to be multiplied by

$5$ to become a perfect square.
Thus, Answer

$= 5$

Perfect square no.

$= 180 \times 5 = 900$
Square root of

$900 = 30$

(iii)

$1008 = 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 7$
Here,

$2$ and

$3$ are in pair, but

$7$ needs a pair to make

$1008$ a perfect square.
Thus,

$1008$ needs to be multiplied by

$7$ to become a perfect square
Hence, Answer

$= 7$
Perfect square no.

$= 1008 \times 7 = 7056$
Square root of

$7056 = 84$

(iv)

$2028 = 2 \times 2 \times 3 \times 13 \times 13$
Here,

$2$ and

$13$ are in pair, but

$3$ needs a pair to make

$2028$ a perfect square.
Thus,

$2028$ needs to be multiplied by

$3$ to become a perfect square.
Hence, Answer

$= 3$
Perfect square no.

$= 2028 \times 3 = 6084$
Square root of

$6084 = 78$

(v)

$1458 = 2 \times 3 \times 3 \times 3 \times 3 \times 3 \times 3$
Here,

$3$ are in pair, but

$2$ needs a pair to make

$1458$ a perfect square.
So,

$1458$ needs to be multiplied by

$2$ to become a perfect square.
Therefore, Answer

$= 2$
Perfect square no.

$= 1458 \times 2 = 2916$
Square root of

$2916 = 54$

(vi)

$768 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3$
Here,

$2$ are in pair, but

$3$ needs a pair to make

$768$ a perfect square.
So,

$768$ needs to be multiplied by

$3$ to become a perfect square.
Hence, Answer

$= 3$
Perfect square no.

$= 768 \times 3 = 2304$

Square root of

$2304 = 48$

Find the least number that must be subtracted from $5607$ so as to get a perfect square. Also, find the square root of the perfect square.

Let us try to find

$\displaystyle \sqrt{5607}$ by long division method.

We get the remainder 131. This shows that

$\displaystyle (74)^{2}$ is less than

$5607$ by

$131$ .
This means if we subtract the remainder from the number we get a perfect square.

Therefore, the required perfect square is

$5067-131=5467$ and

$\displaystyle \sqrt{5476} =74$ .

State whether $7927$ is a perfect square or not.

We need to find square root of

$7927$ .

The unit's place in this number is

$7$ .

In any case, when we take square, we does not get

$7$ at uni'ts place.

Thus

$7927$ is not a perfect square.

Find the square root of $33453$ .

In order to find square root of

$33453$ , we need to split this into groups.

The first group will be

$3$ , second

$34$ and third one is

$53$ .

At first, we need to take square of a number close to equal to

$3$ i.e.,

$1$ .

After subtracting

$1$ from

$3$ , we get

$2$ . Then take next group of digits down.

Then the number will become

$234$ .

The further division is shown in figure.

Thus the square root of

$33453$ is

$182.9$

Find the square root of $22222$

The square root of

$22222$ is shown in figure by long division method.

It comes out to be

$149.06$ .

State whether $23453$ is a perfect square or not.

We need to find square root of

$23453$

But the unit digit in this number is

$3$ .

In any case, we do not get

$3$ in unit's place in any square number at unit's place.

Thus

$23453$ is not a perfect square.

Find the square root of $1058$ .

The unit digit of number 1058 is 8. 1058 is not a perfect square since no square number ends is the digit 8.

Estimate square root of $447$ .

The square root of

$447$ is shown in figure by division method.

Its square root comes out to be

$21.14$ .

Using prime factorization method, find the square root of the following:
(i) $390625$ (ii) $119025$ (iii) $193600$

(i)

$390625$ can be factorized as

$\sqrt {5\times 5\times 5\times5\times 5\times 5\times 5\times 5}$

$=5\times 5\times \times 5\times 5$

$=625$

(ii)

$119025$ can be factorized as

$\sqrt {5\times 5\times 3\times 3\times 23 \times 23}$

$=5\times 3\times 23$

$=345$

(iii)

$193600$ can be factorized as

$\sqrt {5\times 5\times 2\times 2\times 2\times 2\times 2\times 2\times 11\times 11}$

$=5\times 2\times 2\times2\times 11$

$=440$

Find the square root of the following numbers:
(i) $1444$ (ii) $1849$ (iii) $5776$ (iv) $7921$

(i)

$\begin{aligned}{}1444 &= 2 \times 2 \times 19 \times 19\\\sqrt {1444}&= \sqrt {2 \times 2 \times 19 \times 19} \\& = 2 \times 19\\& = 38\end{aligned}$

(ii)

$\begin{aligned}{}1849&=43\times 43\\\sqrt {1849} &= \sqrt {43 \times 43} \\& = 43\end{aligned}$

(iii)

$\begin{aligned}{}5776 &= 2 \times 2 \times 2 \times 2 \times 19 \times 19\\\sqrt {5776} & = \sqrt {2 \times 2 \times 2 \times 2 \times 19 \times 19} \\& = 2 \times 2 \times 19\\ &= 76\end{aligned}$

(iv)

$\begin{aligned}{}7921&=89\times 89\\\sqrt {7921} & = \sqrt {89 \times 89} \\ &= 89\end{aligned}$

Find the smallest number that must be added to get a perfect square.
(i) $2361$ (ii) $4931$ (iii) $18265$ (iv) $390700$

(i)

$2361$

$2500 > 2361$

$\Rightarrow 50 > \sqrt{2361}$

$2401 > 2361$

$\Rightarrow 49 > \sqrt{2361}$

$2304 < 2361$

$\Rightarrow 48 < \sqrt{2361}$

$\therefore 2401$ is the perfect square nearest to

$2361$ .

$\therefore 2401 - 2361 = 40$ is the number to be added.

$\therefore 40$ is the least number to be added to

$2361$ to make it a perfect square.
(ii)

$4931$

$4900 < 4931$

$\Rightarrow 70 < \sqrt{4931}$

$5041 > 4931$

$\Rightarrow 71 > \sqrt{4931}$

$\Rightarrow 5041$ is the perfect square nearest to

$4931$ .

$\Rightarrow 5041 - 4931 = 110$ is the number to be added.

$\therefore 110$ is the least number to be added to

$4931$ to make it a perfect square.
(iii)

$18265$

$18225 < 18265$

$\Rightarrow 135 < \sqrt{18265}$

$18496 > 18265$

$\Rightarrow 136 > \sqrt{18265}$

$\Rightarrow 18496$ is the perfect square nearest to

$18265$ .

$\Rightarrow 18496 - 18265 = 231$ is the number to be added to

$18265$ .

$\therefore 231$ is the least number to be added to

$18265$ to make it a perfect square.
(iv)

$390700$

$390625 < 390700$

$\Rightarrow 625 < \sqrt{390700}$

$391876 > 390700$

$\Rightarrow 626 > \sqrt{390700}$

$\Rightarrow 391876$ is the perfect square nearest to

$390700$ .

$\Rightarrow 391876 - 390700 = 1176$ is the number to be added to

$390700$ .

$\therefore 1176$ is the least number to be added to

$390700$ to make it a perfect square.

Find the square roots of the following numbers correct to two places of decimal:
(i) $1.7$ (ii) $23.1$ (iii) $5$ (iv) $20$ (v) $0.1$

(i)

$1.7$

$\sqrt{1.7} = 1.303$ to three decimal places.

$= 1.30$ correct to two decimal places.
(ii)

$23.1$

$\sqrt{23.1} = 4.806$ to three decimal places.

$= 4.81$ correct to two decimal places.
(i)

$5$

$\sqrt{5} = 2.236$ to three decimal places.

$= 2.24$ correct to two decimal places.
(i)

$20$

$\sqrt{20} = 4.472$ to three decimal places.

$= 4.47$ correct to two decimal places.

Find the least numbers which must be subtracted from the following numbers so as to leave a perfect square:
(i) $2361$ (ii) $4931$ (iii) $18265$ (iv) $390700$

(i)

$2361 < 2500$

$\therefore \sqrt { 2361 } < 50$

$1600 < 2361$

$\therefore 40 < \sqrt { 2361 }$

$\Rightarrow 40 < \sqrt { 2361 } < 50$

$2025 < 2361$

$\Rightarrow 45 < \sqrt { 2361 }$

$2116 < 2361$

$\Rightarrow 46 < \sqrt { 2361 }$

$2209 < 2361$

$\Rightarrow 47 < \sqrt { 2361 }$

$2304 < 2361$

$\Rightarrow 48 < \sqrt { 2361 }$

$\therefore 2401 > 2361$

$\Rightarrow 48 < \sqrt { 2361 } < 49$

$\Rightarrow$ The nearest square number lesser than

$2361$ is

$2304$ .

$\Rightarrow$ The number to be subtracted from

$2361$ to make it a perfect square is

$2361 - 2304 = 57$ .

$\therefore 57$ is the least number to be subtracted from

$2361$ to make it a perfect square.
(ii)

$4931$

$4900 < 4931$

$\Rightarrow 70 < \sqrt { 4931 }$

$5041 > 4931$

$\Rightarrow 71 > \sqrt { 4931 }$

$\Rightarrow$ The nearest square number lesser than

$4931$ is

$4900$ .

$\Rightarrow$ The number to be subtracted from

$4931$ to make it a perfect square is

$4931 - 4900 = 31$ .

$\therefore 31$ is the least number to be subtracted from

$4931$ to make it a perfect square.
(iii)

$18265$

$16900 < 18265$

$\Rightarrow 130 < \sqrt { 18265 }$

$17161 < 18265$

$\Rightarrow 131 < \sqrt { 18265 }$

$17424 < 18265$

$\Rightarrow 132 < \sqrt { 18265 }$

$17689 < 18265$

$\Rightarrow 133 < \sqrt { 18265 }$

$17956 < 18265$

$\Rightarrow 134 < \sqrt { 18265 }$

$18225 < 18265$

$\Rightarrow 135 < \sqrt { 18265 }$

$18496 < 18265$

$\Rightarrow 136 > \sqrt { 18265 }$

$\Rightarrow$ The nearest square number lesser than

$18265$ is

$18225$ .

$\Rightarrow$ The number to be subtracted from

$18225$ to make it a perfect square is

$18265 - 18225 = 40$ .

$\therefore 40$ is the least number to be subtracted from

$18265$ to make it a perfect square.
(iv)

$390700$

$390625 < 390700$

$\Rightarrow 625 < \sqrt { 390700 }$

$391876 < 390700$

$\Rightarrow 625 > \sqrt { 390700 }$

$\Rightarrow$ The nearest square which is lesser than

$390700$ is

$390625$ .

$\Rightarrow$ The number to be subtracted from

$390700$ to make it a perfect square is

$390700 - 390625 = 75$ .

$\therefore 75$ is the least number to be subtracted from

$390700$ to make it a perfect square.

Write true (T) or false (F) for the following statements:
(i) $\sqrt{0.9} = 0.3$
(ii) If $a$ is a natural number, then $\sqrt{a}$ is a rational number.
(iii) If $a$ is negative, then ${a}^{2}$ is also negative.
(iv) If $p$ and $q$ are perfect squares, then $\sqrt{\displaystyle\frac{p}{q}}$ is a rational number.
(v) The square root of a prime number may be obtained approximately, but never exactly.

(i) False, since the square of a number with one decimal will have two decimal places. Thus square of

$0.3$ is

$0.09$ .
(ii) False, since square root of a natural number is not a rational number.
(iii) False, since the square of any negative number will be a positive number.
(iv) True. If

$p$ and

$q$ are perfect squares, then

$\sqrt{\dfrac {p}{q}}$ is a rational number.
(v) True. The square root of a prime number may be obtained approximately, but never exactly.

What is the Pythagorean Triples using the values, $m = 3$ and $n = 5$ ?

Condition for constructing Pythagorean triples:
When m and n are any two positive integers

$(m < n)$ :

$a=n^2-m^2$

$b=2nm$

$c=n^2+m^2$

Then

$a,b$ and

$c$ are Pythagorean triple

Here

$m=3$ and

$n=5$

$\therefore 5^2-3^2$

$\Rightarrow 25-9=16$

$\therefore b=2\times 5\times 3=30$

$\therefore c=5^2+3^2$

$\Rightarrow 25+9=34$

Hence Pythagorean triples are

$16,30,34$ .

Find the Pythagorean Triples using the values, $m = 4$ and $n = 8$ .

Condition for constructing Pythagorean triples:
When

$m$ and

$n$ are any two positive integers

$(m < n)$ :

$a=n^2-m^2$

$b=2nm$

$c=n^2+m^2$

Then

$a,b$ and

$c$ are Pythagorean triple

Here

$m=4$ and

$n=8$

$\therefore 8^2-4^2$

$\Rightarrow 64-16=48$

$\therefore b=2\times 8\times 4=64$

$\therefore c=8^2+4^2$

$\Rightarrow 64+16=80$

Hence, Pythagorean triples are

$48,64,80$ .

What is the Pythagorean Triples using the values, $m = 1$ and $n = 5$ ?

Condition for constructing Pythagorean triples:
When

$m$ and

$n$ are any two positive integers

$(m < n)$ :

$a=n^2-m^2$

$b=2nm$

$c=n^2+m^2$

Then

$a,b$ and

$c$ are Pythagorean triple

Here

$m=1$ and

$n=5$

$\therefore 5^2-1^2$

$\Rightarrow 25-1=24$

$\therefore b=2\times 5\times 1=10$

$\therefore c=5^2+1^2$

$\Rightarrow 25+1=26$

Hence Pythagorean triples are

$24,10,26$ .

Find the Pythagorean Triples using the values, $m = 2$ and $n = 3$ .

When

$m$ and

$n$ are any two positive integers

$(m<n)$ , then

$a=n^2-m^2$

$b=2nm$

$c=n^2+m^2$

Then

$a,b$ and

$c$ are Pythagorean triples

Here

$m=2$ and

$n=3$

$\therefore a=3^2-2^2$

$\Rightarrow 9-4=5$

$\therefore b=2\times 3\times 2\Rightarrow 12$

$\therefore c=3^2+2^2$

$\Rightarrow 9+4=13$

Hence

$5,12,13$ are Pythagorean triples.

Find whether the following lengths are Pythagorean triples: $3, 7$ and $1$ .

The longest side of the triangle is called hypotenuse. According to the Pythagorean theorem the square of the hypotenuse is equal to the sum of the square of the two other sides.

$\therefore 7^2=1^2+3^2$

$\Rightarrow 1^2+3^2=1+9=10$

$\Rightarrow 7^2=49$

$\therefore 49\neq 10$

Hence these length are not Pythagorean triples.

Find whether the following lengths are Pythagorean triples: $6, 1$ and $7$ .

The longest side of the triangle is called hypotenuse. According to the Pythagorean theorem the square of the hypotenuse is equal to the sum of the square of the two other sides.

$\therefore 7^2=1^2+6^2$

$\Rightarrow 1^2+6^2=1+36=37$

$\Rightarrow 7^2=49$

$\therefore 49\neq 37$

Hence these length are not Pythagorean triples.

Check if the following lengths are Pythagorean triples: $8, 15$ and $17$ .

The longest side of the triangle is called hypotenuse .According to the Pythagorean theorem the square of the hypotenuse is equal to the sum of the square of the two other sides.

$\therefore 17^2=15^2+8^2$

$\Rightarrow 15^2+8^2=225+64=289$

$\Rightarrow 17^2=289$

$\therefore 289=289$

Hence these length are Pythagorean triples.

What is the Pythagorean Triples using the values, $m = 0$ and $n = 2$ ?

Condition for constructing Pythagorean triples:
When

$m$ and

$n$ are any two positive integers

$(m < n)$ :

$a=n^2-m^2$

$b=2nm$

$c=n^2+m^2$

Then

$a,b$ and

$c$ are Pythagorean triple

Here

$m=0$ and

$n=2$

$\therefore 2^2-0^2$

$\Rightarrow 4-0=4$

$\therefore b=2\times 2\times 0=0$

$\therefore c=2^2+0^2$

$\Rightarrow 4+0=4$

Hence Pythagorean triples are

$4,0,4$ .

Check if the following lengths are Pythagorean triples: $60, 61$ and $11$ .

The longest side of the triangle is called the hypotenuse. According to the Pythagorean theorem, the square of the hypotenuse is equal to the sum of the squares of the two other sides.

$\therefore 61^2=11^2+60^2$

$\Rightarrow 60^2+11^2=3600+121=3721$

$\Rightarrow 61^2=3721$

$\therefore 3721=3721$

Hence these length are Pythagorean triples.

Determine if the following lengths are Pythagorean triples: $9, 40$ and $41$ .

The longest side of the triangle is called hypotenuse. According to the Pythagorean theorem the square of the hypotenuse is equal to the sum of the square of the two other sides.

$\therefore 41^2=40^2+9^2$

$\Rightarrow 40^2+9^2=1600+81=1681$

$\Rightarrow 41^2=1681$

$\therefore 1681=1681$

Hence these length are Pythagorean triples.

Evaluate using long division method: $\sqrt{6241}$

The following steps to find the square root by long division method
1. Draw lines over pairs of digits from right to left.
2. Find the greatest number whose square is less than or equal to the digits in the first group.
3. Take this number as the divisor and quotient of the first group and find the remainder.
4. Move the digits from the second group besides the remainder to get the new dividend.
5. Double the first divisor and bring it down as the new divisor.
6. Complete the divisor and continue the division.
7. Repeat the process till the remainder becomes zero

Divisor $\downarrow$	Quotient $\downarrow$ 79
7	$\overline{62}$ $\overline{41}$ 49
149	1341 1341
	0 $\leftarrow$ Remainder

$\sqrt{6241} = 79$

Evaluate using long division method: $\sqrt{7056}$

Hence,

$\sqrt{7056}$ is equal to

$84.$

Find the square root of $9.61$ using long division method.

Divisor $\downarrow$	Quotient $\downarrow$ 3.1
3	$\overline{9}$ . $\overline{61}$ 9
6.1	0.61 0.61
	0 $\leftarrow$ Remainder

$\sqrt{9.61} = 3.1$

Find the square root of $79.21$ using long division method.

Divisor $\downarrow$	Quotient $\downarrow$ 8.9
8	$\overline{79}$ . $\overline{21}$ 64
1.69	15.21 15.21
	0 $\leftarrow$ Remainder

$\sqrt{79.21} = 8.9$

What is the square root of $96.04$ using long division method?

Divisor $\downarrow$	Quotient $\downarrow$ 9.8
9	$\overline{96}$ . $\overline{04}$ 81
1.88	15.04 15.04
	0 $\leftarrow$ Remainder

$\sqrt{96.04} = 9.8$

Find the square root of $2.89$ using long division method.

Divisor $\downarrow$	Quotient $\downarrow$ 1.7
1	$\overline{2}$ . $\overline{89}$ 1
2.7	1.89 1.89
	0 $\leftarrow$ Remainder

$\sqrt{2.89} = 1.7$

For each of the following number, find the smallest whole number by which it should be multiplied so as to get a perfect square number. Also find the square root of the square number so obtained.

$(i)\ 252$ $(ii)\ 180$ $(iii)\ 1008$ $(iv)\ 2028$ $(v)\ 1458$ $(vi)\ 768$

$252 = \underline{2 \times 2} \times \underline {3 \times 3} \times 7$

Therefore, we need to multiply

$252$ by

$7$ in order to make it a perfect square.

So,

$252 \times 7 = 1764$

$\therefore \sqrt {1764} = 42$

ii)

$180 = \underline{2 \times 2} \times \underline {3 \times 3} \times 5$

Therefore, we need to multiply

$180$ by

$5$ in order to make it a perfect square.

So,

$180 \times 5 = 900$

$\therefore \sqrt {900} = 30$

iii)

$1008 = \underline{2 \times 2} \times \underline {2 \times 2} \times \underline {3 \times 3} \times 7$

Therefore, we need to multiply

$1008$ by

$7$ in order to make it a perfect square.

So,

$1008 \times 7 = 7056$

$\therefore \sqrt {7056} = 84$

iv)

$2028 = \underline{2 \times 2} \times \underline {13 \times 13} \times 3$

Therefore, we need to multiply

$2028$ by

$3$ in order to make it a perfect square.

So,

$2028 \times 3 = 6084$

$\therefore \sqrt {6084} = 78$

$1458 = \underline{3 \times 3} \times \underline {3 \times 3} \times \underline {3 \times 3} \times 2$

Therefore, we need to multiply

$1458$ by

$2$ in order to make it a perfect square.

So,

$1458 \times 2 = 2916$

$\therefore \sqrt {2916} = 54$

$768 = \underline{2 \times 2} \times \underline {2 \times 2} \times \underline {2 \times 2} \times \underline {2 \times 2} \times 3$

Therefore, we need to multiply

$768$ by

$3$ in order to make it a perfect square.

So,

$768 \times 3 = 2304$

$\therefore \sqrt {2304} = 48$

How many numbers lie between squares of the following numbers ?
(i) $12$ and $13$
(ii) $25$ and $26$
(iii) $99$ and $100$

(i)

$12^{2} = 144$

$13^{2} = 169$

$169 - 144 = 25$
So, there are

$25 - 1 = 24$ numbers lying between

$12^{2}$ and

$13^{2}$

(ii)

$25^{2} = 625$

$26^{2} = 676$

$676-625 = 51$
So, there are

$51 - 1 = 50$ numbers lying between

$25^{2}$ and

$26^{2}$

(iii)

$99^{2} = 9801$

$100^{2} = 10000$

$10000 -9801 = 199$
So, there are

$199 - 1 = 198$ numbers lying between

$99^{2}$ and

$100^{2}$

What will be the unit digit of the squares of the following number ?
(i) 81 (ii) 272 (iii) 799 (iv) 3853 (v) 1234
(vi) 26387 (vii) 52698 (viii) 99880 (ix) 12796 (x) 55555

(i)

$81$

$1^{2} =1$
So, unit digit will be

$1$

(ii)

$272$

$2^{2} =4$
So, unit digit will be

$4$

(iii)

$799$

$9^{2} = 81$
So, unit digit will be

$1$

(iv)

$3853$

$3^{2} =9$
So, unit digit will be

$9$

(v)

$1234$

$4^{2} =16$
So, unit digit will be

$6$

(vi)

$26387$

$7^{2} =49$
So, unit digit will be

$9$

(vii)

$52698$

$8^{2} =64$
So, unit digit will be

$4$

(viii)

$99880$

$0^{2} =0$
So, unit digit will be

$0$

(ix)

$12796$

$6^{2} =36$
So, unit digit will be

$6$

(x)

$55555$

$5^{2} =25$
So, unit digit will be

$5$

Without doing calculation, find the numbers which are surely not perfect squares.
$(i) 153\ (ii) 257 \ (iii) 408 \ (iv) 441$

For any number to be perfect square,

It should have a number from

$\{1,4,9,6,5\}$ at their units place.

Here,

(i)

$153$ ending with

$3$ , so it cannot be a perfect square.

(ii)

$257$ ending with

$7$ , so it cannot be a perfect square.

(iii)

$408$ ending with

$8$ , so it cannot be a perfect square.

(iv)

$441$ ending with

$1$ , so it can be a perfect square.

The square of which of the following number will be odd numbers?
(i) 431
(ii) 2826
(iii) 7779
(iv) 82004

Square of (i) and (iii) will be odd numbers, because an odd number multiplied by another odd number always results in an odd number.

What could be possible one's digit of the square root of each of the following numbers?
(i) 9801
(ii) 99856
(iii) 998001
(iv) 657666025

(i)

$9801$

Observe that the numbers ending with

$1$ and

$9$ are the only numbers whose square numbers end with

$1$

$1^{2} =1$ and

$9^{2} = 81$

Since the last digit of

$9801$ is

$1$ , the possible one's digit of it's square root can be

$1$ or

$9$ .

(ii)

$99856$

Observe that the numbers ending with

$4$ and

$6$ are the only numbers whose square numbers end with

$6$

$4^{2} =16$ and

$6^{2} = 36$

Since the last digit of

$99856$ is

$6$ , the possible one's digit of it's square root can be

$4$ or

$6$ .

(iii)

$998001$

Observe that the numbers ending with

$1$ and

$9$ are the only numbers whose square numbers end with

$1$

$1^{2} =1$ and

$9^{2} = 81$

Since the last digit of

$998001$ is

$1$ , the possible one's digit of it's square root can be

$1$ or

$9$ .

(iv)

$657666025$

Observe that the numbers ending with

$5$ are the only numbers whose square numbers end with

$25$

$5^{2} =25$

Since the last two digits of

$657666025$ are

$25$ , the possible one's digit of it's square root can be

$5$ .

Find the square root of the following number by prime factorisation method
(i) 729
(ii) 400
(iii) 1764
(iv) 4096
(v) 7744
(vi) 9604
(vii) 5929
(viii) 9216
(ix) 529
(x) 8100

(i)

$729 = 3\times 3\times 3\times 3\times 3\times 3 =$

$3^{2} \times 3^{2} \times 3^{2}$

$\sqrt{729} = 27$
(ii)

$400 =$

$2^{2} \times 2^{2} \times 5^{2}$

$\sqrt{400} = 20$
(iii)

$1764 =$

$2^{2} \times 2^{2} \times 7^{2}$

$\sqrt{1764} = 42$
(iv)

$4096 =$

$2^{2} \times 2^{2} \times 2^{2} \times 2^{2} \times 2^{2} \times 2^{2}$

$\sqrt{4096} = 64$
(v)

$7744 =$

$2^{2} \times 2^{2} \times 2^{2} \times 11^{2}$

$\sqrt{7744} = 88$
(vi)

$9604 =$

$2^{2} \times 7^{2} \times 7^{2}$

$\sqrt{9604} = 98$
(vii)

$5929 =$

$11^{2} \times 7^{2}$

$\sqrt{5929} = 77$
(viii)

$9216 =$

$96^{2}$

$\sqrt{9216} = 96$
(ix)

$529 = 23\times 23$

$\sqrt{529} = 23$
(x)

$8100= 90^{2}$

$\sqrt{8100} = 90$ .

Let n be a product of four consecutive positive integers then n is never a perfect square

Let

$N=1+n$ where

$'n'$ is a product of

$'4'$ consecutive no.

$\Rightarrow n=a(a+1)(a+2)(a+3)$

$\Rightarrow N=1+a(a+1)(a+2)(a+3)$

$\Rightarrow N=a^{4}+6a^{3}+11a^{2}+6a+1$

$\Rightarrow N=(a^{2}+3a+1)^{2}$

$\Rightarrow 'N'$ is a perfect square

hence

$n=(N-1)$ can't be a perfect square.

Find square root of each of the following no. by division method
$(i)\ 2304$ $(ii)\ 4489$ $(iii)\ 3481$ $(iv)\ 529$ $(v)\ 3249$ $(vi)\ 1369$ $(vii)\ 5776$ $(viii)\ 7921$ $(ix)\ 576$ $(x)\ 1024$ $(xi)\ 3136$ $(xii)\ 900$

Identify the perfect squares among the following numbers:
$1, 2, 3, 8, 36, 49, 65, 67, 71, 81, 169, 625, 125, 900, 100, 1000, 100000$

Perfect square numbers are the numbers whose square root is a positive integer.

Now, as we can see

$\sqrt{1}=1, \sqrt{36}=6, \sqrt{49}=7, \sqrt{81}=9, \sqrt{169}=13, \sqrt{625}=25, \sqrt{900}=30, \sqrt{100}=10$

So, the perfect squares among the above are:

$1, 36, 49, 81, 169, 625, 900, 100$

Following are not perfect squares:-

$2, 3, 8, 65, 67, 71, 125, 1000, 100000$

Find the number of digits in the square root of each of the following number.
$(i)\ 64$ $(ii)\ 144$ $(iii)\ 4489$ $(iv)\ 27225$ $(v)\ 390625$

If the number of digits of a number is even, then its square root will have half (whole number) of the number of digits.

If the number of digits of a number are odd, then its square root will have half (nearest greater whole number) of the number of digits.

$\sqrt {64}$

The number of digits is 2 i.e. even.

Therefore, there is only 1 digit in the square root of

$64$ .

ii)

$\sqrt {144}$

The number of digits is 3 i.e. odd.

Therefore, there are (1.5 =) 2 digits in the square root of

$144$ .

iii)

$\sqrt {4489}$

The number of digits is 4 i.e. even.

Therefore, there are 2 digits in the square root of

$4489$ .

iv)

$\sqrt {27225}$

The number of digits is 5 i.e. odd.

Therefore, there are (2.5=) 3 digits in the square root of

$27225$ .

$\sqrt {390625}$

The number of digits is 6 i.e. even.

Therefore, there are 3 digits in the square root of

$390625$ .

Find the least number which must be added from each of the following so as to get a perfect square. Also find the square root of the perfect square so obtained.
$(i)\ 525$ $(ii)\ 1750$ $(iii)\ 252$ $(iv)\ 1825$ $(v)\ 6412$

In order to find the least number to be added to the given no.,

we must find a greater perfect square number, closest to the given number.

$525$

The closest perfect square number is

$576$

Difference

$=576-525 = 51$

Hence,

$51$ must be added to

$525$ in order to make it a perfect square.

$\therefore \sqrt {576} = 24$

ii)

$1750$

The closest perfect square number is

$1764$

Difference

$=1764-1750 = 14$

Hence,

$14$ must be added to

$1750$ in order to make it a perfect square.

$\therefore \sqrt {1764} = 42$

iii)

$252$

The closest perfect square number is

$256$

Difference

$=256-252 = 4$

Hence,

$4$ must be added to

$252$ in order to make it a perfect square.

$\therefore \sqrt {256} = 16$

iv)

$1825$

The closest perfect square number is

$1849$

Difference

$=1849 - 1825 = 24$

Hence,

$24$ must be added to

$1825$ in order to make it a perfect square.

$\therefore \sqrt {1849} = 43$

$6412$

The closest perfect square number is

$6561$

Difference

$=6561 - 6412 = 149$

Hence,

$149$ must be added to

$6412$ in order to make it a perfect square.

$\therefore \sqrt {6561} = 81$

Write $3$ -digit numbers ending with $0, 1, 4, 5, 6, 9$ one for each digit, but none of them is a perfect square.

We will first write the three digit numbers between

$14^{2}=196$ and

$15^{2}=225$ ending with

$0,1,4,5,6,9$ where none of them will be a perfect square. And those numbers are as follows:

$200,201,204,205,206$ and

$209$

Also we write the three digit numbers between

$17^{2}=289$ and

$18^{2}=324$ ending with

$0,1,4,5,6,9$ where none of them will be a perfect square. And those numbers are as follows:

$300,301,304,305,306$ and

$309$

A square yard has area $1764$ m $^2$ . From a corner of this yard, another square part of area $784$ m $^2$ is taken out for public utility. The remaining portion is divided in to $5$ equal square parts. What is the perimeter of each of these equal parts?

Let

$a$ be the length of each side of a square yard.

A square yard has area

$1764\ m^2$

$\implies$ Area of yard

$=a^2$

$\implies 1764^2=a^2\implies a=42$

$\therefore$ Length of each side

$(a)=42\ m$

The square

$B$ with dotted lines shows the area taken out for public utility

Let

$b$ be the length of side of a square part

$B$ which is taken out.

$\therefore$ Area of the square part taken out

$=b^2$

$\implies 784 =b^2$

$\implies b=28\ m$

$\therefore$ Length of each side a square

$B$ is

$28\ m$

$A$ denotes the remaining part after B is taken out.

Now, the length of two sides of

$A$ is

$42-28=14\ m$ and other two sides are of length

$42$ .

Area of

$A=$ Total area

$-$ Area of

$B$

$=1764-784=980\ m^2$

To divide

$A$ into

$5$ equal squares, we divide the area by

$5$

$\dfrac{\text{Area of A}}{5}=\dfrac{980}{5}=196\ m^2$

$\therefore$ Area of each smaller squares will be

$196\ m^2$

Let

$c$ be the length of side of each square.

$\implies$ Area of each square

$=c^2$

$\implies 196\ m^2=c^2\implies c=14\ m$

$\therefore$ Length of each side is

$14\ m$

Thus, the perimeter of each smaller squares

$=4\times 14=56\ m$

Hence, the answer is

$56\ m$

Find the largest perfect square factor of each of $11280$

Largest perfect square for

$11280$ will be

$16$ $ as, let us first resolve the given number into its prime factors, we get the prime factors of

$11280$ are

$2\times 2\times 2\times 2\times 5\times 141=11280$

Largest perfect square factor we get:

$2\times 2\times 2\times 2=16$

Find the smallest square no. which is divisible by each of the numbers $4,9$ and $10$ .

$4= 2 \times 2$

$9= 3 \times 3$

$10= 2 \times 5$
L.C.M =

$2^{2} \times 3^{2} \times 5 = 180$

But

$180$ is not a perfect square. Multiply 180 with 5 to get pairs.
Required no.

$= 180 \times 5 = 900$

Find the square root of the following numbers by the factorisation numbers : 1296

Resolving

$1296$ as the product of primes, we get

$\Rightarrow 1296=2^4 \times 3^4$

$\Rightarrow \sqrt{1296}= \sqrt {2^4 \times 3^4}= 2^2 \times 3^2$

$\therefore \sqrt{1296}=36.$

Find the nearest integer to the square root of $728.$

We know,

$26^2 = 676$ and

$27^2 = 729$
Also we have,

$676 < 728 < 729$

$\implies 26^2 < 728 < 27^2$

Difference of

$676$ and

$728$ is

$52$ whereas difference of

$728$ and

$729$ is

$1$ .
So, the square root of

$728$ is nearest to

$27$ .

Hence, the answer is

$27$ .

Find the square root of 19.5364

Step 1 : First we put bars on the integral part (i.e. 19) of the number in the usual manner, and place bars on the decimal part (ie. 5364) on every pair of digits beginning with the first decimal place. We get

$\overline{19}. \overline{53}\overline{64}$

Step 2 : Now the left most bar is 19 and

$16 < 19 < 25$ [Ref. image 1]

$4^2 < 19 < 5^2$ , take

$4$ as divisor.

So the remainder is 3, how we write the number under the next bar to the right of this reminder. So we get new dividend is

$353$

Step 3: Add the divisor 4 and the quotient

$4$ we get

$8$ , to final the new divisor of

$353$ , we choose a number like [Refer image 2]

$81\times 1 = 1$

$82 \times 2 = 164$

$83 \times 3 = 249$

$84 \times 4 = 336$

$85 \times 5 = 425$

So the new divisor is 84

repeat this process we get square root of

$19.5364$

$88\times 1 = 882$

$882 \times 1764$

So the square root of

$19.5364$ is

$4.42$

1081541_559237_ans_353ce79e8cb0499b989524dcd72fe90a.png

Find the square root of $2002225$ .

Clearly, from the above image we get

$\sqrt{2002225}=1415$

Find the nearest integer to the square root of $1729.$

We know,

$41^2 = 1681$ and

$42^2 = 1764$
Also, we have,

$1681 < 1729 < 1764$

$\implies 41^2 < 1729 < 42^2$

Difference of

$1681$ and

$1729$ is

$48$ whereas difference of

$1729$ and

$1764$ is

$35$ .
So, the square root of

$1729$ is nearest to

$42$ .

Hence, the answer is

$42$ .

The nearest integer to the square root of $600$ is ___.

We know,

$24^2 = 576$ and

$25^2 = 625$
Also we have,

$576 < 600 < 625$

$\implies 24^2 < 600 < 25^2$

Difference of

$576$ and

$600$ is

$(600-576)=24$

whereas difference of

$600$ and

$625$ is

$(625-600)=25$ .

Thus, the square root of

$600$ is nearest to

$24$ .

Hence, the answer is

$24$ .

Find the square root of 96721.

1079249_559111_ans_beef9b0cbf2546cc86df6fec14294796.png

Find the square root of $5329$ by division method

Division method for finding square roots

step 1: First place a bar over every pair of digits starting from the unit digit, if the no. of digits is odd then the left most single-digit will also have a bar.

Step 2: Think of the largest number whose square is equal to or just less than the first bar digit, take this number as a divisor and also as the quotient.

Step 3: Next subtract the product of the divisor and the quotient from the first bar digit and bring down the next pair of digits which have a bar to the right side of the reminder, this becomes the new dividend.

Step 4: The new divisor is obtained by (adding the first divisor and the quotient and a digit to the right side of it that we have to choose according to the new dividend, which is chosen in such a way that product of new divisor and this digit is less than or equal to the new dividend )

Step 5: Repeat steps 2,3,4, till all the bar digits have been taken up, new quotient is the required square root of the given no.

square root of the

$\overline{53} \,\overline{29}$

Step 1:

$5329$

Step 2:

$49<53 <64$ [Refer image 1]

$7^2<5^3<8^2$

Step 3: the new dividend is

$429$ [Refer image 2]

Step 4:

$142=\times 2= 284$

$143\times 3 =429$ [refer images 3]

This square root of

$5329$ is

$73$ .

1079258_559368_ans_4a77123cdd3c43459afebae8fbb68dc6.png

A square-field has area 3.32,929 $m^2$ . It has to be fenced using barbed wire, The barbed wire should go round the field 5 times. The barbed wire is available in bundles of 100 m length. The vendor sells the wire only in complete bundles. How many bundles have to be bought?

First we find the perimeter of the square-field by finding its side length.
We use division method.
We obtain

$\sqrt{332929}=577$ Hence the perimeter of the field is

$577\times 4 = 2,308m$ .
The length of the required barbed wire is

$5\times 2308 = 11,540m$ can you notice that

$115\times 100 = 11500 < 11540$ , and

$116\times 100 = 11600 > 11540$ .
Hence the number of bundles to be bought is 116.

Find the square root of the following numbers: $184.96$

Step 1 : Place bar on the integral parts (i.e 184) of the number in the usual and place bar on the decimal. place on the decimal. Place on every pair of digit beginning with the first decimal place then we get

$\bar{1}\overline{84}.\overline{96}$

Step 2 : since

$1^2 \le \bot < 2^2$ [Ref. image 1]

Now new dividend is 084 and we find new divisor in the similar manner as division method.

$22 \times 2 = 4, \, 23\times 3 = 69 , \, 24 \times 4 = 96 > 84$

So the new divisor is 23 we get [Ref. image 2]

New dividend is 1596 to find new divisor

$265 \times 5 = 1325$

$266 \times 6 = 1596$

So we get the remainder is zero.

The square root of 184.96 is 13.6

1081536_559235_ans_67e724b2037f408cabcd9867c406936e.png

Find the least number to be added to get a perfect square: 88417

R.E.F image

Q.Find the least number to be added to get

a perfect square

$88417$

$Sol^{n}$ First we find

$\sqrt{88417}$ , by division method:

$4 < 8 < 9$

$2^2 < 8 < 3^2$

$44\times 4 = 176$

$46\times 6 = 276$

$47\times 7 = 329$

$48\times 8 = 384$

$49\times 9 = 441$

$587\times 7 = 4109$

$588\times 8 = 4704$

We get the reminder

$208$

hence

$88417$ is not a perfect square

since

$(297)^2 < 88417 < (298)^2$

We calculate

$(298)^2 - 88417 = 88807 - 88417 = 387$

hence the least number to be added to

get a perfect square is

$387$

since

$88417+387 = 88804 = (298)^2$

1088098_559379_ans_e37c88d4f8d34ed7bf8f952e5ace42da.PNG

Find the least number to be subtracted from the following number to get a perfect square: 34567

R.E.F image

First we find the square root of

$34567$ by division method

$28\times 8 = 224$

$29\times 9 = 261$

$365 \times 5 = 1825$

$366 \times 6 = 2196$

we get the remainder

$342$ , hence

$34567$ is

not a perfect square, we have to find

the least positive number to be

subtracted from

$34567$ . to got a perfect

square.

Since

$(185)^2 < 34567 < (186)^2$

Thus we subtract

$34567 -(185)^2$

$= 34567 - 34225$

$=342$

The required perfect square is

$=34567 - 342 = 34225$

and the least positive number is

$342$

1081551_559398_ans_d7f97961d00c454d851088f558398d39.PNG

Find the square root of the following number by division method: $186624$

Division method for finding square roots.

Step 1: Fist place a bar over every pair of digits starting from the unit digit, if the no. of digits are add then the left most single digit will also have a bar.

Step 2: Think of the largest number whose square is equal to or just less than the first bar digit. take this number as the divisor and also as the quotient

Step 3: Next subtract the product of the divisor and the quotient from the first bar and digit and bring down the next pair of digit which have bar to the right side of the reminder. Thus becomes the new dividend.

Step 4: Now the new division is obtained by adding the first divisor and the quotient and a digit to the right side of it that we have to choose, according to the new dividend. which is chosen in such a way that product of new divisor and this digit is less than or equal to the new dividend.

Step 5: Repeat steps 2, 3, 4 till the bar digits have been taken up. Now the quotient is the required square root of the given no.

To find the square root of

$186624$

Step 1:

$\overline{18}\,\overline{66} \, \overline{24}$

Step 2:

$4^2 < 18 < 5^2$ [Ref. image 1]

$16 < 18 < 25$

Step 3: New dividend is 266 [Ref. image 2]

Step 4:

$82\times 2 = 164$ [Ref. image 3]

$83 \times 3 = 249$

$84 \times 4=336 > 266$

New divisior is

$83$

New dividend is

$2324$

Next new divisor

$862 \times 2 = 1724$

The square root of

$186624$ is

$432$

Find the least number to be added to get a perfect square: 6200

R.E.F image

Q.Find the least number to be added to get

a perfect square

$6200$

$sol^{n}$ First we find square root of 6200 by division method.

$49 < 62 < 64$

$142 \times 2 =284$

$145 \times 5 =725$

$146 \times 6 =876$

$147 \times 7 =1029$

$148 \times 8 =1184$

$149 \times 9 =1341 > 1300$

Hence.

$(78)^2 < 6200 < (79)^2$

Since remainder is not zero; 6200 is not a perfect square number,

We calculate

$= (79)^2 - 6200 = 6241 - 6200=41$

So the least number to be added is

$41$ , we get

$6200+41 = 6241 = (79)^2$ a perfect square.

1088079_559376_ans_4dafe4e1c8bb45cf863ab8be9604d2f3.PNG

Find the consecutive perfect squares between which the following numbers lie: 56789

R.E.F image

To find the consecutive perfect squares

between which the following number lie: 56789

first we find the square root of 56789 by

division method. (Ref. image)

$43\times 3 = 129$

$44\times 4 = 176$

$45\times 5 = 225$

$466\times 6 = 2796$

$467\times 7 = 3269$

$468\times 8 = 3744$

$469\times 9 = 4221$

Thus

$(238)^2 < 56789 < (239)^2$

hence the consecutive perfect squares are 238 and 239.

1081588_559401_ans_8748bac54b6c4444b19c619d6582796d.png

Find the consecutive perfect squares between which the following numbers lie: 4567

R.E.F image

First we find the square root of

$4567$ by division method.

$126\times 6 = 756$

$127 \times 7 = 889$

$128\times 8 = 1024 > 967$

Since we get the remiander

$78$

hence

$4567$ is not a perfect square,

So the consecutive perfect square between

the number

$4567$ lie is

$(67)^2 < 4567 < (68)^2$

$67$ and

$68$

1081583_559400_ans_ca30df8f582940dd8c2fb85e19d9e737.PNG

Find the square root of the following number by division method: 28224

Division method for finding square roots.

Step 1: Fist place a bar over every pair of digits starting from the unit digit, if the no. of digits are add then the left most single digit will also have a bar.

Step 2: Think of the largest number whose square is equal to or just less than the first bar digit. take this number as the divisor and also as the quotient

Step 5: Repeat steps 2, 3 , 4 till the bar digits have been taken up. Now the quotient is the required square root of the given no.

To find the square root of

$28224$

Step 1:

$\overline{2}\,\overline{82}\,\overline{24}$

Step 2:

$1^2 < 2 < 2^2$

$1 < 2 < 4$

Step 3: The new dividend is

$182$

Step 4:

$22\times 2 = 44$

$23 \times 3 = 69$

$24 \times 4 = 94$

$25 \times 5 = 125$

$26 \times 6 = 156 < 182$

$27\times 7 = 189 > 182$

The new divisor is 26.

Step 5: The next new dividend is

$2624$ and new divisor is

$328$

$322 \times 2 = 644$

$323 \times 3 = 969$

$324 \times 4 = 1296$

$325 \times 5 = 1625$

$328 \times 8 = 2624$

The square root of

$28224$ is

$168$ .

1079264_559371_ans_a654e4b79a8646caa327a1dc887ee482.png

Find the least number to be subtracted from $1234$ to get a perfect square.

So, if

$9$ is subtracted from

$1234$ then the result

$1225$ will be a perfect square of

$35.$

1391409_559387_ans_8a587c384524481392bbddbb85522b22.JPG

Check whether the following numbers form Pythagorean triplet.
$6,8,10$

$\begin{array}{l} Let\, \, 6,8,\, \, and\, \, 10\, \, \, be\, \ the\ \, sides\, \, of\, triangle. \\ Then,\, \, we\, \, know\, \, that, \\ A{ B^{ 2 } }+B{ C^{ 2 } }=A{ C^{ 2 } } \\ \Rightarrow \sqrt { { 8^{ 2 } }+{ 6^{ 2 } } } =AC \\ \Rightarrow \sqrt { 64+36 } =AC \\ AC=10\, \, units \\ Hence,\, \, The\, \, given\ numbers\ are\ \, pythagorean\, \, triplet. \end{array}$

Find the least number to be subtracted from the following numbers to get a perfect square: 109876

R.E.F image

First we find the square root of

$109876$ by

division method.

$63\times 3 = 189$

$64 \times 4 = 256$

$662 \times 2 = 1324$

We get the remainder

$315$

since

$(331)^2 < 109876 < (332)^2$

thus we subtract

$109876 - (331)^2$

$= 109876 - 109561$

$315$ this is the remainder

hence for getting perfect square we have to subtract

$315$ ,

$\Rightarrow 109876 - 315 = (331)^2$

the least number to be subtract is

$315$

1081553_559399_ans_7699eb7f386144509cded81c4db9647d.PNG

Find the consecutive perfect squares between which the following numbers lie: 88888

First we find square root of

$88888$ by division method. (Ref. image)

$44\times 4 = 176$

$49\times 9 = 441$

$586\times 6 = 3516$

$587 \times 7 = 4109$

$588 \times 8 = 4704$

$589 \times 9 = 5301$

thus

$(298)^2 < 88888 < (300)^2$

hence two consecutive perfect square are

$298$ and

$300$

1081603_559402_ans_5b35d53e75fe49a3be9478d7028eb557.PNG

How many digits are there after the decimal points in the following : $(1.234)^2$

We calculate

$(1.234)^{2}=1.522756$

there are 6 digit after decimal points in

$(1.234)^{2}$

Find the square root of the following numbers using division method: 651.7809

R.E.F image

$\sqrt{651.7809} = ?$

$sol^{n}$ by division method.

Step 1: First place bars on the integral parts (i.e. 651)

of the number in the usual manner and

place bars on every pair of digits beginning

with the first decimal place then we get

$\overline{6}\, \overline{51}.\overline{78}\, \overline{09}$

Step 2: since

$2^2 < 6 < 3^2$ [Ref. image 1]

$4 < 6 < 9$

Now new dividend is

$251$ and we find new divisor

in the similar manner as division method.

$42\times 2 = 84, \quad 43\times 3 = 129, \quad 44\times 4 = 176, \quad 45\times 5 = 225,$

$46\times 6 = 276 > 251$

so the new divisor is

$45$ we get [Ref. image 2]

$502\times 2 = 1004$

$503\times 3 = 1509$

$504\times 4 =2016$

$505\times 5 = 2525$

$506\times 6 = 3036$

$5102\times 2 = 10204$

$5103\times 3 = 15309$

hence square root of

$651.7809$ is

$25.53$

1088127_559746_ans_42e765b15b894584af0265d3db3f41d8.png

If $a$ is the side of an equilateral triangle, then its area is given by $\dfrac{\sqrt{3}}{4}a^2$ . Given that the area of an equilateral triangle is $5\sqrt{3}$ units, find its side correct to 4 decimal ploaces

We have the equation

$\dfrac{\sqrt{3}}{4}a^2=5\sqrt{3}$

Hence

$a^2=\dfrac{5\sqrt{3}\times 4}{\sqrt{3}}=20$

We find

$\sqrt{20}$ up to 5 decimal places and round it off to 4 decimal places as shown in the image.

Thus,

$\sqrt{20}=4.47213$

When this is rounded off to 4 decimal places, we get

$4.4721$

Find approximations from below to 4 decimal places to the square root of the following number: 5

1391373_559789_ans_55a9db3679524650938bb764374b3e01.JPG

A square garden has area 900 $m^2$ . Additional land measuring equal area, surrounding it, has been added to it. If the resulting plot is also in the form of a square, what is its side correct to 2 decimal places?

Ares of square garden is 900 m². If additional land, measuring equal area to added to the garden, its area will be

(900+900)² = 1800 m²

Side of the new garden is 1800 m.

A = side x side = (side)² = 42.43m

1391377_559793_ans_99b03e4cfbe741f9a1853170081b6464.JPG

Find the consecutive perfect squares between which the following numbers lie: 123456

R.E.F image

Q. Find the least number to be added to get

a perfect square

$123456$

$sol^{n}$ First we find

$\sqrt{123756}$ by division method

$9 < 12 < 16$

$3^2 < 12 < 4^2$

$62 \times 2 = 124$

$63 \times 3 = 189$

$65 \times 5 = 325$

$66 \times 6 = 396$

$702 \times 2 = 1404$

hence

$(351)^2 < 123456 < (352)^2$

Hence two consecutive perfect square are

$351$ and

$352$

1088109_559403_ans_41cb3cdcb8b24ba682466fe3f2cf2c64.png

Find the square root of 3 and also correct it to 4 decimal places.

Here we find the square root of

$3$ up to

$5$ decimal places and round it off to

$4$ decimal places.
Here the fifth decimal digit is

$5$ and following our convention we round off

$1.73205$ to

$1.7321$ .

Hence,

$1.7321$ is equal to

$\sqrt{3}$ correct to 4 decimal places.

Find the square root of the following number correct to 3 decimal places: 3333

1391369_559786_ans_084690558c464fd7b2c1c41cd0d9cccb.JPG

How many digits are there after the decimal points in the following : $(3.16)^2$

How many digits are there after the decimal point in the following :

$(3.16)^{2}$

since

$(3.16)^{2}=9.9856$

hence there are 4 digit after the decimal point in

$(3.16)^{2}$

Find approximations from below to 4 decimal places to the square root of the following number: 8

1391375_559790_ans_7e454aad7afa4ea597c838a4ffebfe6f.JPG

Explain with reason whether the given number is perfect square or not: 257

Given number is

$257$ .

We know that, a number is a perfect square if its square root is a whole number.

Also, every perfect square has one of the following as the digit at its units place:

$1,\ 4,\ 5,\ 6,\ 9$ or even number of

$0s$ .

It is because:

$1^2=1$

$2^2=4$

$3^2=9$

$4^2=16$

$5^2=25$

$6^2=36$

$7^2=49$

$8^2=64$

$9^2=81$

$10^2=100$

We see that,

$257$ has

$7$ at its units place.

Hence,

$257$ is not a perfect square.

Find the value of
${ \left( 26+15\sqrt { 3 } \right) }^{ \dfrac { 2 }{ 3 } }-{ \left( 26+15\sqrt { 3 } \right) }^{ -\dfrac { 2 }{ 3 } }$

We have

$(26+15\sqrt3)=(8+3\sqrt 3+12\sqrt 3+18)=(2+\sqrt3)^3$

Hence,

$(26+15\sqrt3)^{\dfrac23}-(26+15\sqrt3)^{-\dfrac23}=(2+\sqrt3)^2-(2+\sqrt3)^{-2}$

$=(2+\sqrt3)^2-(2-\sqrt3)^{2}$

$=8\sqrt3$

Which of the following numbers are perfect squares?
$3, 5, 23, 25, 36, 26, 40$

From the given numbers,

$25$ and

$36$ are perfect square because

$25 = 5 \times 5= 5^2$ and

$36 = 6 \times 6 = 6 ^2$

To find the value of $\sqrt [ 4 ]{ -64{ a }^{ 4 } }$ .

$\sqrt [ 4 ]{ -64{ a }^{ 4 } } =\sqrt { \pm 8{ a }^{ 2 }\sqrt { -1 } }$

$2a\sqrt { 2 } \sqrt { \pm \sqrt { -1 } }$
It remains to find the value of

$\sqrt { \pm \sqrt { -1 } }$
Assume

$\sqrt { \pm \sqrt { -1 } } =x+y\sqrt { -1 }$
then

$\pm\sqrt { -1 } ={ x }^{ 2 }-{ y }^{ 2 }+2xy\sqrt { -1 }$

$\therefore { x }^{ 2 }-{ y }^{ 2 }=0$ and

$2xy=\pm 1$
whence

$x=\cfrac { 1 }{ \sqrt { 2 } } ,y=\cfrac { 1 }{ \sqrt { 2 } }$
or

$\quad x=-\cfrac { 1 }{ \sqrt { 2 } } ,y=-\cfrac { 1 }{ \sqrt { 2 } }$

$\therefore \sqrt { +\sqrt { -1 } } =\pm \cfrac { 1 }{ \sqrt { 2 } } (1+\sqrt { -1 } )$
Similarly

$\sqrt { -\sqrt { -1 } } =\pm \cfrac { 1 }{ \sqrt { 2 } } (1-\sqrt { -1 } )$

$\therefore \sqrt { \pm \sqrt { -1 } } =\pm \cfrac { 1 }{ \sqrt { 2 } } (1\pm \sqrt { -1 } )$
and finally

$\sqrt [ 4 ]{ -64{ a }^{ 4 } } =\pm 2a(1\pm \sqrt { -1 } )$

$\sqrt { 4{ x }^{ 4 }-12{ x }^{ 3 }+25{ x }^{ 2 }-24x+16 } =?$

1106170_843883_ans_62c66c21458f4e22b4a4867f3e9d179c.jpg

Solve the following equations.
$\displaystyle\, \frac{x}{\sqrt{1 \, + \, x^2}}\, + \, \frac{x}{\sqrt{1 \, - \, x^2}}\, = \, 0$

$\dfrac{x}{\sqrt{1+x^2}}+\dfrac{x}{\sqrt{1-x^2}}=0$

Here

$x\neq \pm1$

$\implies x\left(\dfrac{1}{\sqrt{1+x^2}}+\dfrac{1}{\sqrt{1-x^2}}\right)=0$

$\implies x=0$

$or$

$\dfrac{1}{\sqrt{1+x^2}}+\dfrac{1}{\sqrt{1-x^2}}=0$

which is not possible as these are positive roots and sum of two positive roots can never be real.

Hence, only solution is

$x=0$

Find the square root of
$5-\sqrt { 10 } -\sqrt { 15 } +\sqrt { 6 }$

$5-\sqrt{10}-\sqrt{15}+\sqrt{6}$

$=5+2\times\left(-\sqrt{\dfrac52}\right)\times1+2\times\left(-\sqrt{\dfrac52}\right)\times\sqrt{\dfrac32}+2\times1\times\left(\sqrt{\dfrac32}\right)$

$=1+\dfrac52+\dfrac32+2\times\left(-\sqrt{\dfrac52}\right)\times1+2\times\left(-\sqrt{\dfrac52}\right)\times\sqrt{\dfrac32}+2\times1\times\left(\sqrt{\dfrac32}\right)$

$=(1+\sqrt{\dfrac32}-\sqrt{\dfrac52})^2$

Hence square root of

$5-\sqrt{10}-\sqrt{15}+\sqrt{6}$ is

$=\pm\left(1+\sqrt{\dfrac32}-\sqrt{\dfrac52}\right)$

Write a Pythagorean triplet whose one member is
$6$
$14$
$16$

$\\(1.)\>6^2+8^2=10^2\\\therefore\>6,8,10\>are\>Pythagorean\>triplet\>\\(2.)\>14^2+48^2=50^2\\hence\>triplet\>is\>14,48,50\\(3.)\>16^2+63^2=65^2Hence\>triplet\>is16,63,65$

Find the square root of a given number by prime factorization method.
2304

2	2304
2	1152
2	576
2	288
2	144
2	72
2	36
2	18
3	9
3	3
	1

$2304=2\times2\times2\times2\times2\times2\times2\times2\times3\times3$

$\\ \sqrt{2304}=2\times2\times2\times2\times3=48$

Which of the following are Pythagorean triplets?
i) (9, 8, 10)
ii) (4, 3, 5)
iii) (6, 8, 10)

$(i) (9,8,10)$

Since these numbers do not follow Pythagoras theorem, we can say that this is not a Pythagoras triplet

$(ii) (4,3,5)$

Since

$5^2=4^2+3^2$

So, this is a Pythagorean triplet.

$(iii) (6,8,10)$

Since

$10^2=8^2+6^2$

so, this is a Pythagorean triplet.

Show that each of the following number is a perfect square. In each case find the number whose square is the given number.
(i) $2025$ (ii) $6561$ (iii) $38416$ (iv) $7921$ (v) $1444$ (vi) $5776$

1310875_1036864_ans_495c32261c284ee994e0e84784c2a6dd.jpeg

Find the square root of a given number by the prime factorization method.
$900$

We will write

$900$ as product of its prime factors,

$900=2 \times2 \times3\times3\times5\times5$

$=2^2\times3^2\times5^2$

$=(2\times3\times5)^2$

$=30^2$

$\Rightarrow \sqrt{900}=30$

Suppose m and n are any two numbers. If ${ m }^{ 2 }-{ n }^{ 2 },2mn$ and ${ m }^{ 2 }+{ n }^{ 2 }$ are the three sides of a triangle, then show that it is a right angled triangle and hence write any two pairs of Pythagorean triplet

${ 3 }^{ 2 }+{ 4 }^{ 2 }={ 5 }^{ 2 }$

${ 5 }^{ 2 }+{ 12 }^{ 2 }={ 13 }^{ 2 }$

Now to show,

${ m }^{ 2 }-{ n }^{ 2 },2mn$ and

${ m }^{ 2 }+{ n }^{ 2 }$ are the sidesof right angled triangle.

Sides of right angled triangle form pythagorean triplet.

$(m^2-n^2)^2+(2mn)^2 = m^4+n^4-2m^2n^2+4m^2n^2$

$=m^4+n^4+2m^2n^2$

$=(m^2+n^2)^2$

$\therefore$ It form right angles triangles,

Examples of triplet:

${ 3 }^{ 2 }+{ 4 }^{ 2 }={ 5 }^{ 2 }$

${ 5 }^{ 2 }+{ 12 }^{ 2 }={ 13 }^{ 2 }$

948992_1016125_ans_e19d0419edd349a8b5be68bd2ea1753d.png

Find the square root of 2025

By log division method

$\therefore$

$\sqrt{2025}=45$
1503761_1035344_ans_28d5330ab65f49b6a6c8f303bc4e9b28.png

$\sqrt{\sqrt{50} + \sqrt{48}} = k(\sqrt{3} + \sqrt{2})$ , then k =

Given that,

$\sqrt{\sqrt{50} + \sqrt{48}} = k(\sqrt{3} + \sqrt{2})$

$k = \dfrac{\sqrt{\sqrt{50} + \sqrt{48}}}{\sqrt{3} + \sqrt{2}}$

$k = \dfrac{\sqrt{5\sqrt{2} + 4\sqrt{3}}}{\sqrt{3} + \sqrt{2}} \times \dfrac{\sqrt{3} - \sqrt{2}}{\sqrt{3} - \sqrt{2}}$

$k = \dfrac{\sqrt{5\sqrt{2} + 4\sqrt{3}} \times (\sqrt{3} - \sqrt{2})}{(\sqrt{3})^2 - (\sqrt{2})^2}$

$\therefore k = \sqrt{5\sqrt{2} + 4\sqrt{3}} \times (\sqrt{3} - \sqrt{2})$

A student writes the formula $\sqrt{ab} \, = \, \sqrt{a} \sqrt{b}$ . Then substitutes a = -1 and b = -1 finds 1 = -Explain where he is wrong.

We first note that the formula

$\sqrt{ab} \, = \, \sqrt{a} \sqrt{b}$ , holds if at least one of a and b are negative.
Since here both a and b are negative, we cannot apply the above formula. Thus if a = -1 and b = -1, we cannot have

$\sqrt{(-1) (-1)} \, = \, \sqrt{-1} \sqrt{-1}$ which gives the absurd result 1 = -1.

solve: $\sqrt {2 + \sqrt {2 + \sqrt {2 + \sqrt {2\,\,\,...........} } } }$

Given,

$\sqrt { 2+\sqrt { 2+ } \sqrt { 2+.. } }$

Let

$x=\sqrt { 2+\sqrt { 2+ } \sqrt { 2+\sqrt { 2+.. } } }$

Squaring both sides

$x^2=2+\sqrt { 2+\sqrt { 2+ } \sqrt { 2+.. } }$

$x^2=2+x$

$x^2-x-2=0$

$x^2-2x+x-2=0$

$x(x-2)+(x-2)=0$

$(x-2)(x+1)=0$

$x=2$ or

$x=-1$

As,

$x>0$ So,

$x=2$

What is the smallest natural number that should be multiplied with $840$ to make it a perfect square?

$840=84\times 10$

$840= 4\times 21\times 2\times 5$

$840={ 2 }^{ 3 }\times 7\times 3\times 5$

To make it a perfect square it should be multiplied with

$2\times 7\times 3\times 5$ so that powers of prime numbers will become even.

$2\times 7\times 3\times 5 = 14\times 15$

$=210$

Find the fifty-sixth square number.

Fifty sixth square number

$\Rightarrow 56\times 56=3136$

Find smallest No: by which $3645$ must be $\div$ to make it a per.square

Smallest number will be 5

So, that 3 6 4 5 is per sq.

$\sqrt{1806336}$ =

We are going to find square root by using prime factorization method

$1806336=2\times 903168$

$=2\times 2\times 451584$

$=2\times 2\times 2\times 225792$

$=2\times 2\times 2\times 2\times 112896$

$=2\times 2\times 2\times 2\times 2\times 56448$

$=2\times 2\times 2\times 2\times 2\times 2\times 28224$

$=2\times 2\times 2\times 2\times 2\times 2\times 2\times 14112$

$=2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 7056$

$=2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 3\times 2352$

$=2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 3\times 3\times 784$

$=2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 3\times 3\times 2\times 392$

$=2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 3\times 3\times 2\times 2\times 196$

$=2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 3\times 3\times 2\times 2\times 2\times 98$

$=2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 3\times 3\times 2\times 2\times 2\times 2\times 49$

$=2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 3\times 3\times 2\times 2\times 2\times 2\times 7\times 7$

$=2^{12}\times 3^2\times 7^2$

$\Rightarrow$

$1806336=2^{12}\times 3^2\times 7^2$

$\Rightarrow$

$\sqrt{1806336}=\sqrt{2^{12}\times 3^2\times 7^2}$

$\Rightarrow$

$\sqrt{1806336}=2^6\times 3\times 7$

$\Rightarrow$

$\sqrt{1806336}=64\times 3\times 7$

$\Rightarrow$

$\sqrt{1806336}=1344$

Find the value: $\sqrt { 8+\sqrt { 773+\sqrt { 116+\sqrt { 2+\sqrt { 520+81 } } } } }$

$\sqrt {8 + \sqrt {773 + \sqrt {116 + \sqrt {2 + \sqrt {520 + \sqrt {81} } } } } }$

$= \sqrt {8 + \sqrt {773 + \sqrt {116 + \sqrt {2 + \sqrt {520 + 9} } } } }$

$= \sqrt {8 + \sqrt {773 + \sqrt {116 + \sqrt {2 + \sqrt {529} } } } }$

$= \sqrt {8 + \sqrt {773 + \sqrt {116 + \sqrt {2 + 23} } } }$

$= \sqrt {8 + \sqrt {773 + \sqrt {116 + \sqrt {25} } } }$

$= \sqrt {8 + \sqrt {773 + \sqrt {116 + 5} } }$

$= \sqrt {8 + \sqrt {773 + \sqrt {121} } }$

$= \sqrt {8 + \sqrt {773 + 11} }$

$= \sqrt {8 + \sqrt {784} }$

$= \sqrt {8 + 28}$

$= \sqrt {36}$

$=6$

$(1035 + x)$ chairs are placed in an auditorium in such way that the number of rows is equal to number of columns. Find the least value of $x$ .

$1035+x$ chairs

given rows

$=$ columns

Let rows be y

$\Rightarrow y^2=1035+x$

It should be perfect square we should find least value of x

We know it should be greater than

$30^2=900$ .

Now

$31^2=961$

$32^2=1024$

$33^2=1089$

$\therefore$ It should be

$1089-1035=54$

Least value of

$x=54$ .

Find whether the square of the following numbers are even or odd?
$431$ .

$431$ is an odd no. and square of an odd number is always odd

Find the square root of the following number by the prime factorisation method.
$5625$

The factors of $5625$ is,

$5625 = 5 \times 5 \times 5 \times 5 \times 3 \times 3$

Take square root both sides,

$\sqrt {5625} = \sqrt {5 \times 5 \times 5 \times 5 \times 3 \times 3}$

$= \sqrt {{5^4} \times {3^2}}$

$= \sqrt {{{\left( {{{\left( 5 \right)}^2} \times 3} \right)}^2}}$

$= 25 \times 3$

$= 75$

Therefore, the square root of 5625 is 75.

Find the square roots of the following numbers by Prime factorization method.
(i)441
(ii)784
(iii)4096
(iv)7056

(a)

$441$

prime factorization:

$441=3\times 3\times 7\times 7$

$\sqrt{441}=\sqrt{3\times 3\times 7\times 7}=3\times 7$

$=21$

(b)

$784=2\times 2\times 2\times 2\times 7\times 7$

$\sqrt{784}=\sqrt{2\times 2\times 2\times 2\times 7\times 7}$

$\sqrt{784}=2\times 2\times 7$

$=28$

(c)

$4096=2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2$

$\sqrt{4096}=\sqrt{2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2}$

$\sqrt{4096}=2\times 2\times 2\times 2\times 2\times 2$

$=2\times 2$

$=64$

(D)

$7056=2\times 2\times 2\times 2\times 3\times 3\times 7\times 7$

$\sqrt{7056}=\sqrt{2\times 2\times 2\times 2\times 3\times 3\times 7\times 7}$

$=2\times 2\times 3\times 7$

$=4\times 21$

$=84$ .

Find whether the square of the following numbers are even or odd?
$9999$

here

$9999=10000-1$

now square of

$a-b$ can be written as

$a^{2}-2ab+b^{2}$

So,

$(9999)^{2}=(10000-1)^{2}$

$=(10000)^{2}-2(10000)(1)+(1)^{2}$

$\rightarrow$ now, any even number square became even number only

$\rightarrow$ Multiplication of any number with even number converts into even it self

$\rightarrow$ here

$(1)^{2}$ will be odd number.

So,

$(9999)^{2}=EVEN+(-EVEN)+ODD$

$\rightarrow$ We can say that

$EVEN+EVEN=EVEN$

$EVEN+ODD=ODD$

$ODD+ODD=EVEN$

So,

$(9999)^{2}=EVEN-EVEN+ODD$

$=EVEN+ODD$

$(9999)^{2}=ODD$

Find the smallest number by which $7776$ is to be divided to get a perfect square.

$7776$

$81\times 96$

$81\times 16\times 6$

It is known that

$81$ and

$16$ are perfect squares.

Hence dividing the given number by

$6$ will result in a perfect square number.

Find the smallest number by which $3645$ must be multiplied to get a perfect square.

$3645$

$=9\times 9\times 9\times 5$

$=(9)^{2}\times (45)\times (45)$

It should be multiplied with

$45$

Find whether the square of the following numbers are even or odd?
$17779$

Here

$17779=(17780-1)$

Now square of

$a-b$ can be written as

$a^{2}-2ab-b^{2}$

So,

$(17779)^{2}=(17780-1)^{2}$

$=(17780)^{2}-2(17780)(1)+(1)^{2}$

Square of an even number will be also even and even multiplied by an even number is also even.

And here

$(1)^{2}$ will be odd number

So,

$(17779)^{2}=$ Even

$-$ Even

$+$ Odd

We know Even

$+$ Even

$=$ Even

and Even

$+$ Odd

$=$ Odd

and Odd

$+$ Odd

$=$ Even

So,

$(17779)^{2}=$ Even

$-$ Even

$+$ Odd

$=$ Even

$+$ Odd

$\therefore\ \ (17779)^{2}=$ Odd

Find whether the square of the following numbers are even or odd?
$8204$

Square of

$8204$ is even because

$8204$ is even and square of an even number is even

Find the square root of the following number by the prime factorisation method.
$1\dfrac{17}{64}$

The given number $1\dfrac{{17}}{{64}}$ can be simplified as $\dfrac{{81}}{{64}}$ .

The factors of $81$ is $3 \times 3 \times 3 \times 3$ and the factors of $64$ is $2 \times 2 \times 2 \times 2 \times 2 \times 2$ . Therefore,

$\dfrac{{81}}{{64}} = \dfrac{{3 \times 3 \times 3 \times 3}}{{2 \times 2 \times 2 \times 2 \times 2 \times 2}}$

Taking square root both sides,

$\sqrt {\dfrac{{81}}{{64}}} = \sqrt {\dfrac{{3 \times 3 \times 3 \times 3}}{{2 \times 2 \times 2 \times 2 \times 2 \times 2}}}$

$= \sqrt {\dfrac{{{3^4}}}{{{2^6}}}}$

$= \sqrt {{{\left( {\dfrac{{{3^2}}}{{{2^3}}}} \right)}^2}}$

$= \dfrac{{{3^2}}}{{{2^3}}}$

$= \dfrac{9}{8}$

Therefore, the square root of $1\dfrac{{17}}{{64}}$ is $\dfrac{9}{8}$

Find the square root of the following number by the prime factorisation method.
$4096$

The factors of $4096$ is,

$4096 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2$

Take square root both sides,

$\sqrt {4096} = \sqrt {2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2}$

$= \sqrt {{2^{12}}}$

$= \sqrt {{{\left( {{{\left( 2 \right)}^6}} \right)}^2}}$

$= {2^6}$

$= 64$

Therefore, the square root of 4096 is 64.

Find the largest number of $6$ digits which is a perfect square.

The largest $6$ digit which is a perfect square is $998001=$ $(999)^2$

The largest $6$ digit number is $999999$ .While it is not a perfect square, when 1 is added to it we get $1000000$ which is smallest seven digit number and also perfect square $=(1000)^2$

So the smallest 6 digit number which is perfect square $=(1000–1)^2=(999)^2=998001$ Ans

Find the square root of the following number by factorisation.
$256$

1180621_1068276_ans_2b769205857e464abfad0ebcc6b560ac.jpg

Find the square root of the following number by the prime factorisation method.
$1936$

The factors of $1936$ is,

$1936 = 2 \times 2 \times 2 \times 2 \times 11 \times 11$

Take square root both sides,

$\sqrt {1936} = \sqrt {2 \times 2 \times 2 \times 2 \times 11 \times 11}$

$= \sqrt {{2^4} \times {{11}^2}}$

$= \sqrt {{{\left( {{{\left( 2 \right)}^2} \times 11} \right)}^2}}$

$= 4 \times 11$

$= 44$

Therefore, the square root of 1936 is 44.

Find the square root of the following number by the prime factorisation method.
$7744$

The factors of $7744$ is,

$7744 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 11 \times 11$

Take square root both sides,

$\sqrt {7744} = \sqrt {2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 11 \times 11}$

$= \sqrt {{2^6} \times {{11}^2}}$

$= \sqrt {{{\left( {{{\left( 2 \right)}^3} \times 11} \right)}^2}}$

$= 8{\kern 1pt} \times 11$

$= 88$

Therefore, the square root of 7744 is 88.

Find the square root of the following number by the prime factorisation method.
$6\dfrac{1}{4}$

The given number $6\cfrac{1}{4}$ can be simplified as $\cfrac{{25}}{4}$ .

The factors of 25 is $5 \times 5$ and the factors of 4 is $2 \times 2$ . Therefore,

$\cfrac{{25}}{4} = \cfrac{{5 \times 5}}{{2 \times 2}}$

Taking square root both sides,

$\sqrt {\cfrac{{25}}{4}} = \sqrt {\cfrac{{5 \times 5}}{{2 \times 2}}}$

$= \sqrt {\cfrac{{{5^2}}}{{{2^2}}}}$

$= \sqrt {{{\left( {\cfrac{5}{2}} \right)}^2}}$

$= \cfrac{5}{2}$

Therefore, the square root of $6\cfrac{1}{4}$ is $\cfrac{5}{2}$ .

Find the square root of the following number by the prime factorisation method.
$4356$

The factors of $4356$ are,

$4356 = 2 \times 2 \times 3 \times 3 \times 11 \times 11$

Take square root on both sides,

$\sqrt {4356} = \sqrt {2 \times 2 \times 3 \times 3 \times 11 \times 11}$

$= \sqrt {{2^2} \times {3^2} \times {{11}^2}}$

$= \sqrt {{{\left( {2 \times 3 \times 11} \right)}^2}}$

$= 2 \times 3 \times 11$

$= 66$

Therefore, the square root of $4356$ is $66$ .

Find the square root of the following number by the prime factorisation method.
$1521$

The factors of $1521$ is,

$1521 = 3 \times 3 \times 13 \times 13$

Take square root both sides,

$\sqrt {1521} = \sqrt {3 \times 3 \times 13 \times 13}$

$= \sqrt {{3^2} \times {{13}^2}}$

$= \sqrt {{{\left( {3 \times 13} \right)}^2}}$

$= 39$

Therefore, the square root of 1521 is 39.

Find the square root of the following number by the prime factorisation method.
$5929$

The factors of $5929$ is,

$5929 = 7 \times 7 \times 11 \times 11$

Take square root both sides,

$\sqrt {5929} = \sqrt {7 \times 7 \times 11 \times 11}$

$= \sqrt {{7^2} \times {{11}^2}}$

$= \sqrt {{{\left( {7 \times 11} \right)}^2}}$

$= 7 \times 11$

$= 77$

Therefore, the square root of 5929 is 77.

Three-eighth of the students of a class opted for visiting an old age home. Sixteen students opted for having a nature walk. Square root of total number of students in the class opted for tree plantation in the school. The number of students who visited the old age home is same as the number of students who went for a nature walk and did tree plantation. Find the total number of students. What values are inculcated in students through such activities?

1333766_1063651_ans_377ca0cacea94d7e9ec33fa82aa384d2.jpg

Find the square root of the following number by factorisation.
$13225$

1171041_1068287_ans_70c8ff931d634541bef45a7c05e445f0.jpg

Find the square root of the following number by the prime factorisation method.
$\dfrac{289}{144}$

The factors of 289 is $17 \times 17$ and the factors of 144 is $2 \times 2 \times 2 \times 2 \times 3 \times 3$ . Therefore,

$\cfrac{{289}}{{144}} = \cfrac{{17 \times 17}}{{2 \times 2 \times 2 \times 2 \times 3 \times 3}}$

Taking square root both sides,

$\sqrt {\cfrac{{289}}{{144}}} = \sqrt {\cfrac{{17 \times 17}}{{2 \times 2 \times 2 \times 2 \times 3 \times 3}}}$

$= \sqrt {\cfrac{{{17^2}}}{{{2^4} \times {3^2}}}}$

$= \sqrt {{{\left( {\cfrac{{17}}{{{2^2} \times 3}}} \right)}^2}}$

$= \cfrac{{17}}{{{2^2} \times 3}}$

$= \cfrac{{17}}{{12}}$

Therefore, the square root of $\cfrac{{289}}{{144}}$ is $\cfrac{{17}}{{12}}$ .

Estimate the value of the following numbers to the nearest whole number
(i) $\sqrt{97}$
(ii) $\sqrt{250}$
(iii) $\sqrt{780}$

(i)

The value of $\sqrt {97}$ is $9.84 \approx 10$ . Therefore, the nearest whole number of $\sqrt {97}$ is 10.

(ii)

The value of $\sqrt {250}$ is $15.81 \approx 16$ . Therefore, the nearest whole number of $\sqrt {250}$ is 16.

(iii)

The value of $\sqrt {780}$ is $27.92 \approx 28$ . Therefore, the nearest whole number of $\sqrt {780}$ is 28.

Find the square root of the following number by factorisation.
$10404$

$10404 = 2 \times 2 \times 3 \times 3 \times 17 \times 17$

$= 2 \times 3 \times 17 \times 2 \times3\times 17$

$10404 = 102 \times 102=102^2$

$\therefore \sqrt{10404} = 102$

1424231_1068279_ans_23f0d6f2ff9242059d60689a4cc288fa.png

Find the square root of the following number by factorisation.
$196$

1180620_1068274_ans_20b9e27aee914a95af627d07d0b05a01.jpg

Write a Pythagorean triplet whose one member is $14$ .

$\textbf{Step 1: Find m and substitute in the general formula of Pythagorean triplet form.}$

$\text{For any natural number, the Pythagorean triplet is in the form 2m, m$^2$ - 1, m$^2$ + 1}$

$\text{Let, 2m = 14}$

$\Rightarrow \text{m = $\dfrac{14}{7}$}$

$\Rightarrow \text{m = 7}$

$\text{........eqn(1)}$

$\text{If we assume m$^2$ + 1 = 14}$

$\Rightarrow \text{m$^2$ = 14-1}$

$\Rightarrow \text{m$^2$ = 13}$

$\text{Since 13 is not a perfect square number, so m$^2$ $\neq$ 14}$

$\text{........eqn(2)}$

$\text{If we assume m$^2$ - 1 = 14}$

$\Rightarrow \text{m$^2$ = 14+1}$

$\Rightarrow \text{m$^2$ = 15}$

$\text{Since 15 is not a perfect square number, so m$^2$ $\neq$ 15}$

$\text{........eqn(3)}$

$\text{From eqn (1), (2) and (3) we have m = 7}$

$\text{So, now m$^2$ + 1 = 7$^2$ + 1}$

$\text{= 49 +1 }$

$\text{= 50 }$

$\text{So, now m$^2$ - 1 = 7$^2$ - 1}$

$\text{= 49 - 1 }$

$\text{= 48 }$

$\textbf{Hence the Pythagorean triplet is 14 ,48 and 50 }$

Find the square root by long-division method:
$7921$

Performing long division of the given number

$7921$ , as shown above.

Hence, by long division method,

$\sqrt{7921}=89$ .

2085560_1074018_ans_89f0e59076a7411eb78473c1d5ab8e76.png

Find the square root of the following number by long division method.
$1024$

1099764_1074024_ans_f305ae9601c14cecb2f5887efa4928f3.jpg

Find the square root of the following number by division method.
$1369$

1099707_1074010_ans_2c8161e35c474dbdbd765fe9b73c37ce.jpg

Find the square root of the following number by division method.
$3249$

___________

$5 | 3249$

$5 | 25$

|______

$107 | 749$

$7 | 749$

|_______

$0$

$So, √3249 = 57$

Write a Pythagorean triplet whose one member is $16$ .

2175043_1069963_ans_b5ac96f1a5c7483a96853c1877eaa412.jpg

Find the square root of the following number by long division method.
$576$

1099749_1074020_ans_d392f74d89ff4923a2ba4d73a2ca8754.jpg

Find the square root of the following number by Division method.
$3481$

Consider first two digits and then the next two digits.If there are three digits in the given number then first consider the first digit and then the take two digits together

5 | 3481 | 5x

10x | 981 |

Now the number 981 has 1 in units place so take a number which has 1 at the end (ex.1*1=9 and 9*9=81)

So put 9 in the place of x and multiply the x value with 10x(that is 981) u will get the remainder as 0

Therefore the square root of 3481 is 5x*5x i.e. 59*59=3481

Find the square root of the following number by long division method.
$529$

1099650_1074006_ans_6fb0be4a3d4d4f23b192cf103626cf50.jpg

Find the square root of following surds:
$a+\sqrt{a^{2}-b^{2}}$

Square root of Sure:

$a+\sqrt{a^2-b^2}$

$\Rightarrow \sqrt{a+\sqrt{a^2-b^2}}$

$=\dfrac{1}{\sqrt{2}}\sqrt{2a+2\sqrt{a+b}\sqrt{a-b}}$

$=\dfrac{1}{\sqrt{2}}\sqrt{(\sqrt{a+b}+\sqrt{a-b})}$

$=\pm\dfrac{1}{\sqrt{2}}(\sqrt{a+b}+\sqrt{a-b})$ .

1356702_1128439_ans_77953b2cbd4547969501a176d35280f1.jpg

Find the square root of the following number by Division method.
$4489$

Solution:

Consider first two digits and then the next two digits.If there are three digits in the given number then first consider the first digit and then the take two digits together

$6 | 4489 | 6x$

$36$

$12\times | 889 |$

Now the number $889$ has $9$ in units place so take a share number which has $9$ at the end (ex. $3\times 3=9 and 7\times 7=49$ )

So put $7$ in the place of $x$ and multiply the $x$ value with $12\times (that is 127) u$ will get the remainder as $0$

Therefore the square root of $4489$ is $6x\times 6x$ i.e. $67\times 67=4489$

Find the square root of the following number by long division method.
$3136$

1099778_1074026_ans_246a83cda9e54dd59f00473df1c53f77.jpg

Find the $\,\sqrt {39204}$

1332512_1079734_ans_53943cbf67df4a669568316f90783bd7.png

Find the value of $\sqrt {73441}$ and evaluate $\sqrt {734.41} \times \sqrt {7.3441}$

Now,

$\sqrt{734.41}=\sqrt{\dfrac{73441}{100}}=\dfrac{271}{10}=27.1$

and

$\sqrt{7.3441}=\sqrt{\dfrac{73441}{10000}}=\dfrac{271}{100}=2.71$

therefore

$\sqrt{734.41} \times \sqrt{7.3441}=27.1 \times 2.71=73.441$

988604_1084128_ans_058624ff383f4434a0efc90832f30cbb.png

The students of $8^{th}$ class of a school donate $Rs.2401$ in all for Prime Minister National Relief Fund. Each student donates as many rupees as the no. of students in the class. Find the no. of students in the class.

Let, number of students in the class be

$x$ .

So, each student donates

$Rs.x$

$\therefore$ Total donation

$=Rs.(x\times x)=Rs.{ x }^{ 2 }$

$\therefore x^2=2401$

$\Rightarrow x=\sqrt{2401}$

$\Rightarrow x=49$

Hence, total number of students in the class is

$49$ .

What will be the unit digit of the squares of the following numbers?
$(i) 81 \quad \quad(ii) 272 \quad(iii) 799 \quad(iv) 243 \quad(v) 26387 \quad(vi) 52698 \quad(vii) 2796 \quad(viii) 55555$

(i)

$\because 1\times 1=1$

$\therefore$ The units digit of

$(81)^{2}$ will be

$1$

(ii)

$\because 2\times 2=4$

$\therefore$ The unit digit of

$(272)^{2}$ will be

$4$

(iii) Since

$9\times 9=81$

$\therefore$ The unit is digit of

$(799)^{2}$ will be

$1$

(iv) Since

$3\times 3=9$

$\therefore$ The units digit of

$(3583)^{2}$ will be

$9$

(v) Since

$4\times 4=16$

$\therefore$ The units digit of

$(1234)^{2}$ will be

$6$

(vi) Since

$7\times 7=49$

$\therefore$ The units digit of

$(26387)^{2}$ will be

$9$

(vii) Since

$8\times 8=64$

$\therefore$ The units digit is

$(52698)^{2}$ will be

$4$

(viii) Since

$0\times 0=0$

$\therefore$ The units digits of

$(99880)^{2}$ will be

$0$

(ix) Since

$6\times 6=36$

$\therefore$ The units digit of

$(12796)^{2}$ will be

$6$

(x) Since

$\times 5=25$

$\therefore$ Units digit of

$(55555)^{2}$ will be

$5$

Find the square root of the following number by long division method.
$900$

1099794_1074028_ans_99b663960fed45d6b1310b396a1dc56b.png

Find the value of $\sqrt {110889}$ and hence find the value of $\sqrt {11.0889} + \sqrt {1108.89}$

Now,

$\sqrt{11.0889}=\sqrt{\dfrac{110889}{10000}}=\dfrac{333}{100}=3.33$

and

$\sqrt{1108.89}=\sqrt{\dfrac{110889}{100}}=\dfrac{333}{10}=33.3$

therefore

$3.33+33.3=36.63$

988591_1084125_ans_67338ef42e094c32ae356b2301389381.png

Mr. Khanna has $1200$ plants. He wants to plant these in such a way that the number of rows and the number of column remain the same.
a) What is the minimum number of plants he needs more for this purpose ?
b) Which value is depicted by Mr. Khanna ?

Mr. Khanna has

$1200$ plants.

Let the number of rows = number of columns

$=x$

Then total number of plants

$=x\times x=x^{2}$

So, the number of plants should be a perfect square.

Now, the square-root of

$1200$ is,

$\sqrt{1200}\approx 34.64$

So,

$1200$ is not a perfect square.

The whole number larger than

$34.64$ is

$35$

And,

$(35)^{2}=1225$

So, number of plants he needs more

$=1225-1200=25$

And the number of rows = number of columns

$=35$

What will be the unit digit of the square of the following numbers?
$81$

Unit digit of

${ 81 }^{ 2 } \Rightarrow$ Unit digit of

${ 1 }^{ 2 }=1$

Find the square root by long-division method:
$5776$ .

$\sqrt { 5776 } =76$
1109568_1108965_ans_7570a80da7324ef2913d615e8e81a23d.jpg

Find the value of $\sqrt{2401}$ using prime factorization method.

Equcation for number

$2401$ factorization is:

$7\times7\times7\times7$

$\therefore$ square root of 2401 is 49.

Find the square root of following surd :
$7+\sqrt{48}$

$7+\sqrt{48}=7+\sqrt{3*4*4}=7+4\sqrt3$

$\sqrt{112}$ ?

We can rewrite

$112$ as

${ 4 }^{ 2 }\ast 7$

$\sqrt { 112 } =\sqrt { { 4 }^{ 2 }\ast 7 }$

The result can be shown in both exact and decimal forms.
Exact Form=

$4\sqrt { 7 }$
Decmal form=

$10.58300$

Nearest whole number is 11

Find the square root of:
$245$ correct to two places of decimal

Let us use long division method to find the square root of

$245$ correct to two places of decimal.

Hence

$\sqrt{245}=15.65$ correct to two decimal places.

992077_1092324_ans_db52194fe3b84fad87824cf6a775bb07.png

Find the least number which must be added to $4931$ to make it a perfect square ?

${ 70 }^{ 2 }=4900$

${ 71 }^{ 2 }=5041$

$5041-4931=110$
so

$110$ must be added to

$4931$ to get a perfect Square

Find the square root of $256$ .

$=256=16\times 16=16^2$

$=\sqrt { 16^2}=16$

Find the square root of $1024$ . Hence find the value of $\sqrt{10.24}+\sqrt{0.1024}+\sqrt{10240000}$ .

$\left.\begin{matrix} 2\\2\\2\\2\\2\\2\\2\\2\\2\\\\ \end{matrix}\right|\begin{matrix}1024 \\512\\256\\128\\64\\32\\16\\8\\4\\2 \end{matrix}$

$\Rightarrow 1024=2^{10}$

$\therefore$

$\sqrt{1024}=2^{5}$

$=32$

$\therefore$

$\sqrt{10.24}+\sqrt{0.1024}+\sqrt{10240000}$

$=\sqrt{\dfrac{1024}{100}}+\sqrt{\dfrac{1024}{10000}}+\sqrt{1024\times 10^{4}}$

$=\dfrac{32}{10}+\dfrac{32}{100}+32(100)$

$-(32)\left(\dfrac{10+1+10000}{100}\right)$

$=32\times \dfrac{10011}{100}$

$=3203.52$

What will be the unit digit of the square of the following numbers?
$1234$

Since,

${ 4 }^{ 2 }$ =

$16$ , so,

${ 1234 }^{ 2 }$ will have

$6$ at unit’s place

What will be the unit digit of the square of the following numbers?
$55555$

Since,

${ 5 }^{ 2 }$ =

$5$ ,so,

${ { 55555 }^{ 2 } }$ will have

$5$ at unit’s place

What will be the unit digit of the square of the following numbers?
$26387$

Since,

${ 7 }^{ 2 }$ =

$49$ ,so,

${ { 26387 }^{ 2 } }$ will have

$9$ at unit’s place

Find the smallest number by which $98$ should be multiplied to make it a perfect square.

Factorize $4802$ then make pairing.
$98=2\ast 7\ast 7$
so, smallest no is $2$ that should be multiples to get perfect square.
so perfect square will be $196$ .
square root will be $14$ .

What will be the unit digit of the square of the following numbers?
$52698$

Since,

${ 8 }^{ 2 }$ =

$64$ ,so,

${ { 52698 }^{ 2 } }$ will have

$4$ at unit’s place

What will be the unit digit of the square of the following numbers?
$99880$

Since,

${ 0 }^{ 2 }$ =

$0$ ,so,

${ { 99880 }^{ 2 } }$ will have

$0$ at unit’s place

Write a phythogorean triphlets whose one of the member is 26.

1333639_1124906_ans_d336c19ad4324344af6885405dec3d5b.jpg

What will be the unit digit of the square of the following numbers?
$799$

Since,

${ 9 }^{ 2 }=81,So,799^2$ will have

$1$ at unit’s place

What will be the unit digit of the square of the following numbers?
$3853$

Since

${ 3 }^{ 2 }=9$ . So,

${ 3853 }^{ 2 }$ will have

$9$ at unit’s place.

Find the square root of the following in the form of a binomial surd.
$12+2\sqrt{35}$ .

Let

$x = \sqrt{12+2\sqrt{35}} = \sqrt n + \sqrt m$

$12+2\sqrt{35} = n + m + 2\sqrt{nm}$

Hence,

$n+m = 12 , nm = 35$

$n^2-12n+35 =0$

$n = 7$ or

$n = 5$

Similarily,

$m = 5$ or

$m=7$

Find the square root of: $33.64$

1354259_1134157_ans_bd71e182e5224654afcfdd123999af3a.jpg

Find the square root of the following in the form of a binomial surd.
$2+\dfrac{2}{3}\sqrt{5}$ .

Let

$x = \sqrt{2+\cfrac{2}{3}\sqrt{5}} = \sqrt n + \sqrt m$

$2+\cfrac{2}{3}\sqrt{5} = n + m + 2\sqrt{nm}$

Hence,

$n+m = 2 , nm = \cfrac{5}{9}$

$9n^2-18n+5 =0$

$n = \cfrac{5}{3}$ or

$n = \cfrac{1}{3}$

Similarily,

$m = \cfrac{5}{3}$ or

$m=\cfrac{1}{3}$

Find the square root of following surds:
$8-3\sqrt{7}$

Square root of the surd

$\Rightarrow \sqrt{8-3\sqrt{7}}$

$=\dfrac{1}{\sqrt{2}}\sqrt{16-6\sqrt{7}}$

$=\dfrac{1}{\sqrt{2}}\sqrt{(\sqrt{7})^2+3^2-2(3)(\sqrt{7})}$

$=\pm\dfrac{1}{\sqrt{2}}(\sqrt{7}-3)$ .

What is the closest perfect square root of following numbers?
(i) $248$ (ii) $1234$ (iii) $1678$ .

1352062_1133088_ans_7e48a2d4a5214f8084f5c8ab7fb12ebe.jpg

Solve $\sqrt{\dfrac{2\times (9\times 10^9)(1.6\times 10^{-19})(47)(1.6\times 10^{-19})}{(1.67\times 10^{-27})(2.5\times 10^{-14})}}$

$=\sqrt{\dfrac{2\times 9\times 10^9\times 1.6\times 10^{-19}\times 47\times 1.6\times 10^{-19}}{1.67\times 10^{-27}\times 2.5\times 10^{-14}}}$

$=\sqrt{\dfrac{2\times 9\times (1.6)^2 \times 47}{(1.67)\times (2.5) }\times 10^9 + 19 - 19 + 27 + 14}$

$=1.6\times 1.3 \sqrt{\dfrac{2\times 47\times 10^{12}}{1.67\times 2.5}}$

$=1.6\times 3\times 10^2 \sqrt{\dfrac{2\times 47}{1.67\times 2.5}}$

$=48\times 10^5\sqrt{22.52}$

$=48\times 10^5 \times 4.75$

$= 227.79\times 10^5$

Do the following numbers make Pythagorean triplets?
(i) $8,\ 10,\ 17$
(ii) $8,\ 10,\ 12$

As Pythagoras theorem says

$c^2=a^2+b^2$ where

$c$ is the longest side of the triangle, called the hypotenuse

$a$ and

$b$ are other sides. So if this condition satisfies the number then they will make Pythagorean triplets.

$(i)\ 8,10,17$

So,

$17^2=289$

$8^2+10^2=64+100$

$=164$

Which means

$17^2\ne 8^2+10^2$

So, these numbers will not make Pythagorean triplets.

$(ii) \ 8,10,12$

$12^2=144$

$8^2+10^2=64+100$

$=164$

Which means

$12^2\ne 8^2+10^2$

So, these numbers will not make Pythagorean triplets.

How many square tiles of side $9 \,cm$ will be needed to fit in a square floor of a bathroom of side $720 \,cm$ ? Find the cost of tilling at the rate of $Rs. 7.50$ per tile.

Side of square of tile

$=9cm$

Area of a square tile

$=Side \times Side$

$=(9\times 9)\ cm^2$

$=81\ cm^2$

Side of a square floor

$=20\ cm$

Area of square floor

$=Side \times Side$

$=(20\times 20)\ cm^2$

$=400\ cm^2$

No. of tiles required to lift in a floor

$=\dfrac {Area\ of\ square\ floor}{Area\ of\ square\ tile}$

$=\dfrac {400}{81}$

$=4.93$

$\cong 5$

Cost of one tile

$=Rs. 7.50$

Cost of

$5$ tiles

$=Rs. (7.50\times 5)$

$=Rs.37.50$

$\therefore \$ No. of tiles required

$=5$

Cost of tiling

$=Rs. 37.50$

Find the least number of $4$ digits which is a perfect square. Find the square roots of this number.

$32^{2} = 1024 \rightarrow$ least number

$31^{2} = 961$

$\therefore \sqrt{1024} = \pm 32$

1121228_1165047_ans_6b38af537d0844c89e2c0bdc86d2b198.jpeg

Solve : $\sqrt{\frac{1+e^{x}}{1-e^{x}}}$

We have,

$\sqrt{\dfrac{1+{{e}^{x}}}{1-{{e}^{x}}}}$

$=\sqrt{\dfrac{1+{{e}^{x}}}{1-{{e}^{x}}}\times \dfrac{1-{{e}^{x}}}{1-{{e}^{x}}}}$

$=\sqrt{\dfrac{1+{{e}^{x}}}{1-{{e}^{x}}}\times \dfrac{1-{{e}^{x}}}{1-{{e}^{x}}}}\,\,\,\,\,\,\,\,On\,rationlize$

$=\sqrt{\dfrac{{{1}^{2}}-{{e}^{2x}}}{{{\left( 1-{{e}^{x}} \right)}^{2}}}}$

$=\dfrac{\sqrt{1-{{e}^{2x}}}}{\left( 1-{{e}^{x}} \right)}$

Hence, this is the answer.

Value $\phi \sqrt \infty$

Solution :-

$\Rightarrow \phi \sqrt{\infty }$

$\Rightarrow \infty$ Ans (square root of infinity is infinity)

1171661_1172578_ans_0a262387929847bf9f643457c0503b25.jpg

A group of boys collected 256 marbles for some project. The contribution of each boy was equal to the number of boys in the group.How many boys were in the group?

Let the number of boys were

$x$ .

Contribution of each boy =

$x \ marbles$

Therefore,

$x^2=256$

$x=16$

Hence, there were 16 boys in the group.

Find the square roots of the following perfect squares using the long division method :
a) $357096609$
b) $159.7696$

18897

-----------------------------------------

1 0357096609

----------------------------------

28 257

224

----------------------------------

368 3309

2944

----------------------------------

365

3769 36566

33921

----------------------------------

2645

37787 264509

264509

----------------------------------

$\therefore \sqrt{357096609} = 18897$

12.6400237

-----------------------------------------

1 0159.7696

----------------------------------

059

22 44

----------------------------------

246 1576

1476

----------------------------------

100

25.24 10096

10096

-----------------------------------

060

50.564

----------------------------------

25.282 9.436

943.6

935.3616169000001

25.2800437 --------------------------------

8.238383099999965

$\therefore \sqrt{159.7696} = 12.64$

Estimate each root, to the nearest whole number.
(a) $\sqrt{171}$
(b) $\sqrt{35}$
(c) $\sqrt{407}$
(d) $\sqrt{292}$

$(a)\sqrt{171}=13.07$ nearest whole number is

$13$

$(b)\sqrt{35}=5.916$ nearest whole number is

$6$

$(c)\sqrt{407}=20.17$ nearest whole number is

$20$

$(d)\sqrt{292}=17.08$ nearest whole number is

$17$

Solve the term:-
$1^{2},\ 3^{2},\ 5^{2},\ 7^{2},\ ......$

The given series is

$1^2,3^2,5^2,7^2....$

So the general term of series is

$(2n-1)^2$

If $\sqrt{6}=2.55$ , then the value of $\sqrt{\dfrac{2}{3}}+3\sqrt{\dfrac{3}{2}}$ is

$\sqrt { 6 } =2.55$

$\sqrt { \dfrac { 2 }{ 3 } } +\sqrt [ 3 ]{ \dfrac { 3 }{ 2 } }$

$=\dfrac { \sqrt { 6 } }{ 3 } +\dfrac { 3\sqrt { 6 } }{ 2 }$

$=\dfrac { 11\sqrt { 6 } }{ 6 } =\dfrac { 11\times 2.55 }{ 6 } =\dfrac { 9.35 }{ 2 } =4.675$ (Answer)

What is a Pythagorean triplet? Give some examples.

$\begin{array}{l} Pythagorean\, triple\, \, are\, the\, combination\; \, of\, \, three\, \, positive\, \, \\ { { int } }egers\, \, a,\, b,\, c\, \, such\, that\, one\, \, of\, \, the\, \, angle\, \, is\, \, right\, angle\, \, 'and\, \, it\, follow\, the\, \, prperty \\ { a^{ 2 } }+{ b^{ 2 } }+{ c^{ 2 } } \\ let \\ a=3 \\ b=4 \\ so,\, becomes\, \, 5. \end{array}$
1203180_1277893_ans_7e021de1906143cfb64dbf6ddcb3535d.PNG

1203180_1277893_ans_7e021de1906143cfb64dbf6ddcb3535d.PNG

If $'n'$ is odd no & $n> 1$ , them prove that $\left(n,\dfrac{n^2-1}{2},\dfrac{n^2+1}{2}\right)$ is a Pythagorean, triplet. Write two Pythagorean triplet taking suitable value of $'n'$ .

$\left(\dfrac{n^2-1}{2}\right)^2=\dfrac{n^4+1-2n^2+n^2}{4}$

$\Rightarrow \dfrac{n^4+1-2n^2+4n^2}{4}=\left(\dfrac{n^2+1}{2}\right)^2$

$\Rightarrow \left(\dfrac{n^2-1}{2}\right)^2+n^2=\left(\dfrac{n^2+1}{2}\right)^2$

$\Rightarrow$ These form Pythagoras triplet.

For e.g., for

$n=2$ , &

$n=3$

$\left(\dfrac{3}{2}\right)^2+2^2=\left(\dfrac{5}{2}\right)^2$

$4^2+3^2=5^2$

$\left(\dfrac{3}{2}, 2, \dfrac{5}{2}\right)$

$(4, 3, 5)$ .

1165333_1242358_ans_90f09b88b96545ca9fc2e51597f545ff.jpg

How many non-perfect square numbers lie between $n^2$ and $(n+1)^2$ ?

Both

$n^2$ and

$(n+1)^2$ are perfect square numbers and they are consecutive perfect squares.

$\Rightarrow$ All the numbers between them are non-perfect square.

Numbers between

$n^2$ and

$(n+1)^2$ are

$=(n+1)^2-n^2-1$

$=n^2+2n+1-n^2-1$

$=2n$

$\Rightarrow$ There are

$2n$ non-perfect square numbers.

Find the square root of $97344$ by prime factorization method.

The given number is

$97344$

By prime factorisation, we get,

$97344=2\times 2\times 2\times 2\times 2\times 2\times 3\times 3\times 13\times 13$

$\therefore \sqrt{97344}=\sqrt{2\times 2\times 2\times 2\times 2\times 2\times 3\times 3\times 13\times 13}$

$\therefore \sqrt{97344}={2\times 2\times 2\times 3\times 13}$

$=312$

Hence, the square root is

$312$

Find the value of $\sqrt{8 + \sqrt{289}}$ :

We have,

$\sqrt{8 + \sqrt{289}}$

$\Rightarrow \sqrt{8 +17}$

$\because \sqrt{289}=17$

$\Rightarrow \sqrt{25}$

$\Rightarrow 5$

Hence, this is the answer.

In $n$ is an odd number and n > 1, then prove that $\left(n,\dfrac{n^{2}-1}{2},\dfrac{n^{2}+1}{2}\right)$ is a Pythagorean triplet. Write two Pythagorean triplet making suitable value of $n$ .

$\textbf{Hint: A Pythagorean triplet is a set of integers a, b, c such that}$

$\ a^{2}+b^{2}=c^{2}$

Solution:

$\textbf{Step 1: Add the square of the first two numbers}$

Given that

$\ n, \ \dfrac {n^{2}-1}{2},\dfrac {n^{2}+1}{2}$ are three numbers

Adding the square of first two numbers

$n^{2}+\left (\dfrac {n^{2}-1}{2}\right)^2$

$=n^{2}+\dfrac {n^{4}-2n^{2}+1}{4}$

$\ \left[\because \left ( a-b \right)^{2}=a^{2}-2ab+b^{2} \right]$

$=\dfrac {4n^{2}+n^{4}-2n^{2}+1}{4}$

$=\dfrac {n^{4}+2n^{2}+1}{4}$

$=\dfrac {(n^{2})^2+2(n^{2})(1)+(1)^2}{4}$

$=\dfrac {(n^{2}+1)^{2}}{(2)^{2}}=\left (\dfrac {n^{2}+1}{2}\right)^{2}$

$\ \left[\because \left ( a+b \right)^{2}=a^{2}+2ab+b^{2} \right]$

$\therefore \ n^{2}+\left (\dfrac {n^{2}-1}{2}\right)^{2}=\left (\dfrac {n^{2}+1}{2}\right)^{2}$

It is a pythagoraean triplet as the sum of the square of two number is equal to the square of third number.

$\textbf{Hence,}$

$\left (n, \ \dfrac {n^{2}-1}{2},\dfrac {n^{2}+1}{2} \right)$

$\textbf{is a pythagorean triplet.}$

$\textbf{Step 2: Put any two value of n in}$

$\left (n, \ \dfrac {n^{2}-1}{2},\dfrac {n^{2}+1}{2} \right)$

Put

$n=5$ in

$\left (n, \ \dfrac {n^{2}-1}{2},\dfrac {n^{2}+1}{2} \right)$ , we get

$\left (5, \ \dfrac {5^{2}-1}{2},\dfrac {5^{2}+1}{2} \right)$ =

$\left (5, \ 12,\ 13\right)$

Now, put

$n=7$ in

$\left (n, \ \dfrac {n^{2}-1}{2},\dfrac {n^{2}+1}{2} \right)$ , we get

$\left (7, \ \dfrac {7^{2}-1}{2},\dfrac {7^{2}+1}{2} \right)$ =

$\left (7, \ 24,\ 25\right)$

$\textbf{Hence, }$

$\left (5, \ 12,\ 13\right)$

$\textbf{and}$

$\left (7, \ 24,\ 25\right)$

$\textbf{are the required Pythagorean triplet}$

what will be the units digits of the square of the following number?
39, 297, 5125, 7286

$39^{2}= 39 \times 39 =$ unit digits of the squares

$=1$

$297^{2}= 297 \times 297=$ unit digit is

$9$

$5125^2= 5125 \times 5125=$ unit digit is

$5$

$7286^2=7286 \times 7286=$ unit digit is

$6$

Find the least number that must be added to 1500 so as to get a perfect square. Also find the square root of the perfect square.

1182576_1224121_ans_21f779a15c374d1ea66620a883b26a54.jpg

Check whether the following numbers form Pythagorean triplet
$9,10,11$

$\begin{array}{l} By\, \, { { using } }\, \, pytogerous\, \, therorem \\ As\, \, we\, \, know \\ \Rightarrow A{ B^{ 2 } }+B{ C^{ 2 } }=A{ C^{ 2 } } \\ \Rightarrow { 10^{ 2 } }+{ 9^{ 2 } }=A{ C^{ 2 } } \\ \Rightarrow \sqrt { 100+81 } =AC \\ \Rightarrow \sqrt { 181 } =AC \\ 13.4\, \, unit \\ Hence,\, \, 9,10\, \, and\, \, 11\, \, are\, \, not\, \, pythogerous\, \, theorem\, \, triflet. \end{array}$
1204957_1297700_ans_657b2f0290364031958966a60dfb7d4a.PNG

1204957_1297700_ans_657b2f0290364031958966a60dfb7d4a.PNG

Check whether the following numbers form Pythagorean triplet:-
$2,3,4$

Let,

$a=2$

$b=3$

$c=4$

Now,

$\begin{aligned}{}{a^2} + {b^2}& = {2^2} + {3^2}\\ &= 4 + 9\\ &= 13\end{aligned}$

And,

$c^2=16$

As,

$a^2+b^2\ne c^2$ then

$2,3,4$ is not a Pythagorean triplet.

Is $2352$ a perfect square? If not, find the smallest number that should be multiplied to $2352$ to make a perfect square. Find the square root of the new numbers.

$\text{Given}\ \text{number}\ \text{is}\ \Rightarrow \ 2352$

$\text{It }\!\!'\!\!\text{ s}\ \text{Factor}$

$2352=2\times 2\times 2\times 2\times 3\times 7\times 7$

$2352={{2}^{4}}\times {{7}^{2}}\times 3$

$\text{It }\!\!'\!\!\text{ s}\ \text{not}\ \text{perfect}\ \text{square}\text{.}$

$\text{So,}$

$\text{The}\ \text{smallest}\ \text{number}\ \text{3}\ \text{is}$

$\text{multiplied}\ \text{by}\ \text{2352}\ \text{is}\ \text{make}$

$\text{perfect}\ \text{square}\text{.}$

$\text{Hence,}\ \text{3}\ \text{is}\ \text{the}\ \text{answer}\text{.}$

Solve: $\sqrt{27^{x+1}}=9^{x-2}$

Given,

$\sqrt{27^{x+1}}=9^{x-2}$

Now squaring both sides we get,

$3^{(3x+3)}=3^{2\times 2(x-2)}$

or,

$3x+3=4(x-2)$

or,

$3x+3=4x-8$

or,

$x=11$ .

Find the square root of $12+2\sqrt{35}$ .

$\begin{array}{l} 12+2\sqrt { 35 } \\ \Rightarrow { \left( { \sqrt { 7 } +\sqrt { 5 } } \right) ^{ 2 } } \\ \Rightarrow \sqrt { 12+2\sqrt { 35 } } \\ \Rightarrow \sqrt { 7 } +\sqrt { 5 } \end{array}$

Hence, the square root is

$\sqrt 7 + \sqrt 5$

Check whether the following numbers form a Pythagorean triplet.
$8,15,17$

Let

$B=8, C=15$ be the smaller numbers and

$A=17$ be the larger number in the Pythagorean triplet.

$\begin{array}{l} By\, \, { { using } }\, \, pythagorus\, \, therorem \\ As\, \, we\, \, know \\ \Rightarrow { B^{ 2 } }+{ C^{ 2 } }={ A^{ 2 } } \\ \Rightarrow { 8^{ 2 } }+{ 15^{ 2 } }={ A^{ 2 } } \\ \Rightarrow \sqrt { 64+225 } =A \\ \Rightarrow \sqrt { 289 } =A \\ A=17\, \, unit = Given \ value \ of \ A\\ Hence,\, \, 8,15\, \, and\, \, 17\, \, are\, \, pythagorean\, \, triplet. \end{array}$

Find the pythagorean triplet whose smallest number is $16$

For every natural number

$m>2,\,\,\left(2m,\,{m}^{2}-1,\,{m}^{2}+1\right)$ is a pythagorean triplet and the smallest number is

$2m$

Given:

$2m=16\Rightarrow m=\dfrac{16}{2}=8$

$\Rightarrow {m}^{2}-1={8}^{2}-1=64-1=63$

and

${m}^{2}+1={8}^{2}+1=64+1=65$

$\therefore$ The triplet is

$\left(16,\,63,\,65\right)$

Solve:
$\sqrt{6\sqrt{6\sqrt{6}}}$

Let

$y=\sqrt {6\sqrt{6\sqrt {6}}}$

$y=\sqrt {6\sqrt{6\times 2.45}}$

$y=\sqrt {6\sqrt{14.7}}$

$y=\sqrt {6\times 3.83}$

$y=\sqrt {23}$

$y=4.79$

Hence, this is the answer.

Expression the following as the product of expression through prime factorization
$36$

Given that,

Prime factorization is

$36$

$36 = 2 \times 2 \times 3 \times 3$

$= {2^2} \times {3^2}$

Thus,

We get factor

${2^2} \times {3^2}$ .

Expression the following as the product of expression through prime factorization
$1125$

We have,

$\begin{matrix} \dfrac { { 1125 } }{ 3 } =375 \\ \dfrac { { 375 } }{ 3 } =325 \\ \dfrac { { 325 } }{ 5 } =65 \\ \dfrac { { 65 } }{ 5 } =13 \\ \end{matrix}$

Then,

We get

$1125 = 3 \times 3 \times 5 \times 5 \times 13 = {3^2} \times {5^2}\times13^1$

Express the following as the product of expression through prime factorization
$288$

We have,

$\begin{matrix} \dfrac { { 288 } }{ 2 } =144 \\ \dfrac { { 144 } }{ 2 } =72 \\ \dfrac { { 72 } }{ 2 } =36 \\ \dfrac { { 36 } }{ 2 } =18 \\ \dfrac { { 18 } }{ 2 } =9 \\ \dfrac { 9 }{ 3 } =3 \\ \end{matrix}$

Then,

We get

$288 = 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3={2^5} \times {3^2}$

Which of the following triplets are Pythagorean ?
$( 3,4,5 ) , ( 6,7,8 ) , ( 10,24,26 ) , ( 2,3,4 ).$

Let the smallest even number be

$2$ m

(i)

$(3, 4, 5)$

$2m=4$

$\Rightarrow m=2$

$m^2-1=4-1=3$

$m^2+1=4+1=5$

$\therefore (3, 4, 5)$ is a Pythagorean triplet.

(ii)

$(10, 24, 26)$

$2m=10$

$\Rightarrow m=5$

$m^2-1=25-1=24$

$m^2+1=25+1=26$

$\therefore (10, 24, 26)$ is not a Pythagorean triplet.

(iii)

$(6, 7, 8)$

$2m=6$

$\Rightarrow m=3$

$m^2-1=9-1=8$

$m^2+1=9+1=10$

$\therefore (6, 7, 8)$ is not a Pythagorean triplet.

(iv)

$(2, 3, 4)$

$2m=2$

$\Rightarrow m=1$

$m^2-1=1-1=0$

$m^2+1=1+1=2$

$\therefore (2, 3, 4)$ is not a Pythagorean triplet.

Square root of $375$ is -

$\begin{array}{l} We\, have \\ \sqrt { 375 } \\ =\sqrt { 5\times 5\times 5\times 3 } \\ =5\sqrt { 15 } \end{array}$

Find the least number that should be subtracted from $1,300$ to get perfect square.

$\sqrt {1300} = 36.05$

Observe that the square root of

$1300$ is slightly more than

$36$ .

and

$(36)^2$ =

$1296$

$\therefore$ The smallest number to be subtracted from

$1300$ to make it a perfect square = 1300 - 1296 =

$4$ .

by what smallest number should $10584$ be multiplied so that product may be a perfect square ? also find the square root of the new number.

Prime factorisation of

$10584$ is

$10584={ 2 }^{ 3 }\times { 3 }^{ 2 }\times { 7 }^{ 2 }$

Now for the above to be a perfect square we can see that we need to multiply

$2\times 3=6$

$\therefore$ the least number to be multiplied to

$10584$ to make it a perfect square is

$6$

$\therefore 10584\times 6={ 2 }^{ 3 }\times { 3 }^{ 3 }\times { 7 }^{ 2 }\times 2\times 3={ 2 }^{ 4 }\times { 3 }^{ 4 }\times { 7 }^{ 2 }$

The square root of this new number is

$\sqrt { { 2 }^{ 4 }\times { 3 }^{ 4 }\times { 7 }^{ 2 } } \\ ={ 2 }^{ 2 }\times { 3 }^{ 2 }\times 7\\ =4\times 9\times 7\\ =63\times 4\\ =192$

Find the square root of the following decimal fraction :
$0.09$

$\sqrt{0.09}=\sqrt{\dfrac{9}{100}}$

$=\sqrt{\dfrac{{3}^{2}}{{10}^{2}}}$

$=\sqrt{{\left(\dfrac{3}{10}\right)}^{2}}$

$=\dfrac{3}{10}$

$=0.3$

Simplify:
$3\sqrt {45}-\sqrt {125}+\sqrt {200}-81\sqrt {30}$

$3\sqrt {45} - \sqrt {125} + \sqrt {200} - 81\sqrt {30} = 3\sqrt {9 \times 5} - \sqrt {25 \times 5} + \sqrt {100 \times 2} - \sqrt {30}$

$= 3 \times 3\sqrt 5 - 5\sqrt 5 + 10\sqrt 2 - \sqrt {30}$

$= 9\sqrt 5 - 5\sqrt 5 + 10\sqrt 2 - \sqrt {30}$

$= 4\sqrt 5 + 10\sqrt 2 - \sqrt {30}$

Simplify :-
$\sqrt { 9\times { 6 }^{ 3 }\times 24 }$

$\sqrt{9\times{6}^{3}\times 24}$

$=\sqrt{3\times 3\times{6}^{3}\times 6\times 2\times 2}$

$=\sqrt{{3}^{2}\times{6}^{4}\times {2}^{2}}$

$=3\times {6}^{2}\times 2$

$=3\times 36\times 2$

$=216‬$

$625$ is a perfect square number.
If above statement is true, type $1$ otherwise $0$ .

We are going to find prime factors of

$625$ .

$625=5\times 125$

$=5\times 5\times 25$

$=5\times 5\times 5\times 5$

By grouping the prime factors in equal pairs we get,

$=(5\times 5)(5\times 5)$

By observation, none of the prime factors are left out.

$625=5^2\times 5^2$

$\Rightarrow$

$\sqrt{625}=\sqrt{5^2\times 5^2}$

$\Rightarrow$

$\sqrt{625}=5\times 5$

$\Rightarrow$

$\sqrt{625}=25$

$\therefore$

$625$ is a perfect square.

Which of the following numbers are perfect square?
$2500$

1873016_1671268_ans_b0d443237d9541c2a73813cfad5c0f01.jpg

$576$ is a perfect square number.
If above statement is true, type $1$ otherwise $0$ .

We are going to find prime factors of

$576$ .

$576=2\times 288$

$=2\times 2\times 144$

$=2\times 2\times 2\times 72$

$=2\times 2\times 2\times 2\times 36$

$=2\times 2\times 2\times 2\times 2\times 18$

$=2\times 2\times 2\times 2\times 2\times 2\times 9$

$=2\times 2\times 2\times 2\times 2\times 2\times 3\times 3$

By grouping the prime factors in equal pairs we get,

$=(2\times 2)\times(2\times 2)\times (2\times 2)\times (3\times 3)$

By observation, none of the prime factors are left out.

$576=2^6\times 3^2$

$\Rightarrow$

$\sqrt{576}=\sqrt{2^6\times 3^2}$

$\Rightarrow$

$\sqrt{576}=2^3\times 3$

$\Rightarrow$

$\sqrt{576}=24$

$\therefore$

$576$ is a perfect square.

Find the square root of $1296$ by prime factorization method.

Hint: Do prime factorisation of the given number to find its square root

Prime factors of 1296 = 2 × 2 × 2 × 2 × 3 × 3 × 3× 3

Square root of 1296 = 2 × 2 × 3 × 3 = 36

square root of 1296 = 36

Simplify : $\sqrt{41-\sqrt{21+\sqrt{19-\sqrt{9}}}}$

$\sqrt{41-\sqrt{21+\sqrt{19-\sqrt{9}}}}$

$=\sqrt{41-\sqrt{21+\sqrt{19-3}}}$

$=\sqrt{41-\sqrt{21+\sqrt{16}}}$

$=\sqrt{41-\sqrt{21+4}}$

$=\sqrt{41-\sqrt{25}}$

$=\sqrt{41-5}$

$=\sqrt{36}$

$=6$

Evaluate: $(6+\sqrt {27})-(3+\sqrt 3)+(1-2\sqrt 3)$

1327948_1495155_ans_726603e6cbc94c8a9456b9f6d8c74019.JPG

$941$ is a perfect square number.
If above statement is true, type $1$ otherwise $0$ .

We know that

$941$ itself is a prime factor.

$\therefore$

$941$ is not a perfect square.

Find the Pythagorean triplets from among the following set of numbers.
$3, 4, 5$ .

Yes.
Explanation:
Yes. A pythagorean triple is a sequence of integer numbers that solve the Pythagora's theorem.It means that three numbers a,b,c are a pythagorean triples when

$\sqrt {a^2+b^2} =c$
or, just to remove the square root and write it in a more elegant format

$a^2+b^2=c^2 .$
With 3,4,5 we have that

$3^2+4^2=5^2 \\ 9+16=25 \\ 25=25$

$484$ is a perfect square number.
If above statement is true, type $1$ otherwise $0$ .

We are going to find prime factors of

$484$ .

$484=2\times 242$

$=2\times 2\times 121$

$=2\times 2\times 11\times 11$

By grouping the prime factors in equal pairs we get,

$=(2\times 2)(11\times 11)$

By observation, none of the prime factors are left out.

$484=2^2\times 11^2$

$\Rightarrow$

$\sqrt{484}=\sqrt{2^2\times 11^2}$

$\Rightarrow$

$\sqrt{484}=2\times 11$

$\Rightarrow$

$\sqrt{484}=22$

$\therefore$

$484$ is a perfect square.

$961$ is a perfect square number.
If above statement is true, type $1$ otherwise $0$ .

We are going to find prime factors of

$961$ .

$961=31\times 31$

By grouping the prime factors in equal pairs we get,

$=(31\times 31)$

By observation, none of the prime factors are left out.

$961=31^2$

$\Rightarrow$

$\sqrt{961}=\sqrt{31^2}$

$\Rightarrow$

$\sqrt{961}=31$

$\therefore$

$961$ is a perfect square.

Which of the following numbers are perfect square?
$11, 12, 16, 32, 36, 50, 64, 79, 81, 111, 121$

$\Rightarrow$

$11$ is a prime number by itself.

So, it is not a perfect square.

$\Rightarrow$

$12=2\times 2\times 3$

$\sqrt{12}=\sqrt{2^2\times 3}$

$\sqrt{12}=2\sqrt{3}$

$\therefore$

$12$ is not perfect square.

$\Rightarrow$

$16=2\times 2\times 2\times 2\times 2$

$16=2^4$

$\sqrt{16}=2^2=4$

$\therefore$

$16$ is a perfect square.

$\Rightarrow$

$32=2\times 2\times 2\times 2\times 2$

$32=2^5$

$\sqrt{32}=\sqrt{2^5}$

$\sqrt{32}=4\sqrt{2}$

$\therefore$

$32$ is not perfect square.

$\Rightarrow$

$36=2\times 2\times 3\times 3$

$\sqrt{36}=\sqrt{2^2\times 3^2}$

$\sqrt{36}=6$

$\therefore$

$36$ is a perfect square.

$\Rightarrow$

$50=2\times 5\times 5$

$\sqrt{50}=\sqrt{5^2\times 2}$

$\sqrt{50}=5\sqrt{2}$

$\therefore$

$50$ is not perfect square.

$\Rightarrow$

$79$ is a prime number.

So, it is not a perfect square.

$\Rightarrow$

$81=3\times 3\times 3\times 3$

$\sqrt{81}=\sqrt{3^4}$

$\sqrt{81}=9$

$\therefore$

$81$ is a perfect square.

$\Rightarrow$

$111$ is a prime number.

So, it is not a perfect square.

$\Rightarrow$

$121=11\times 11$

$\sqrt{121}=\sqrt{11^2}$

$\sqrt{121}=11$

$\therefore$

$121$ is a perfect square.

Using prime factorization method, find which of the following numbers are perfect square?
$189, 225, 2048, 343, 441, 2916, 11025, 3549$

The given numbers are

$189,\ 225,\ 2048,\ 343,\ 441,\ 2916,\ 11025,\ 3549$ .

$(1)$ Representing

$189$ as the product of its prime factors,

$189=3^2\times 3\times 7$

Since, it does not have equal pairs of factors, it is not a perfect square.

$\therefore$

$189$ is not a perfect square.

$(2)$ Representing

$225$ as the product of its prime factors,

$225=(5\times 5)\times (3\times 3)$

Since

$225$ has equal pairs of factors, it is a perfect square.

$\therefore\ 225$ is a perfect square.

$(3)$ Representing

$2048$ as the product of its prime factors,

$2048=(2\times 2)\times (2\times 2)\times (2\times 2 )\times (2\times 2)\times (2\times 2)\times 2$

Since

$2048$ does not have equal pairs of factors, it is not a perfect square.

$\therefore\ 2048$ It is a not a perfect square.

$(4)$ Representing

$343$ as the product of its prime factors,

$343=(7\times 7)\times 7$

Since

$343$ does not have equal pairs of factors, it is not a perfect square.

$\therefore \ 343$ is not a perfect square.

$(5)$ Representing

$441$ as the product of its prime factors,

$441=(7\times 7)\times (3\times 3)$

Since

$441$ has equal pairs of factors, it is a perfect square.

$\therefore \ 441$ is a perfect square.

$(6)$ Representing

$2916$ as the product of its prime factors,

$2916=(3\times 3)\times (3\times 3)\times (3\times 3)\times (2\times 2)$

Since

$2961$ has equal pairs of factors, it is a perfect square.

$\therefore \ 2916$ is a perfect square.

$(7)$ Representing

$11025$ as the product of its prime factors,

$11025=(3\times 3)\times (5\times 5)\times (7\times 7)$

Since

$11025$ has equal pairs of factors, it is a perfect square.

$\therefore\ 11025$ is a perfect square.

$(8)$ Representing

$3549$ as the product of its prime factors,

$3549=(13\times 13)\times 3\times 7$

Since

$3549$ does not have equal pairs of factors, it is not a perfect square.

$\therefore \ 3549$ It is a not a perfect square.

Hence, out of all the given numbers,

$225,\ 441,\ 2916 \ and\ 11025$ are perfect squares.

Find the number whose square is the given following number:
$4761$

The given number is

$4761$

We are going to find square root by prime factorization method:

$4761=3\times 1587$

$=3\times 3\times 529$

$=3\times 3\times 23\times 23$

$=3^2\times 23^2$

$\Rightarrow$

$4761=3^2\times 23^2$

$\Rightarrow$

$\sqrt{4761}=\sqrt{3^2\times 23^2}$

$\Rightarrow$

$\sqrt{4761}=3\times 23$

$\Rightarrow$

$\sqrt{4761}=69$

Hence,

$4761$ is a square of

$69$ .

The following number is not perfect square. Give reason.
$45743$

1873020_1671367_ans_2ceda6c0dbfa4b23b6c65effd4e2c0b1.jpg

Find the number whose square is the given number.
$1156$

We are going to find square root by using prime factorization method

$\Rightarrow$

$1156=2\times 578$

$=2\times 2\times 289$

$=2\times 2\times 17\times 17$

$=2^2\times 17^2$

$\Rightarrow$

$1156=2^2\times 17^2$

$\Rightarrow$

$\sqrt{1156}=\sqrt{2^2\times 17^2}$

$\Rightarrow$

$\sqrt{1156}=2\times 17$

$\therefore$

$\sqrt{1156}=34$

Hence,

$1156$ is the square of

$34.$

Find the number whose square is the given number.
$14641$

We are going to find square root by prime factorization method :

$14641=11\times 1331$

$=11\times 11\times 121$

$=11\times 11\times 11\times 11$

$=11^4$

$\Rightarrow$

$14641=11^4$

$\Rightarrow$

$\sqrt{14641}=\sqrt{11^4}$ (appied square root on both sides)

$\Rightarrow$

$\sqrt{14641}=11^2$

$\Rightarrow \sqrt{14641}=121$

$\therefore 14641$ is a square of

$121$

Find the greatest number of two digits which is a perfect square.

The greatest two digit number

$=99$

$9$

$9)\overline{\,\,\,\,\,99\,\,\,\,\,\,}$

$-\,\,81$

$\overline{\,\,\,\,\,\,\,\,\,\,\,\,\,\,}$

$18$

Here, we can see that

$99$ is not a perfect square as its leave

$18$ as a remainder.

So if

$18$ will be added to

$99$ than it will become a

$3$ digit number.

$\therefore$ To find the

$2$ digit perfect square we will subtract

$18$ from

$99$

$\Rightarrow$

$99-18=81$

$\therefore$ The greatest number of two digits which is a perfect square is

$81$

The following number is a perfect square. Give reason.
$1547$

1873018_1671366_ans_c11dec314a4d441c8099b7d0be6d10c6.jpg

Find the least number of three digits which is perfect square.

The smallest three digit number

$=100$

$\sqrt{100}=10$

$\therefore$ We can see

$100$ is a perfect square.

Hence, least number of three digit which is perfect square is

$100$

Find the number whose square is the given number:
$2025$

The given number is

$2025.$

We are going to find square root of the given number by prime factorisation method :

$\Rightarrow$

$2025=3\times 675$

$=3\times 3\times 225$

$=3\times 3\times 5\times 45$

$=3\times 3\times 5\times 5\times 9$

$=3\times 3\times 5\times 5\times 3\times 3$

$=3^4\times 5^2$

$\Rightarrow$

$2025=3^4\times 5^2$

$\Rightarrow$

$\sqrt{2025}=\sqrt{3^4\times 5^2}$

$\Rightarrow$

$\sqrt{2025}=3^2\times 5$

$\Rightarrow$

$\sqrt{2025}=45$

Hence,

$2025$ is the square of

$45.$

The square of which of the following numbers would be an odd number?
$3456$

1873640_1671389_ans_c59ae8f4407749cba1a706748ac3417f.jpg

$9327$

1873026_1671374_ans_9fe71e35da75439a9cb08078cd573527.jpg

The square of which of the following numbers would be an odd number?
$5559$

1873645_1671390_ans_e381734ba6744fd8885fd3e5a7fb3993.jpg

$743522$

1873039_1671378_ans_5b3057164a704fd0a82b1855a03f637c.jpg

The square of the following number would be an odd number?
$42008$

1873647_1671393_ans_1c7dc9d654f643408ddc33a1a2e19dc8.jpg

The number $1027$ is a perfect square.
Type 1 if above statement is true, otherwise 0.

The unit digit of a perfect square can be only 0, 1, 4, 5, 6 or 9.

The square of a number having:

$1.$ 1 or 9 at the units place ends in 1.

$2.$ 2 or 8 at the units place ends in 4.

$3.$ 3 or 7 at the units place ends in 9.

$4.$ 4 or 6 at the units place ends in 6.

$5.$ 5 at the units place ends in 5.

Since

$1027$ ends with

$7,$ hence, it is not a perfect square.

Check whether the following triplet is Pythagorean?
$(10, 24, 26 )$

1874079_1671436_ans_26e275d04ca6456990fedb489e9dcbc7.jpg

Check whether the following triplet is Pythagorean?
$(14, 48, 51)$

1874076_1671434_ans_37db5c8286be4385b8e0fb249fc41960.jpg

Check whether the following triplet is Pythagorean?
$(16, 63, 65 )$

1874155_1671438_ans_f00d23deea464aeb9a1ab4f783afac03.jpg

By just examining the units digits, can you tell whether the following can be whole square?
$1024$

1876530_1671472_ans_f9a599753a574c35a923c150134a053d.jpg

Find the square root of the following number by prime factorization:
$11664$

By using prime factorization method we are going to find square root of

$11664$

$11664=2\times 5832$

$=2\times 2\times 2916$

$=2\times 2\times 2\times 1458$

$=2\times 2\times 2\times 2\times 729$

$=2\times 2\times 2\times 2\times 3\times 243$

$=2\times 2\times 2\times 2\times 3\times 3\times 81$

$=2\times 2\times 2\times 2\times 3\times 3\times 3\times 27$

$=2\times 2\times 2\times 2\times 3\times 3\times 3\times 3\times 9$

$=2\times 2\times 2\times 2\times 3\times 3\times 3\times 3\times 3\times 3$

$11664=2^4\times 3^6$

$\Rightarrow$

$\sqrt{11664}=\sqrt{2^4\times 3^6}$

$\Rightarrow$

$\sqrt{11664}=2^2\times 3^3$

$\Rightarrow$

$\sqrt{11664}=108$

Find the square root of the following number by prime factorization:
$47089$

By using prime factorization method we are going to find square root of

$47089$

$47089=7\times 6727$

$=7\times 7\times 961$

$=7\times 7\times 31\times 31$

$47089=7^2\times 31^2$

$\Rightarrow$

$\sqrt{47089}=\sqrt{7^2\times 31^2}$

$\Rightarrow$

$\sqrt{47089}=7\times 31$

$\Rightarrow$

$\sqrt{47089}=217$

Find the square root of the following number by prime factorization:
$7056$

By using prime factorization method we are going to find square root of

$7056$

$7056=2\times 3528$

$=2\times 2\times 1764$

$=2\times 2 \times 2\times 882$

$=2\times 2 \times 2\times 2\times 441$

$=2\times 2 \times 2\times 2\times 3\times 147$

$=2\times 2 \times 2\times 2\times 3\times 3\times 49$

$=2\times 2 \times 2\times 2\times 3\times 3\times 7\times 7$

$7056=2^4\times 3^2\times 7^2$

$\Rightarrow$

$\sqrt{7056}=\sqrt{2^4\times 3^2\times 7^2}$

$\Rightarrow$

$\sqrt{7056}=2^2\times 3\times 7$

$\therefore$

$\sqrt{7056}=84$

Find the square root of the following number by prime factorization:
$196$

By using prime factorization method we are going to find square root of

$196$

$196=2\times 98$

$=2\times 2\times 49$

$=2\times 2\times 7\times 7$

$196=2^2\times 7^2$

$\sqrt{196}=\sqrt{2^2\times 7^2}$

$\sqrt{196}={2\times 7}$

$\therefore$

$\sqrt{196}=14$

Find the square root of the following number by prime factorization:
$8281$

By using prime factorization method we are going to find square root of

$8281$

$8281=7\times 1183$

$=7\times 7\times 169$

$=7\times 7\times 13\times 13$

$8281=7^2\times 13^2$

$\Rightarrow$

$\sqrt{8281}=\sqrt{7^2\times 13^2}$

$\Rightarrow$

$\sqrt{8281}7\times 13$

$\therefore$

$\sqrt{8281}=91$

Using the prime factorization method, find if the following number is a perfect square:
$441$

Given

$441$ ,
A perfect square can always be expressed as a product of pairs of equal factors.
Now resolve

$441$ into prime factors, we get

$441=49\times 9=7\times 7\times 3\times 3=7\times 3\times 7\times 3=21\times 21=(21)^2$
Hence,

$21$ is the number whose square is

$441$

$\therefore 441$ is a perfect square.

Give reason to show that none of the numbers given below is a perfect square:
$2500000$

Given

$2500000$
We know that property

$2$ of perfect square i.e., number ending in an add number of zeros is never a perfect square.

$2500000$ has

$5$ zero in the end.
Therefore the given number

$2500000$ is not a perfect square.

Find the square root of the following number by prime factorization:
$4096$

By using prime factorization method we are going to find square root of

$4096$

$4096=2\times 2048$

$=2\times 2\times 1024$

$=2\times 2\times 2\times 512$

$=2\times 2\times 2\times 2\times 256$

$=2\times 2\times 2\times 2\times 2\times 128$

$=2\times 2\times 2\times 2\times 2\times 2\times 64$

$=2\times 2\times 2\times 2\times 2\times 2\times 2\times 32$

$=2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 16$

$=2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 8$

$=2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 4$

$=2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2$

$\Rightarrow$

$4096=2^{12}$

$\Rightarrow$

$\sqrt{4096}=\sqrt{2^{12}}$

$\Rightarrow$

$\sqrt{4096}=2^{6}$

$\therefore$

$\sqrt{4096}=64$

Find the square root of the following number by prime factorization:
$1764$

By using prime factorization method we are going to find square root of

$1764$

$1764=2\times 882$

$=2\times 2\times 441$

$=2\times 2\times 3\times 147$

$=2\times 2\times 3\times 3\times 49$

$=2\times 2\times 3\times 3\times 7\times 7$

$1764=2^2\times 3^2\times 7^2$

$\Rightarrow$

$\sqrt{1764}=\sqrt{2^2\times 3^2\times 7^2}$

$\Rightarrow$

$\sqrt{1764}=2\times 3\times 7$

$\therefore$

$\sqrt{1764}=42$

Find the square root of the following number by prime factorization:
$1156$

By using prime factorization method we are going to find square root of

$1156$

$1156=2\times 578$

$=2\times 2\times 289$

$=2\times 2\times 17\times 17$

$\Rightarrow$

$1156=2^2\times 17^2$

$\Rightarrow$

$\sqrt{1156}=\sqrt{2^2\times 17^2}$

$\Rightarrow$

$\sqrt{1156}=2\times 17$

$\therefore$

$\sqrt{1156}=34$

Find the square root of the following number by prime factorization:
$441$

By using prime factorization method we are going to find square root of given number :

$441=3\times 147$

$=3\times 3\times 49$

$=3\times 3\times 7\times 7$

$=3^2\times 7^2$

$\sqrt{441}$

$=\sqrt{3^2\times 7^2}$

$\therefore$

$\sqrt{441}=3\times 7$

$\therefore$

$\sqrt{441}=21$

Find the square root of the following number by prime factorization:
$529$

By using prime factorization method we are going to find square root of

$529$ .

$529=23\times 23$

$529=23^2$

$\sqrt{529}=\sqrt{23^2}$

$\therefore$

$\sqrt{529}=23$

Find the least number of $4$ digits which is a perfect.

1874372_1671846_ans_8148ffa603074b25b5dec0eb0bc25b3b.jpg

Write the prime factorization of the following number and hence find its square root:
$7056$

$7056 = 2 \times 2 \times 2 \times 2 \times 3\times 3\times 7 \times 7$

So, the square root of

$7056$ is

$84$ .

Find the square root of the following by prime factorization:
$27225$

$27225 = 3 \times 3 \times 5 \times 5 \times 11 \times 11$

So, the square root of

$27225$ is

$165$

Find the square root of the following by long division method:
$97344$

1868120_1671795_ans_ce1f90278855442f8718dd860fd869cc.jpg

Find the square root of the following by long division method:
$12544$

1868128_1671793_ans_93f3ef2df56d4391bf3cd43c24a9251c.jpg

Write the prime factorization of the following number and hence find its square root:
$7744$

$7744 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 11 \times 11$

So, the square root of

$7744$ is

$88$ .

Find the square root of the following by prime factorization:
$24336$

By using prime factorization method we are going to find square root of

$24336$

$24336=2\times 12168$

$=2\times 2\times 6084$

$=2\times 2\times 2\times 3042$

$=2\times 2\times 2\times 2\times 1521$

$=2\times 2\times 2\times 2\times 3\times 507$

$=2\times 2\times 2\times 2\times 3\times 3\times 169$

$=2\times 2\times 2\times 2\times 3\times 3\times 13\times 13$

$24336=2^4\times 3^2\times 13^2$

$\Rightarrow$

$\sqrt{24336}=\sqrt{2^4\times 3^2\times 13^2}$

$\Rightarrow$

$\sqrt{24336}=2^2\times 3\times 13$

$\Rightarrow$

$\sqrt{24336}=156$

Write the prime factorization of the following number and hence find its square root:
$9604$

$9604 = 2 \times 2 \times 7 \times 7 \times 7 \times 7$

So, the square root of

$9604$ is

$98$ .

Using the prime factorization method, find if the following number is a perfect square:
$9075$

Given

$9075$ ,

A perfect square can always be expressed as a product of pairs of equal factors.

Now resolve

$9075$ into prime factors, we get

$9075=25\times 363$

Again resolve

$363$ into prime factors we get,

$9075=5\times 5\times 11\times 11\times 3=5\times 11\times 5\times 11\times 3=55\times 55\times 3$

In prime factorization of

$9075,\ 3$ is present only once.

$\therefore 9075$ is not a perfect square.

Using the prime factorization method, find if the following number is a perfect square:
$1176$

Given

$1176$ ,
A perfect square can always be expressed as a product of pairs of equal factors.
Now resolve

$1176$ into prime factors, we get

$11767\times 7\times 3\times 2\times 2\times 2$

there is odd numbers of

$2\ and\ 3$ in prime factorization of

$1176$
Hence,

$1176$ is not a perfect square.

Which of the following squares are squares of even number?
$900$

Given

$900$
We know that from the property of the perfect square, the square of an even number is always even.
So the given number

$900$ is even, therefore it is square of even number.

Which of the following squares are squares of even number?
$324$

Given

$324$
We know that from the property of perfect square, the square of an even number is always even.
So the given number

$324$ is even, therefore it is square of even number.

Using the prime factorization method, find if the following number is a perfect square:
$11025$

Given

$11025$ ,
A perfect square can always be expressed as a product of pairs of equal factors.
Now resolve

$11025$ into prime factors, we get

$11025=7\times 7\times3\times 3\times 5\times 5$

$=7\times 3\times 5\times 7\times 3\times 5=105\times 105=(105)^2$
Hence,

$105$ is the number whose square is

$11025$

$\therefore 11025$ is a perfect square.

Using the prime factorization method, find if the following number is a perfect square:
$4225$

Given

$4225$ ,
A perfect square can always be expressed as a product of pairs of equal factors.
Now resolve

$4225$ into prime factors, we get

$4225=5\times 5\times 13\times 13=5\times 13\times 5\times 13$

$=65\times 65=(65)^2$
Hence,

$65$ is the number whose square is

$4225$

$\therefore 4225$ is a perfect square.

Using the prime factorization method, find if the following number is a perfect square:
$5625$

Given

$5625$ ,
A perfect square can always be expressed as a product of pairs of equal factors.
Now resolve

$5625$ into prime factors, we get

$5625=3\times 3\times 5\times 5\times 5\times 5=3\times 5\times 5\times 3\times 5\times 5$

$=75\times 75=(75)^2$
Hence,

$755$ is the number whose square is

$5625$

$\therefore 5625$ is a perfect square.

Which of the following are squares of even number?
$441$

Given

$441$

We know that from the property of perfect square, the square of an even number is always even.

But the given number

$441$ is odd, therefore it is not a square of even number.

Using the prime factorization method, find if the following number is a perfect square:
$576$

Given

$576$ ,
A perfect square can always be expressed as a product of pairs of equal factors.
Now resolve

$576$ into prime factors, we get

$576=2\times 2\times 2\times 3\times 2\times 2\times 2\times 3=24\times 24=(24)^2$
Hence,

$24$ is the number whose square is

$576$

$\therefore 576$ is a perfect square.

Which of the following are squares of even number?
$625$

Given

$625$
We know that from the property of perfect square, the square of an even number is always even.
But the given number

$625$ is odd, therefore it is not a square of even number.

Using the prime factorization method, find if the following number is a perfect square:
$1089$

Given

$1089$ ,
A perfect square can always be expressed as a product of pairs of equal factors.
Now resolve

$1089$ into prime factors, we get

$1089=3\times 3\times 11\times 11=3\times 11\times 3\times 11$

$=33\times 33=(33)^2$
Hence,

$33$ is the number whose square is

$4225$

$\therefore 1089$ is a perfect square.

Find the square root of each of the following numbers by using the method of prime factorization:
$9216$

Given

$9216$

A square can always be expressed as a product of pairs of equal factors

Resolve the given number into prime factors

We get

$9216=2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 3\times 3$

$\surd 9216=2\times 2\times 2\times 2\times 2\times 3=96$

Find the square root of each of the following numbers by using the method of prime factorization:
$225$

Given

$225$
A square can always be expressed as a product of pairs of equal factors
Resolve the given number into prime factors
We get

$225=3\times 3\times 5\times 5$

$\surd 225=3\times 5=15$

Find the square root of each of the following numbers by using the method of prime factorization:
$1296$

Given

$1296$
A square can always be expressed as a product of pairs of equal factors
Resolve the given number into prime factors
We get

$1296=2\times 2\times 2\times 2\times 3\times 3\times 3\times 3$

$\surd 1296=2\times 2\times 3\times 3=36$

Find the square root of each of the following numbers by using the method of prime factorization:
$441$

Given

$441$
A square can always be expressed as a product of pairs of equal factors
Resolve the given number into prime factors
We get

$441=3\times 3\times 7\times 7$

$\surd 441=3\times 7=21$

By what least number should the given number be multiplied to get a perfect square number ? Also find the number, whose square is the new number.
$9075$

Given number is

$9075$ ,

Prime factorising

$9075$ , we get

$9075=3\times 5\times 5\times 11\times 11=(3 \times 5^2 \times 11^2)$

Here, the factor

$3$ doesn't have a pair.

Hence, to get a perfect square we have to multiply the number by

$3$

Then we get,

$9075\times 3=3\times 3\times 5\times 5\times 11\times 11$

$=(3^2\times 5^2\times 11^2)$

Square root of the new number

$=\sqrt{3^2\times 5^2\times 11^2}$

$=3\times5\times11$

$=165$

Hence, the new number is square of

$165$ .

By what least number should the given number be multiplied to get a perfect square number ? In the case, find the number whose square is the new number.
$2925$

Given

$2925$ ,
Resolve

$2925$ into prime factors, we get

$2925=3\times 3\times 5\times 5\times 13=(3^2 \times 5^2 \times 13)$
Now to get perfect square we have to multiply the above equation by

$13$
Then we get,

$2156=3\times 3\times 5\times 5\times 13\times 13$
New number

$=(9\times 25\times 169)$

$=(3^2\times 5^2\times 13^2)$
Taking squares as common from the above equation we get

$\therefore$ New number

$=(3\times 5\times 13)^2$

$=(195)^2$
Hence, the new number is square of

$195$

By what least number should the given number be multiplied to get a perfect square number ? In the case, find the number whose square is the new number.
$3332$

Given

$3332$ ,
Resolve

$3332$ into prime factors, we get

$3332=2\times 2\times 7\times 7\times 17=(2^2 \times 7^2 \times 17)$
Now to get perfect square we have to multiply the above equation by

$17$
Then we get,

$3332=2\times 2\times 7\times 7\times 17\times 17$
New number

$=(4\times 49\times 289)$

$=(2^2\times 7^2\times 17^2)$
Taking squares as common from the above equation we get

$\therefore$ New number

$=(2\times 7\times 17)^2$

$=(238)^2$
Hence, the new number is square of

$238$

By what least number should the given number be multiplied to get a perfect square number ? In the case, find the number whose square is the new number.
$2156$

Given

$2156$ ,
Resolve

$2156$ into prime factors, we get

$2156=2\times 2\times 7\times 7\times 11=(2^2 \times 7^2 \times 11)$
Now to get perfect square we have to multiply the above equation by

$11$
Then we get,

$2156=2\times 2\times 7\times 7\times 11\times 11$
New number

$=(4\times 49\times 121)$

$=(2^2\times 7^2\times 11^2)$
Taking squares as common from the above equation we get

$\therefore$ New number

$=(2\times 7\times 11)^2$

$=(154)^2$
Hence, the new number is square of

$154$

Find the smallest number by which $252$ must be multiplied to get a perfect square. Also, find the square root of the perfect square so obtained.

Given number

$=252$

Resolve the given number into prime factors we get

$252=2\times 2\times 3\times 3\times 7$

To get perfect square we have to multiply by

$7$

Then we get

$252=2\times 2\times 3\times 3\times 7\times 7$

New number

$=252\times 7=1764$

Therefore

$\surd 1764=2\times 3\times 7=42$

By what least number should the given number be multiplied to get a perfect square number ? In the case, find the number whose square is the new number.
$3675$

Given

$3675$ ,
Resolve

$3675$ into prime factors, we get

$3675=3\times 5\times 5\times 7\times 7$
Now to get perfect square we have to multiply the above equation by

$3$
Then we get,

$3675=3\times 3\times 5\times 5\times 7\times 7=(3\times 5^2\times 7^2)$
New number

$=(9\times 25\times 49)$

$=(3^2\times 5^2\times 7^2)$
Taking squares as common from the above equation we get

$\therefore$ New number

$=(3\times 5\times 7)^2$

$=(105)^2$
Hence, the new number is square of

$105$

Find the square root of each of the following numbers by using the method of prime factorization:
$11025$

Given

$11025$

A square can always be expressed as a product of pairs of equal factors

Resolve the given number into prime factors

We get

$11025=3\times 3\times 5\times 5\times 7\times 7$

$\surd 11025=3\times 5\times 7=105$

Find the square root of each of the following numbers by using the method of prime factorization:
$17424$

Given

$17424$
A square can always be expressed as a product of pairs of equal factors
Resolve the given number into prime factors
We get

$17424=2\times 2\times 2\times 2\times 3\times 3\times 11\times 11$

$\surd 17424=2\times 2\times 3\times 11=132$

Write a Pythagorean triplet whose smallest number is
$6$

Given

$6$ is the smallest number in Pythagorean triplet

But we have for every number

$m > 1$ , we have

$(2m, m^2-1, m^2+1)$ as a Pythagorean triplet

$2m=6$

$m=6/2=3$

Then

$m^2=9$

$m^2-1=9-1=8$

$m^2+1=9+1=10$

Therefore the Pythagorean triple is

$=(2m,m^2-1, m^2+1)=(6,8,10)$

Find the square root of each of the following numbers by using the method of prime factorization:
$15876$

Given

$15876$

A square can always be expressed as a product of pairs of equal factors

Resolve the given number into prime factors

We get

$15876=2\times 2\times 3\times 3\times 3\times 3\times 7\times 7$

$\surd 15876=2\times 3\times 3\times 7=126$

By what least number should the given number be divided to get a perfect square number?
In the case, find the number of square is the new number
$9075$

Given

$9075$ ,

Resolve

$9075$ into prime factors, we get

$9075=3\times 5\times 5\times 11\times 11=(3 \times 5^2 \times 11^2)$

Now to get perfect square we have to divide the above equation by

$3$

Then we get,

$\dfrac{9075}{3}=5\times 5\times 11\times 11$

New number

$=(25\times 121)$

$=(5^2\times 11^2)$

Taking squares as common from the above equation we get

$\therefore$ New number

$=(5\times 11)^2$

$=(55)^2$

Hence, the new number is square of

$55$

By what least number should the given number be divided to get a perfect square number?
In the case, find the number of square is the new number
$3380$

Given

$3380$ ,

Resolve

$3380$ into prime factors, we get

$3380=2\times 2\times 5\times 13\times 13=(2^2 \times 5 \times 13^2)$

Now to get perfect square we have to divide the above equation by

$5$

Then we get,

$3380/5=2\times 2\times 13\times 13$

New number

$=(4\times 169)$

$=(2^2\times 13^2)$

Taking squares as common from the above equation we get

$\therefore$ New number

$=(2\times 13)^2$

$=(26)^2$

Hence, the new number is square of

$26$

By what least number should the given number be divided to get a perfect square number?
In the case, find the number of square is the new number
$4851$

Given

$4851$ ,

Resolve

$4851$ into prime factors, we get

$4851=3\times 3\times 7\times 7\times 11=(3^2 \times 7^2 \times 11)$

Now to get perfect square we have to divide the above equation by

$11$

Then we get,

$\dfrac{4851}{11}=3\times 3\times 7\times 7$

New number

$=(9\times 49)$

$=(3^2\times 7^2)$

Taking squares as common from the above equation we get

$\therefore$ New number

$=(3\times 7)^2$

$=(21)^2$

Hence, the new number is square of

$21$

Write a Pythagorean triplet whose smallest number is
$16$

Given

$16$ is the smallest number in Pythagorean triplet

But we have for every number

$m > 1$ , we have

$(2m, m^2-1, m^2+1)$ as a Pythagorean triplet

$2m=16$

$m=16/2=8$

Then

$m^2=64$

$m^2-1=64-1=63$

$m^2+1=64+1=65$

Therefore the Pythagorean triple is

$=(2m,m^2-1, m^2+1)=(16,63,65)$

By what least number should $2475$ be multiplied to get a perfect square number ? In the case, find the number whose square is the new number.

Given

$2475$ ,

Resolve

$2475$ into prime factors, we get

$2475=3\times 3\times 5\times 5\times 11$

$=(3^2 \times 5^2 \times 11)$

Now to get perfect square we have to multiply the above equation by

$11$

Then we get,

$2475=3\times 3\times 5\times 5\times 11\times 11$

New number

$=(9\times 25\times 121)$

$=(3^2\times 5^2\times 11^2)$

Taking squares as common from the above equation we get

$\therefore$ New number

$=(3\times 5\times 11)^2$

$=(165)^2$

Hence, the new number is square of

$165.$

By what least number should the given number be multiplied to get a perfect square number? Also, find the number whose square is the new number.
$7623$

Given:

$7623$ ,

Resolving

$7623$ into prime factors, we get

$7623=3\times 3\times 7\times 11\times 11=(3^2 \times 11^2 \times 7)$

Now to get a perfect square we have to multiply the above equation by

$7$ .

Then we get,

New number

$=3\times 3\times 11\times 11\times 7\times 7=(9\times 49\times 121)$

$=(3^2\times 7^2\times 11^2)$

Taking squares as common from the above equation we get,

$\therefore$ New number

$=(3\times 7\times 11)^2$

$=(231)^2$

Hence, the new number is the square of

$231.$

By what least number should the given number be divided to get a perfect square number?
In the case, find the number of square is the new number
$1575$

Given

$1575$ ,

Resolve

$1575$ into prime factors, we get

$1575=3\times 3\times 5\times 5\times 7=(3^2 \times 5^2 \times 7)$

Now to get perfect square we have to divide the above equation by

$7$

Then we get,

$1575/7=3\times 3\times 5\times 5$

New number

$=(9\times 25)$

$=(3^2\times 5^2)$

Taking squares as common from the above equation we get

$\therefore$ New number

$=(3\times 5)^2$

$=(15)^2$

Hence, the new number is square of

$15$

By what least number should the given number be multiplied to get a perfect square number ? In the case, find the number whose square is the new number.
$3380$

Given

$3380$ ,

Resolve

$3380$ into prime factors, we get

$3380=2\times 2\times 5\times 13\times 13=(2^2 \times 13^2 \times 5)$

Now to get perfect square we have to multiply the above equation by

$5$

Then we get,

$3380=2\times 2\times 5\times 5\times 13\times 13$

New number

$=(4\times 25\times 169)$

$=(2^2\times 5^2\times 13^2)$

Taking squares as common from the above equation we get

$\therefore$ New number

$=(2\times 5\times 13)^2$

$=(130)^2$

Hence, the new number is square of

$130$

Evaluate:
$\surd 6241$

Given

$\surd 6241$ , using long division method we get
Therefore

$\surd 6241=79$
1620750_1771878_ans_8003c459e1d5483cbbd018d0830aa490.PNG

Evaluate:
$\sqrt{ 1444}$

Given

$\surd 1444$ , using long division method we get

Therefore

$\surd 1444=38$

1620723_1771873_ans_8600df74e811438d89a59d98ba44d362.PNG

Write a Pythagorean triplet whose number is
$15$

$15$
Take

$2n=15$
So

$n=n/2$ is not possible

$n^2-1=15$
By further calculation

$n^2=15+1=16=4^2$

$n=4$
So we get

$2n=2\times 4=8$

$n^2-1=15$

$n^2+1=4^2+1=16+1=17$
Therefore, the triplets are

$8, 15$ and

$17$ .

By what least number should the given number be divided to get a perfect square number?
In the case, find the number of square is the new number
$7776$

Given

$7776$ ,

Resolve

$7776$ into prime factors, we get

$7776=2\times 2\times 2\times 2\times 2\times 3\times 3\times 3\times 3\times 3=(2^2 \times 2^2 \times 3^2\times 3^2\times 2\times 3)$

Now to get perfect square we have to divide the above equation by

$2$ x

$3$

Then we get,

$7776/6=2\times 2\times 2\times 2\times 3\times 3\times 3\times 3$

New number

$=(4\times 4\times 9\times 9)$

$=(2^2\times 2^2\times 3^2\times 3^2)$

Taking squares as common from the above equation we get

$\therefore$ New number

$=(2\times 2\times 3\times 3)^2$

$=(36)^2$

Hence, the new number is square of

$36$

By what least number should the given number be divided to get a perfect square number?
In the case, find the number of square is the new number
$4056$

Given

$4056$ ,

Resolve

$4056$ into prime factors, we get

$4056=2\times 2\times 2\times 3\times 13\times 13=(2^2 \times 13^2 \times 3\times 2)$

Now to get perfect square we have to divide the above equation by

$3$ x

$2$

Then we get,

$4056/6=2\times 2\times 13\times 13$

New number

$=(4\times 169)$

$=(2^2\times 13^2)$

Taking squares as common from the above equation we get

$\therefore$ New number

$=(2\times 2\times 13\times 13)^2$

$=(26)^2$

Hence, the new number is square of

$26$

By what least number should the given number be divided to get a perfect square number?
In that case, find the new number which is a square number.
$8820$

Given

$8820$ ,

Resolve

$8820$ into prime factors, we get

$8820=2\times 2\times 3\times 3\times 5\times 7\times 7=(2^2 \times 3^2 \times 7^2\times 5)$

Now to get perfect square we have to divide the above equation by

$5$ as rest powers of all prime numbers are even, the resulting number will be a square number.

Then we get,

$\dfrac{8820}{5}=2\times 2\times 3\times 3\times 7\times 7=1764$

By what least number should the given number be divided to get a perfect square number?
In the case, find the number of square is the new number
$4500$

Given

$4500$ ,

Resolve

$4500$ into prime factors, we get

$4500=2\times 2\times 3\times 3\times 5\times 5\times 5=(2^2 \times 3^2 \times 5^2\times 5)$

Now to get perfect square we have to divide the above equation by

$5$

Then we get,

$4500/5=2\times 2\times 3\times 3\times 5\times 5$

New number

$=(4\times 9\times 25)$

$=(2^2\times 3^2\times 5^2)$

Taking squares as common from the above equation we get

$\therefore$ New number

$=(2\times 3\times 5)^2$

$=(30)^2$

Hence, the new number is square of

$30$

Evaluate:
$\surd 4489$

Given

$\surd 4489$ , using long division method we get
Therefore

$\surd 4489=67$
1620730_1771874_ans_eaf00ade016a4033be88b94321e5c6b4.png

Evaluate:
$\surd 10404$

Given

$\surd 10404$

$10404=2\times 2\times 3\times 3\times 17\times 17$

$10404=2^2\times 3^2\times 17^2$

$\surd 10404=2\times 3\times 17$
Therefore

$\surd 10404=102$

Evaluate: $\sqrt {9025}$

$Ref$ Image

Given:

$\sqrt{ 9025}$ ,

Using the long division method we get the square root of

$9025=95$
Therefore

$\sqrt{ 9025}=95$

1620770_1771882_ans_46fddf5b6ed64f18b9700767aecf7d82.png

Evaluate:
$\surd 7056$

Given

$\surd 7056$ , using long division method we get
Therefore

$\surd 7056=84$
1620762_1771880_ans_fe6347cb0c9542da8463c895e31e395c.PNG

Evaluate:
$\sqrt {14161}$

We need to find the square root of

$14161$ .

The number can be represented as the product of its prime factors as:

$14161=7\times 7\times 17\times 17$

$\Rightarrow 14161=7^2\times 17^2$

$\Rightarrow 14161=(7\times 17)^2$

Hence,

$\sqrt {14161}=7\times 17$

$\Rightarrow \sqrt {14161}=119$

Hence,

$\sqrt {14161}=119$ .

Fill in the blanks to make the statements true.
There are ________ perfect squares between 1 and 100.

$\sqrt{1} = 1$ since

$1^2 = 1$

$\sqrt{4} = 2$ since

$2^2 = 4$

$\sqrt{9} = 3$ since

$3^2 = 9$

$\sqrt{16} = 4$ since

$4^2 = 16$

$\sqrt{25} = 5$ since

$5^2 = 25$

$\sqrt{36} = 6$ since

$6^2 = 36$

$\sqrt{49} = 7$ since

$7^2 = 49$

$\sqrt{64} = 8$ since

$8^2 = 64$

$\sqrt{81} = 9$ since

$9^2 = 81$

$\sqrt{100} = 10$ since

$10^2 = 100$

Between

$1$ and

$100$ is

$4,9,16,25,36,49,64$ and

$81$
There are

$\underline{8}$ perfect squares between

$1$ and

$100$ .

The unit digit in the square of $1294$ is _______.

The unit digit of

$1294$ is

$4.$

And,

$4^2=16$

Hence, unit digit of square of

$1294$ will be

$6.$

Fill in the blanks:
The square of an even number is.......... .

Even
According to the property of perfect squares, the square of an Even number is always even.

As an even number must be of form

$2n$

$(2n)^2=4n^2\\=2(2n^2)\\=Even$

Fill in the blanks:
The square of an odd number is.......... .

Odd
According to the property of perfect square, i.e., the square of an Odd number is always odd.

As an odd number must be of form

$2n+1$

$(2n+1)^2=4n^2+4n+1=2(2n^2+2n)+1=2k+1=ODD$

Using prime factorisation, find the square roots of
a) 11025
b) 4761

a) Factors of

$11025$

$11025 = \underline{3\times 3}\times \underline{5 \times 5} \times \underline{7 \times 7}$

$\sqrt{11025} = 3 \times 5 \times 7$

$= 105$

Therefore, square root of

$11025 = 105$

b) Factors of

$4761$

$\sqrt{4761} = 3\times 3\times 23\times 23$

$\sqrt{4761} = 3\times 23 = 69$

Therefore, square root of

$4761 = 69$

Find the square root of the following by long division method.
a) 1369
b) 5625

Show that $500$ is not a perfect square.

The factors of

$500$ are,

$\begin{aligned}{}500 &= 2 \times 2 \times 5 \times 5 \times 5\\& = {2^2} \times {5^2} \times 5\\\sqrt {500} & = 2 \times 5\sqrt 5 \\& = 10\sqrt 5 \end{aligned}$

As one

$5$ is left in the root,

$500$ is not a perfect square.

Write the Pythagorean triplet whose one of the numbers is 4.

For any natural number greater than

$1, (2m, m^2-1, m^2+1)$ is Pythagorean triplets.
So, if one number is 2m, then another two numbers will be

$m^2-1$ and

$m^2+1$ .
Given, one number = 4
Then Pythagorean triplets:
2m=4 or m = 2
So,

$m^2 - 1 = (2)^2 -1 = 4 - 1= 3$

$m^2 + 1 = (2)^2 +1 = 4 + 1= 5$
Now,

$(3)^2 + (4)^2 = (5)^2$
Or 9 + 16 = 25

By what smallest number should 216 be divided so that the quotient is a perfect square? Also find the square root of the quotient.

$216 = \underline{2\times 2}\times2 \times \underline{3 \times 3} \times 3$
Here, pair of 2 and 3 is not completed.
So, 216 is not a Perfect Square.
The number should be divided by

$2 \times 3$ to make it a perfect square.

$216 \div 6 = 36$

Factors of 36 =

$2 \times 2 \times 3 \times 3$ which is the perfect square of

$2\times3 = 6$ .
Now, 36 is a perfect square. Hence, the smallest number 6 should divide 216 so that the quotient is a perfect square.

1653233_1792387_ans_2f6e4af0dc594d2fa8ffc4ff30ae659f.PNG

Are 176 a perfect square? If not, find find the smallest number by which it should be multiplied to get a perfect square.

What is the least number that should be subtracted from 1385 to get a perfect square? Also find the square root of the perfect square.

The square root of the perfect square:
The least number that should be subtracted from 1385 to get a perfect square is 16.
The required perfect square number = 1385 - 16 = 1369
So,

$\sqrt{1369} = \sqrt{37 \times 37} = 37$
1653244_1792450_ans_b04ce1c3130a421faaf4f2f4a67b37b0.PNG

A perfect square number has four digits, none of which is zero. The digits from left to right have values that are: even, even. odd, even. Find the number.

The number that satisfies the given condition is

$8836$ and is a perfect square of

$94.$

Find the greatest number of three digits which is a perfect square.

The greatest number of three digits = 999
Now, square root of 999
To get a perfect square, you have to subtract 38 from 999.
Therefore, 999 -38 = 961
Square root of 961= 31
1650205_1792490_ans_bbcce4f84c844a28955e3d4c3324517e.PNG

1650205_1792490_ans_bbcce4f84c844a28955e3d4c3324517e.PNG

Find the decimal fraction which when multiplied by itself $84.64$ .

Given, the product is

$84.64$
Let the number is

$z$ .
Now, the product =

$z \times z = (z)^2$
So,

$(z)^2 = 84.64$
or

$z = \sqrt{84.64}$
Now, we calculate the square root of

$84.64$

$\therefore \ \ \sqrt{84.64} = 9.2$
The required decimal fraction is

$9.2$

1650169_1792522_ans_2c7c32efd771496fa2c764e336925055.PNG

Find the least number of four digits that is a perfect square.

The least number of four digits = 1000
By long division method, square root of 1000
So, the least number of four digits is 1000.
i.e. a perfect square is 1000 + 24 = 1024.
1650212_1792481_ans_f254ce3ee17649ffa20067207480073e.PNG

1650212_1792481_ans_f254ce3ee17649ffa20067207480073e.PNG

Find the least square number which is exactly divisible by 3, 4, 5, 6 and 8.

The least square number which is exactly divisible by

$3,\ 4,\ 5,\ 6$ and

$8 =$ L.C.M of

$3,\ 4,\ 5,\ 6$ and

$8$

L.C.M of

$3,\ 4,\ 5,\ 6$ and

$8 = \underline{2 \times 2} \times 2 \times 3 \times 5 = 120$

Pair of

$2,\ 3$ and

$5$ is not completed. To make it a perfect square, the number should be multiplied with

$2 \times 3 \times 5 = 50$

Now,

$120 \times 50 = 3600$

The least square number which is exactly divisible by

$3,\ 4,\ 5,\ 6$ and

$8$ is

$3600.$

1650195_1792496_ans_297a1646ffab445589072758dc2e7764.png

A decimal number is multiplied by itself. If the product is 51.84, find the number.

Given, the product is 51.84
Let the number is z.
Now, the product =

$z \times z = (z)^2$
So,

$(z)^2 = 51.64$
or

$z = \sqrt{51.84}$
Now, we calculate the square root of 51.84

$\therefore \ \ \sqrt{51.84} = 7.2$
The required number is 7.2
1649617_1792510_ans_ead1f0c52704485fa6f1caa2ef8aa812.PNG

What is the least number that should be added to 6200 to make it a perfect square?

Given, the number = 6200
Hence,

$(78)^2 = 6084 < 6200$
Subsequently, perfect square =

$(79)^2 = 6241$
So, the least number is (6241 - 6200 = 41). This number should be added to 6200 to get a perfect square. Also, the perfect square number = 6241
Therefore,

$\sqrt{6241} = \sqrt{41 \times 41}= 41$
1653250_1792463_ans_1493450902234c3c991641a54be611c9.PNG

Find the smallest square number divisible by each one of the numbers 8, 9 and 10.

The least square number which is exactly divisible by 8, 9, and 10 = L.C.M. of 8, 9, and 10.
Hence,L.C.M =

$2 \times 2 \times 2 \times 3 \times 3 \times 5 = 360$
As we calculate,

$2 \times 2 \times 2 \times 3 \times 3 \times 5$
2 and 5 have incomplete pairs.
So, to make if perfect square, we have to multiple it by
2

$\times$ 5 = 10
Now, 360

$\times$ 10 = 3600
Hence, the answer is 3600.

What will be the unit digit of the squares of the following numbers?
$951$

$951$
The unit digit of the square is

$1$ .

What will be the unit digit of the squares of the following numbers?
$7625$

$7625$
The unit digit of the square is

$5$ .

The following numbers are obviously not perfect. Give reason.
$567$

In the given number
If the square of a number does not have

$2, 3, 7, 8$ or

$0$ as its unit digit, the square

$567, 2453, 5208, 46292$ and

$74000$ cannot be the perfect squares as they have

$7, 2, 8, 2$ digits at the unit place.

What will be the unit digit of the squares of the following numbers?
$5124$

$5124$
The unit digit of the square is

$6$ .

What will be the unit digit of the squares of the following numbers?
$502$

$502$
The unit digit of the square is

$4$ .

What will be the unit digit of the squares of the following numbers?
$329$

$329$
The unit digit of the square is

$1$ .

What will be the unit digit of the squares of the following numbers?
$12796$

$12796$
The unit digit of the square is

$6$ .

What will be the unit digit of the squares of the following numbers?
$95628$

$95628$
The unit digit of the square is

$4$ .

What will be the unit digit of the squares of the following numbers?
$99880$

$99880$
The unit digit of the square is

$0$ .

What will be the unit digit of the squares of the following numbers?
$68327$

$68327$
The unit digit of the square is

$9$ .

What will be the unit digit of the squares of the following numbers?
$643$

$643$
The unit digit of the square is

$9$ .

Find the square root of each of the following by division method:
$2401$

$\surd{2401}=49$
By division method
1702532_1806446_ans_a6a3b1d9e03a45ab9f08f1d62098de34.png

The following numbers are obviously not perfect. Give reason.
$74000$

In the given number
If the square of a number does not have

$2, 3, 7, 8$ or

$0$ as its unit digit, the square

$567, 2453, 5208, 46292$ and

$74000$ cannot be the perfect squares as they have

$7, 2, 8, 2$ digits at the unit place.

Find the square root of each of the following by division method:
$4489$

$\surd{4489}=67$
By division method
1702480_1806449_ans_05607cc9a0f4409bac1571cf035ec899.png

The following numbers are obviously not perfect. Give reason.
$2453$

In the given number
If the square of a number does not have

$2, 3, 7, 8$ or

$0$ as its unit digit, the square

$567, 2453, 5208, 46292$ and

$74000$ cannot be the perfect squares as they have

$7, 2, 8, 2$ digits at the unit place.

The square of which of the following numbers would be an odd number or an even number? Why?
$8267$

We know that
The square of an odd number is odd and a square of an even number is even.
So

$573$ and

$8267$ are odd numbers and their squares will be an odd number.

$4096$ and

$37916$ are even numbers and their square will also be even number.

The following numbers are obviously not perfect. Give reason.
$46292$

In the given number
If the square of a number does not have

$2, 3, 7, 8$ or

$0$ as its unit digit, the square

$567, 2453, 5208, 46292$ and

$74000$ cannot be the perfect squares as they have

$7, 2, 8, 2$ digits at the unit place.

The square of which of the following numbers would be an odd number or an even number? Why?
$37916$

We know that
The square of an odd number is odd and a square of an even number is even.
So

$573$ and

$8267$ are odd numbers and their squares will be an odd number.

$4096$ and

$37916$ are even numbers and their square will also be even number.

The square of the following number would be an odd number or an even number? Why?
$573$

We know that,
The square of an odd number is odd and square of an even number is even.

The given number

$573$ is an odd number, so its square will be an odd number.

Hence, the square of

$573$ is an odd number.

The square of which of the following numbers would be an odd number or an even number? Why?
$4096$

We know that
The square of an odd number is odd and a square of an even number is even.
So

$573$ and

$8267$ are odd numbers and their squares will be an odd number.

$4096$ and

$37916$ are even numbers and their square will also be even number.

Find the smallest four- digit number which is a perfect square.

It is given that
Smallest four-digit number

$=1000$
We know that
By taking square root, we find that

$39$ is left.
If we subtract any number from

$1000$ we get

$3$ digit number
Take

$32^{2}=1024$
Here

$1024-1000=24$ is to be added to get a perfect square of least

$4$ digit number
Therefore, the required

$4$ digit smallest number is

$1024$ .

1704288_1806551_ans_c248837f0dc64e2b88c68f0dc85f5afb.png

Find the least number which must be added to each of the following numbers to make them a perfect square. Also, find the square root of the perfect square number so obtained:
$8000$

$8000$
We know that
By taking square root

$89^{2}$ is less than

$8000$
So by taking

$90^{2}$

$8100-8000=100$ less
Adding

$100$ we get a square of

$90$ which is

$8100$

1704267_1806549_ans_2b355a8b4a8b44b0afd59c3bc479a6b9.png

Find the square root of the following numbers correct to two decimal places:
$2$

$\surd 2=1.41$
It can be written as
1703461_1806502_ans_b85f512b995245819f74749d78c815ba.png

Find the square root of the following numbers by prime factorization method:
$17 \dfrac{13}{ 36}$

$17 \dfrac{13}{ 36} = (17 \times 36 +13) / 36$
By further calculation

$=(612+13) / 36$

$=625/36$
By squaring we get

$\sqrt{\dfrac{625}{36}} = \dfrac{\sqrt{625}}{\sqrt{36}}$
We know that
So we get

$=(5 \times 5)/(2 \times 3)$

$=25/6$

$=4\dfrac{1}{6}$
1705585_1806558_ans_aaf0287774704b0f9a4d56a5a58f20b5.PNG

Find the square root of the following numbers by prime factorization method:
$1.96$

$1.96 = 196/100$
By squaring we get

$\sqrt{\dfrac{196}{100}} = \dfrac{\sqrt{196}}{\sqrt{100}}$
We know that
So we get

$=(2 \times 7)/(2 \times 5)$

$=14/10$

$=1.4$

1705596_1806559_ans_ad1c4c7dea674098908e27011431efef.PNG

Find the square root of the following numbers by prime factorization method:
$9\dfrac{67}{121}$

$9\dfrac{67}{121} = (9 \times 121 +67) / 121$
By further calculation

$(1089+67) / 121$

$=1156/121$
By squaring we get

$\sqrt{\dfrac{1156}{121}} = \dfrac{\sqrt{2\times2\times17\times17}}{\sqrt{11\times11}}$
We know that
So we get

$=(2 \times 17)/11$

$=34/11$

$=31/11$

Find the square root of the following numbers by prime factorization method:
$8836$

We know that
square root of

$6241$
It can be written as

$\sqrt{8836}=\sqrt{2 \times 2 \times 47 \times 47 }$
So we get

$2 \times 47$

$=94$

1704239_1806545_ans_9e5947150ede411e909033dc0ce418a3.png

In a right triangle $ABC, \angle B=90^{o}$
If $AC=37\ cm, BC=35\ cm$ , find $AB$

Given that,

In right triangle

$ABC$

$\angle B=90^{o}, \ AC=37\ cm, \ BC=35\ cm$
Using Pythagoras theorem

$AC^{2}=AB^{2}+BC^{2}$
Substituting the values

$37^{2}=AB^{2}+35^2$

$1369=AB^{2}+1225$

$AB^{2}=1369-1225$

$AB^{2}=144$

$AB=\sqrt{144}$

$AB=\pm 12$

$AB$ is the side of a triangle, which cannot be negative.

Hence,

$AB=12\ cm$ .

1705617_1806567_ans_09ff26461366454697c5f2b7c405e7d6.PNG

Show that each of the following numbers is a perfect square. Also, find the number whose square is the given number.
$3025$

We know that
It can be written as

$3025=5 \times 5 \times 11 \times 11$
Here
After pairing the same prime factors, no factor is left
Therefore,

$3025$ is a perfect square of

$5 \times 11 =55$
1705774_1806685_ans_06822d77723946ed98df1ce489e7c01d.PNG

Which of the following natural numbers are perfect squares? Give reasons in support of your answer.
$1024$

We know that
It can be written as

$1024=2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2$
Here
After pairing the same prime factors, there is no factor left
Therefore,

$1024$ is a perfect square.
1705747_1806673_ans_3395d11afd9f47a7b00c74ab47aaecdf.PNG

check if the following natural number is a perfect square? Give reasons in support of your answer.
$729$

We know that
It can be written as

$729=3 \times 3 \times 3 \times 3 \times 3 \times 3$
Here

$729$ is the product of pair of equal prime factors
Therefore,

$729$ is a perfect square.
1705735_1806670_ans_432c34c88f8f45a4952ed557d48f33df.png

Show that each of the following numbers is a perfect square. Also, find the number whose square is the given number.
$1764$

We know that
It can be written as

$1764=2 \times 2 \times 3 \times 3 \times 7 \times 7$
Here
After pairing the same prime factors, no factor is left
Therefore,

$1764$ is a perfect square of

$2 \times 3 \times 7 =42$
1705768_1806684_ans_c80d2ff1e1234a12a586a868c9129bb4.PNG

Show that each of the following numbers is a perfect square. Also, find the number whose square is the given number.
$1296$

We know that
It can be written as

$1296=2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3 \times 3$
Here
After pairing the same prime factors, no factor is left
Therefore,

$1296$ is a perfect square of

$2 \times 2 \times 3 \times 3 \times=36$
1705759_1806680_ans_b9bc5929f95b45c4a6d72bce8b337f48.PNG

Find the smallest number by which $2592$ be multiplied so that the product is a perfect square.

$\textbf{Step-1: Find the prime factorisation of given expression & ungrouped pair.}$

$\text{We have given 2592}$

$\therefore$

$\text{Prime factorisation of}$

$2592 = 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3 \times 3$

$\text{Now,}$

$\text{On grouping the factors in pairs, we notice that one 2 is left out.}$

$\text{So, we should multiply 2592 with 2, so that prime factor 2 gets grouped in pairs.}$

$\textbf{Hence, 2592 should be multiplied by 2 to make it a perfect square.}$

Find the smallest perfect square divisible by $3, 4, 5$ and $6.$

L.C.M. of

$3, 4, 5, 6$

$= 2 \times 2 \times 3 \times 5 = 60$
In which

$3$ and

$5$ are not in pairs
L.C.M.

$= 2 \times 3 \times 2 \times 5 = 60$
We should multiple it by

$3 \times 5$ i.e. by

$15$
Required perfect square

$= 60 \times 15 = 900$
1780267_1837497_ans_8f5cb2ec6d4141a893e9d99864893e49.png

Find the square root of:
$4761$

Required square root

$= 69$

1780272_1837499_ans_ef6b17e822be44de84d3c2d4c3f0d47a.png

Identify , with reason , if the following is a pythagoran triplet .
$(4,9,12)$

In the triplet ( 4,9,12)

$4^{2}=16,9^{2}=81,12^{2}=144$ and

$16 + 81 =97 \neq 144$

The square of the largest number is not equal to the sum of the square of the other two numbers

$\therefore (4,9,12)$ is not a pythagorean triplet.

Identify , with reason , if the following is a pythagoran triplet .
$(3,5, 4)$

In the triplet ( 3,5,4)

$3^{2}=9,5^{2}=25,4^{2}=16$ and

$9+16 = 25$

The square of the largest number is equal to the sum of the square of the other two numbers.

$\therefore (3,5,4)$ is a pythagorean triplet .

Identify, with reason, if the following is a Pythagorean triplet.
$(5,12,13)$

In the triplet

$( 5,~12,~13)$

$5^{2}=25,12^{2}=144,13^{2}=169$

$\implies 5^2+12^2=13^2$

$\therefore (5,12,13)$ is a pythagorean triplet.

By splitting into prime factors, find the square root of:
$396900$

$\sqrt{396900}$

$\sqrt{2 \times 2 \times \overline{3 \times 3} \times \overline{3 \times 3} \times \overline{5 \times 5 \times 7 \times 7}}$
(Splitting the terms)

$= 2 \times 3 \times 3 \times 5 \times 7 = 630$
Taking L.C.M.
1779670_1837481_ans_c6a134054aa9420ebea4bb8d02b12aa0.png

Find the square root of:
$15129$

Required square root

$= 123$

1780278_1837501_ans_e9364727ddec49f1850937f0fd4d36c7.png

Show that each of the following numbers is a perfect square. Also, find the number whose square is the given number.
$3969$

We know that
It can be written as

$3969=3 \times 3 \times 3 \times 3$
Here
After pairing the same prime factors, no factor is left
Therefore,

$3969$ is a perfect square of

$3 \times 3 \times 7 =63$
1705781_1806686_ans_5472cc548a8e4d868540520868feb610.PNG

Find the smallest number by which $10368$ be divided so that the result is a perfect square. Also, find the square root of the resulting numbers.

$10368 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3 \times 3$
On grouping the prime factors of

$10368;$ one factor i.e.

$2$ is left which cannot be paired with equal factor.
Taking L.C.M.

$\therefore$ The given number should be divided by

$2$
Now

$\sqrt{ \dfrac{10368}{2}}$

$= \sqrt{ \dfrac{2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3 \times 3}{2}}$
(Splitting the terms)

$= 2 \times 2 \times 2 \times 3 \times 3 = 72$
1779675_1837488_ans_1f5c22d74813454d8b0585996c37df42.png

Divide, and write the answer in simplest form.
$\sqrt{310}\div\sqrt{5}$

$\sqrt{310}\div\sqrt{5}$

$=\dfrac{\sqrt{310}}{\sqrt{5}}=\sqrt{\dfrac{310}{5}}=\sqrt{62}$

Divide and write in simplest form.
$\sqrt{54}\div\sqrt{27}$

$\sqrt{54}\div\sqrt{27}$

$=\dfrac{\sqrt{54}}{\sqrt{27}}=\sqrt{\dfrac{54}{27}}=\sqrt{2}$

Identify , with reason , if the following is a pythagoran triplet .
$(24,70,74)$

In the triplet

$(24,70,74)$

$24^{2}=576,70^{2}=4900,74^{2}=5476$ and

$576 + 4900 =5476$

The square of the largest number is equal to the square of the other two numbers.

$\therefore ( 24, 70 , 74)$ is a pythagorean triplet

Write the digit in units place when the following numbers are squared.
$4, 5, 9, 24, 17, 76, 34, 52, 33, 2319, 18, 3458,$

Number

$4, 5, 9, 24, 17, 76, 34, 52, 33, 2319, 18, 3458, 3453.$

Digit in the units place when
squared

$6,5,1,6,9,6,6,4,9,1,4,4,9$

Identify , with reason , if the following is a pythagoran triplet .
$(10,24,27)$

In the triplet

$(10, 24,27)$

$10^{2}=100,24^{2}=576,27^{2}=729$ and

$100 + 576 =676 \neq 729$

The square of the largest number is not equal to the sum of the square of the other two number

$\therefore ( 10, 24 ,27 )$ is not a pythagorean triplet.

Identify , with reason , if the following is a pythagoran triplet .
$(11,60,61)$

In the triplet

$(11,60,61)$

$11^{2}=121,60^{2}=3600,61^{2}=3721$ and

$121 + 3600=3721$

The square of the largest number is equal to the sum of the square of the other two numbers.

$\therefore (11,60,61)$ is a pythagorean triplet.

Find the sum of the digits of
$(111111111)^{2}$

Observe the pattern

$11^{2}$

$=$

$121$

$111^{2}$

$=$

$12321$

$1111^{2}$

$=$

$1234321$

$11111^{2}$

$=$

$123454321$

$111111^{2}$

$=$

$12345654321$

$1111111^{2}$

$=$

$1234567654321$

$11111111^{2}$

$=$

$123456787654321$

$111111111^{2}$

$=$

$12345678987654321$
Sum of the digits in

$111111111^{2}$ is

$1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1 = 81$

Can $1010$ be written as a difference of two perfect squares ?

$a^{2} - b^{2} = 1010$ for any two integers

$a$ and

$b$ then either 'a' and 'b' are odd or both even. Hence

$a^{2} - b^{2}$ is divisible by

$4$ . But

$1010$ is not divided by

$4$ . Hence

$1010$ is not the difference of two perfect squares .

What is the least perfect square which leaves the remainder $1$ when divided by $7$ as well as by $11$ ?

The least number divisible by both

$7$ and

$11$ is

$7 \times 11 = 77$ . When

$1$ is added to

$77$ we get

$78$ . But

$78$ is not a perfect square.

$(77 \times 2) + 1 = 154 + 1 = 155$ is not a perfect square. In the same way, continue

we find that

$(77 \times 15) + 1 = 1155 + 1 = 1156$ in a perfect square.

$\therefore$ The number required is

$1156 = 342$

What are the remainders when a perfect cube is divided by $7$ ?

When the prefect cubes

$8 , 27 , 64 , 125 , 216 , 343 , 512 , 729 , 1000$ are divided by

$7$ the remainders respectively are

$1 , 6 , 1 , 6 , 6 , 0 , 1 , 1 , 6$ .

Hence the remainders are

$0 , , 1$ or

$6$ .

Find the value of $\sqrt{30}$ upto $4$ places of decimal.

$\sqrt{30}=\sqrt{36-6}=(36-6)^{1/2}$

$=(36-6)^{1/2}\Bigg[1-\dfrac{6}{36}\Bigg]^{1/2}=6\bigg[1-\dfrac{1}{6}\bigg]^{1/2}$

$=6\Bigg[1+\dfrac{1}{2}\times\bigg(\dfrac{-1}{6}\bigg)+\dfrac{1}{2}\bigg(\dfrac{1}{2}-1\bigg)\times\bigg(\dfrac{-1}{6}\bigg)^2 \times \dfrac{1}{2!}+\dfrac{1}{2}\bigg(\dfrac{1}{2}-1\bigg)\bigg(\dfrac{1}{2}-2\bigg)\times\bigg(\dfrac{-1}{6}\bigg)^3 \times \dfrac{1}{3!}+.... \Bigg]$

$=6\Bigg[1-\dfrac{1}{12} + \dfrac{1}{2} \times \bigg(\dfrac{-1}{2}\bigg)\times \dfrac{1}{36} \times \bigg(\dfrac{1}{2}\bigg)+ \dfrac{1}{2} \times \bigg(\dfrac{-1}{2}\bigg)\bigg(\dfrac{-3}{2}\bigg)\bigg(\dfrac{-1}{216}\bigg) \times \dfrac{1}{6}+...\Bigg]$

$=6\Bigg[1-\dfrac{1}{12} - \dfrac{1}{288} - \dfrac{1}{3456} +\,...\Bigg]$

$= 6 [1 - 0 08333 - 0 - 00347 - 0 - 00029 + …]$

$= 6 [1 - 0 - 08709] = 6[0 - 91291]$

$= 5.47746 = 5. 4775$ (upto four places).

$\sqrt{0.0037}$

Let

$y = \sqrt{x}, X = 0.0036, y = 0.06$

$\Delta x = 0.0037 - 0.0036 = - 0.001$

$\because y = \sqrt{x}$

Diff. w.r.t. x

$\frac{dy}{dx} = \frac{1}{2\sqrt{x}}$

$\therefore dy = (\frac{dy}{dx}) \Delta x$

$= \frac{1}{2\sqrt{x}} \times 0.0001$

$\Rightarrow dy = \frac{0.0001}{2\sqrt{0.0036}} = \frac{0.0001}{2 \times 0.06}$

$\Rightarrow dy = \frac{0.0001}{0.12} = \frac{1}{1200}$

$\Rightarrow dy = 0.000833 = \Delta y$

$(\because \Delta y \cong dy)$

$\therefore \sqrt{0.0037} = y + \Delta y$

$= 0.06 + 0.000833$

$= 0.060833 = 0.0608$

Hence, approximate value of

$\sqrt{0.0037}$ is

$0.0608$ .

Write all the numbers from $400$ to $425$ which end in $2 , 3 , 7$ or $8$ . Check if any of these in a perfect square.

$402 , 403 , 407 , 408 , 412 , 413 , 417 , 418 , 422 , 423$
None of these is a perfect square.

What is the objective of the school administration behind such an arrangement?

Total students

$104+96=200$

$\dfrac{104}{8}=13$ ;

$\dfrac{96}{8}=12$

Maximum no of parallel rows

$=12+13=25$

No. of students in each row is

$8$ .

$\Rightarrow$ The objective of school behind such an arrangements would be to easily arrange the students in rows such that no adjacent rows have students belonging to one class.

Find the square root of perfect square number $10329796$ .

Let point

$P(-1,6)$ divides the line segment joining the points

$A(-3,10)$ and

$B(6,-8)$ internally in the ratio

$m_{1}$

$m_{2}$
By section formula

$x=\dfrac{m_{1} x_{2}+m_{2} x_{1}}{m_{1}+m_{2}}$

$-1=\dfrac{m_{1}(6)+m_{2}(-3)}{m_{1}+m_{2}}$

$\Rightarrow-m_{1}-m_{2}=6 m_{1}-3 m_{2}$

$\Rightarrow-m_{1}-6 m_{1}=-3 m_{2}+m_{2}$

$\Rightarrow-7 \mathrm{m}_{1}=-2 \mathrm{m}_{2}$

$\Rightarrow \dfrac{m_{1}}{m_{2}}=\dfrac{-2}{-7}=\dfrac{2}{7}$

$\Rightarrow \mathrm{m}_{1}: \mathrm{m}_{2}=2: 7$
Hence, correct choice is

$\mathrm{A}$

A natural number $n$ is chosen strictly between two consecutive perfect squares. The smaller of these two squares is obtained by subtracting $k$ from $n$ and the larger one is obtained by adding $l$ to $n$ . Prove that $n kl$ is a perfect square.

Let

$u$ be a natural number such that

$u^2 < n < (u + 1)^2$ . Then

$n - k = u^2$ and

$n + l = (u + 1)^2$ . Thus

$n kl = n (n u^2) ((u + 1)^2 n)$

$= n n(u + 1)^2 + n^2 + u^2 (u + 1)^2 nu^2$

$= n^2 + n(1 (u + 1)^2) u^2) + u^2(u + 1)^2$

$= n^2 + n(1 2u^2 2u 1) + u^2(u + 1)^2$

$= n^2 2nu(u + 1) + (u(u + 1))^2$

$= (n - u(u +1)^2$ .

Find the square root of the following.
$10.0489$

Simplifying $\sqrt {10.0489}$ .

$\sqrt {10.0489} = \sqrt {\dfrac{{100489}}{{10000}}}$

The factors of 100489 is $317 \times 317$ and the factors of 10000 is $2 \times 2 \times 2 \times 2 \times 5 \times 5 \times 5 \times 5$ .

So,

$\sqrt {\dfrac{{100489}}{{10000}}} = \sqrt {\dfrac{{{{317}^2}}}{{{2^4} \times {5^4}}}}$

$= \sqrt {{{\left( {\dfrac{{317}}{{{2^2} \times {5^2}}}} \right)}^2}}$

$= \frac{{317}}{{4 \times 25}}$

$= \frac{{317}}{{100}}$

$= 3.17$

Therefore, the square root of $\ {10.0489}$ is $3.17$ .

Find the $\sqrt {39204}$ .

2041691_1080433_ans_ae0d15c80d764fd79b4b7d97ecaa2616.jpg

Prove that $\dfrac{1}{3 -\sqrt{8}} - \dfrac{1}{\sqrt{8} - \sqrt{7}} + \dfrac{1}{\sqrt{7} - \sqrt{6}} - \dfrac{1}{\sqrt{6} - \sqrt{5}} + \dfrac{1}{ \sqrt{5} - 2} = 5$

$\dfrac{1}{\sqrt{a+1}+\sqrt{a}}$

$=\dfrac{\sqrt{a+1}+ \sqrt {a}}{a+1-a}$

$=\sqrt{a+1}-\sqrt{a}$

$\Rightarrow \dfrac{1}{3-\sqrt{a}}- \dfrac{1}{\sqrt{8}-\sqrt{7}} + \dfrac{1}{\sqrt{7}-\sqrt{6}} +\dfrac{1}{\sqrt{5}-2} - \dfrac{1}{\sqrt{6}-\sqrt{5}}$

$=3+\sqrt{8}-\sqrt{8}-\sqrt{7}+\sqrt{7}+\sqrt{6}+\sqrt{5}+2-\sqrt{6}-\sqrt{5}$

$=3+2$

$=5$

What will be the unit digit of the square of the following numbers?
$272$

Since,

$2^2=4$ ,therefore,square of

$272$ will have

$2$ at it's unit place.

So,

$272^2$ will have

$4$ at unit’s place

What will be the unit digit of the square of the following numbers?
$12796$

Since,

${ 6 }^{ 2 }$ =

$6$ ,so,

${ { 12796 }^{ 2 } }$ will have

$6$ at unit’s place

Find the square root of following surd :
$6+4\sqrt{2}$

Let

$\sqrt{6+4\sqrt{2}}=\pm (\sqrt{a}+\sqrt{b})$

Squaring both sides,

$6+4\sqrt{2}=a+b+2\sqrt{ab}$

Comparing both sides,

$a+b=6$ ........

$(1)$

$4\sqrt{2}=2\sqrt{ab}$

$8=ab\Rightarrow a=\dfrac{8}{b}$ ............

$(2)$

$\therefore$ From

$(1)$ and

$(2)$

$\dfrac{8}{b}+b=6$

or,

$8+b^{2}=6b$

or,

$b^{2}-6b+8=0$

or,

$b^{2}-4b-2b+8=0$

or,

$b(b-4)-2(b-4)=0\Rightarrow (b-4)(b-2)=0$

$\therefore b=2, 2$

$\therefore a=2, 4$

$\therefore$ Squaring root is

$\pm(\sqrt{2}+\sqrt{4})=\pm(2+\sqrt{2})$ Ans

Find the square root of following surds:
$5+\sqrt{21}$

Let

$\sqrt{5+\sqrt{21}}=\pm(\sqrt{a}+\sqrt{b})$

Squaring both sides,

$5+\sqrt{21}=a+b+2\sqrt{ab}$

$a+b=5$ .............

$(1)$ and

$2\sqrt{ab}=\sqrt{21}$

$4ab=21\Rightarrow a=\dfrac{21}{4b}$

or,

$\dfrac{21}{4b}+b=5$

or,

$4b^{2}-20b+21=0$

or,

$4b^{2}-6b-14b+21=0$

or,

$2b(2b-3)-7(2b-3)=0$

or,

$(2b-3)(2b-7)=0$

$\therefore b=\dfrac{3}{2}$

$b=\dfrac{7}{2}$

$a=\dfrac{7}{2}$

$a=\dfrac{3}{2}$

$\therefore \sqrt{5+\sqrt{21}}=\pm\left(\sqrt{\dfrac{3}{2}}+\sqrt{\dfrac{7}{2}}\right)$ Ans

Solve:
$\sqrt{\left(\dfrac{3}{4}\right)^{1-3x}}=2\dfrac{10}{27}$

or,

$1-3x=-6$

or,

$3x=7$

or,

$x=\dfrac{7}{3}$ .

1202621_1281807_ans_21576b2455124680a1308c5dbca7d70f.jpg

By what smallest number must $180$ be multiplied so that it becomes a perfect square? Also, find the square root of the number so obtained.

$180=2\times 2\times 3\times 3\times 5$

So, here is only 5 which is not in pair so, we can say that it multiplied by 5

Square root will be

$2\times 3\times 5=30$

Find the least number which must be added to each of the following numbers to make them a perfect square. Also find the square root's of the perfect square number so obtained.
$1750$

$\begin{array}{l} { 41^{ 2 } }<1750 \\ next\, perfect\, \, square\, is\, \\ \Rightarrow { 42^{ 2 } }=1764 \\ hence,\, \, number\, \, to\, \, be\, added\, \, is\, \, \\ \Rightarrow { 42^{ 2 } }-1750=1764-1750=14 \\ \Rightarrow therefore,\, \, the\, \, perfect\, \, square\, so,\, \, obtained\, \, is \\ 1750+14=1764 \\ and\, \, its\, \, square\, \, root=\sqrt { 1764 } =42 \end{array}$

Square root of $\sqrt{5-\sqrt{21}}$ .

Let

$\sqrt{5-\sqrt{21}}=\pm(\sqrt{x}-\sqrt{y})......(1)$

Squaring both sides,

$\Rightarrow 5-\sqrt{21}=x+y-2\sqrt{xy}$

Comparing both sides,

$\Rightarrow x+y=5........(2)$

and

$2\sqrt{xy}=\sqrt{21}$

Squaring both sides,

$\Rightarrow 4xy=21$

$\Rightarrow xy=\dfrac{21}{4}......(3)$

Now, squaring both sides of eq

$(1)$

$\left( x+y\right)^2=25$

$\Rightarrow x^2+y^2+2xy=25$

$\Rightarrow x^2+y^2+2\times\dfrac{21}{4}=25$

$\Rightarrow x^2+y^2=25-\dfrac{21}{2}$

$\Rightarrow x^2+y^2=\dfrac{29}{2}.......(4)$

Now,

$\left( x-y\right)^2= x^2+y^2-2xy=\dfrac{29}{2}-\dfrac{21}{2}=\dfrac{8}{2}=4$

$\Rightarrow x-y=2......(5)$

Solving

$(2)$ and

$(5)$

$x=\dfrac{7}{2},y=\dfrac{3}{2}$

From

$(1)$

$\therefore \sqrt{5-\sqrt{21}}=\pm\left(\sqrt{\dfrac{7}{2}}-\sqrt{\dfrac{3}{2}}\right)$

Hence, solved.

$\sqrt{244}=?$

1311765_1587550_ans_09a1f27d35e1450993b28eca373cccb1.jpg

Verify that the following numbers represent Pythagorean triplet.
$27, 36, 45$ .

2039878_1677891_ans_d32432c2abeb4972a22677b92ef69817.jpeg

Find the value of :
$\sqrt{256cm^2}$

2045317_1570196_ans_5d6ce1e95f864a8f843964fd0267e253.jpg

Find the approximate value of $\sqrt { 8.95 }$

2044449_1564910_ans_ac8b1630714c4725b5435247cb30a0c7.jpg

Find the least 8 digit number which is a perfect square.

1311665_1586742_ans_6277e9cec03a4f96bee349bed9f7eced.jpg

Verify that the following numbers represent Pythagorean triplet.
$12, 35, 37$ .

2039876_1677889_ans_bd5aba3506b6481da0884d86f07a00b2.jpeg

Verify that the following numbers represent Pythagorean triplet.
$7, 24, 25$ .

2039877_1677890_ans_d7e8b4e12f2d422bba7d63dac61b32d4.jpeg

Find the smallest number by which $2925$ must be multiplied to get a perfect square. Also, find the square root of the perfect square so obtained.

$\textbf{Step 1 : First find the Smallest number }$

$\text{Given number}$

$=2925$

$\text{Resolve the given number into prime factors we get}$

$2925=3\times 3\times 5\times 5\times 13$

$\text{To get a perfect square we have to multiply by}$

$13$

$\text{Hence ,the Smallest number =13}$

$\textbf{Step 2 : Find the square root of the perfect square}$

$2925=3\times 3\times 5\times 5\times 13$

$\text{New number after multiplying with 13}$

$=2925\times 13=38025$

$\therefore \sqrt {38025}=3\times 5\times 13=195$

$\textbf{Hence, the Square root of the perfect square is =195}$

Estimate square root : $\sqrt {13}$

1993368_1744902_ans_f8375710d7584ccabab4ca8b861ec501.jpg

Verify that the following numbers represent Pythagorean triplet.
$15, 36, 39$ .

2039879_1677892_ans_f868f7621ef94532af20185e69240874.jpeg

Squares And Square Roots - Class 8 Maths - Extra Questions

Find the square root of the following numbers correct to two decimal places:5.4625.462

Find the square root of the following decimal numbers by division method:18.404118.4041

Find the square root of the following numbers by prime factorization method:18491849

Find the square root of the following numbers correct to two decimal places:645.8645.8

Find the square root of the following numbers correct to two decimal places:107.45107.45

Find the square root of the following numbers by prime factorization method:62416241

Find the square root of the following decimal numbers by division method:5.7744095.774409

Find the square root of the following numbers by prime factorization method:441441

Find the square root of the following numbers by prime factorization method:784784

Find the square root of the following numbers by prime factorization method:43564356

Find the square root of the following numbers correct to two decimal places:33

Find the square root of the following numbers by prime factorization method:90259025

Find the square root of the following number by prime factorization method:82818281

Which of the following natural numbers are perfect squares? Give reasons in support of your answer.54885488

Which of the following natural numbers are perfect squares? Give reasons in support of your answer.243243

Find the square root of the following numbers by prime factorization method:0.00640.0064

Show that each of the following numbers is a perfect square. Also, find the number whose square is the given number.70567056

Find numbers from 100100 to 400400 that end with 0,1,4,5,60, 1, 4, 5, 6 or 99, which are perfect squares.

Find the square root of the following numbers by factorization 256256

Find the smallest positive integer with which one has to multiply 847847 to get a perfect square.

Find the square root by prime factorization method 11561156

Find the square root of 1040410404 by prime factorization method.

Simplify: √169×√361\sqrt{169} \times \sqrt{361}

Simplify: √1764−√1444\sqrt{1764} - \sqrt{1444}

Simplify: √1360+9\sqrt{1360 + 9}

Find the square root by prime factorization method 1322513225

Simplify √2704+√144+√289\sqrt{2704} + \sqrt{144} + \sqrt{289}

Simplify: √225−√25\sqrt{225} - \sqrt{25}

Simplify √100+√36\sqrt{100} + \sqrt{36}

Find the largest perfect square factor of 4848

Find the largest perfect square factor in the number 13521352.

Find the smallest positive integer with which one has to multiply 450450 to get a perfect square.

Find the smallest positive integer with which one has to multiply to 14451445 to get a perfect square.

Find the smallest positive integer with which 13521352 be multiplied to get a perfect square.

Find the nearest integer to the square root of 232232.

Find the nearest integer to the square root of 824.824.

Find the largest perfect square factor in 729729

If the area of a square is 9090 cm2^2, what is its side-length, rounded to the nearest integer?

A square piece of land has area 112 m2^2. What is the closest integer which approximates the perimeter of the land?

Find the square root of 0.017424.

Find the square root of the following numbers by the factorisation numbers : 82944

Explain with reason whether the given number is perfect square or not: 24332433

Find whether the square of the given number is even or odd: 8204

Explain with reason whether the given number is perfect square or not: 60986098

What will be the units digit of the square of the given number:87428742

Find the square root of 12961296 by Prime Factorization

How many numbers lie between the square of the given numbers: 2525 and 2626

Find the square roots of the given number by Prime factorization method: 441

Explain with reason whether the given number is perfect square or not: 4592

Find the square root of24+4\sqrt { 15 } -4\sqrt { 21 } -2\sqrt { 35 } 24+4\sqrt { 15 } -4\sqrt { 21 } -2\sqrt { 35 }

Express the following square numbers as the sum of two odd numbers.i)8181 ii)144144

A society collected \text{Rs. }9,216\text{Rs. }9,216 wherein, each member contributed as many rupees as there were members in the society. Find the number of members.

Find the smallest whole number by which 14581458 should be multiplied, so as to get a perfect square number.

For the following number, find the smallest whole number by which it should be multiplied so as to get a perfect square number.252

Find the least number that must be added to 13001300 so as to get a perfect square. Also find the square root of the perfect square.

Find the square root of each of the following by Long division method:(i) 23042304 (ii) 44894489(iii) 34813481 (iv) 529529(v) 32493249 (vi) 13691369 (vii) 57765776 (viii) 79217921 (ix) 576576 (x) 31363136

Find the square root of the following decimal numbers:(i) 2.562.56 (ii) 7.297.29(iii) 51.8451.84 (iv) 42.2542.25 (v) 31.3631.36 (vi) 0.29160.2916(vii) 11.5611.56 (viii) 0.0018490.001849

Find the square root of the following numbers by long division method: 5314414401604

Find the square root of each of the following numbers by using the method of prime factorization:729729

Find the square root of each of the following numbers by using the method of prime factorization:40964096

Find the square root of each of the following numbers by using the method of prime factorization:70567056

Find the square root of each of the following numbers by using the method of prime factorization:20252025

Find the square root of each of the following numbers by using the method of prime factorization:81008100

Write a Pythagorean triplet whose smallest number is2020

Write a Pythagorean triplet whose smallest number is1414

Find the least square number which exactly divisible by each of the numbers 8,12,158,12,15 and 2020

Find the largest number of 33 digits which is perfect square.

Evaluate:\surd 576\surd 576

Write a Pythagorean triplet whose number is 6363

Find the square root of the following decimal numbers by division method:42.2542.25

Find the square root of each of the following by division method:106929106929

Write a Pythagorean triplet whose number is 88

Find the square root of 1764 by the prime factorisation method.

Evaluate :\sqrt { 92416 } \sqrt { 92416 }

Find the unit digit in the square of the numbers 8181 and 7979.

Show that the following number is not perfect square:2245322453

The square of which of the following numbers would be an odd number?731731

Show that the following number is not a perfect square. Give reason.89488948

Show that the following number is not perfect square:40584058

Find the square root of the following numbers correct to two decimal places:
$5.462$

Find the square root of the following decimal numbers by division method:
$18.4041$

Find the square root of the following numbers by prime factorization method:
$1849$

Find the square root of the following numbers correct to two decimal places:
$645.8$

Find the square root of the following numbers correct to two decimal places:
$107.45$

Find the square root of the following numbers by prime factorization method:
$6241$

Find the square root of the following decimal numbers by division method:
$5.774409$

Find the square root of the following numbers by prime factorization method:
$441$

Find the square root of the following numbers by prime factorization method:
$784$

Find the square root of the following numbers by prime factorization method:
$4356$

Find the square root of the following numbers correct to two decimal places:
$3$

Find the square root of the following numbers by prime factorization method:
$9025$

Find the square root of the following number by prime factorization method:
$8281$

Which of the following natural numbers are perfect squares? Give reasons in support of your answer.
$5488$

Which of the following natural numbers are perfect squares? Give reasons in support of your answer.
$243$

Find the square root of the following numbers by prime factorization method:
$0.0064$

Show that each of the following numbers is a perfect square. Also, find the number whose square is the given number.
$7056$

Find numbers from $100$ to $400$ that end with $0, 1, 4, 5, 6$ or $9$ , which are perfect squares.

Find the square root of the following numbers by factorization $256$

Find the smallest positive integer with which one has to multiply $847$ to get a perfect square.

$1156$

Find the square root of $10404$ by prime factorization method.

Simplify: $\sqrt{169} \times \sqrt{361}$

Simplify: $\sqrt{1764} - \sqrt{1444}$

Simplify: $\sqrt{1360 + 9}$

Find the square root by prime factorization method $13225$

Simplify $\sqrt{2704} + \sqrt{144} + \sqrt{289}$

Simplify: $\sqrt{225} - \sqrt{25}$

Simplify $\sqrt{100} + \sqrt{36}$

Find the largest perfect square factor of $48$

Find the largest perfect square factor in the number $1352$ .

Find the smallest positive integer with which one has to multiply $450$ to get a perfect square.

Find the smallest positive integer with which one has to multiply to $1445$ to get a perfect square.

Find the smallest positive integer with which $1352$ be multiplied to get a perfect square.

Find the nearest integer to the square root of $232$ .

Find the nearest integer to the square root of $824.$

Find the largest perfect square factor in $729$

If the area of a square is $90$ cm $^2$ , what is its side-length, rounded to the nearest integer?

A square piece of land has area 112 m $^2$ . What is the closest integer which approximates the perimeter of the land?

Explain with reason whether the given number is perfect square or not: $2433$

Explain with reason whether the given number is perfect square or not: $6098$

What will be the units digit of the square of the given number:
$8742$

Find the square root of $1296$ by Prime Factorization

How many numbers lie between the square of the given numbers: $25$ and $26$

Find the square root of
$24+4\sqrt { 15 } -4\sqrt { 21 } -2\sqrt { 35 }$

Express the following square numbers as the sum of two odd numbers.
i) $81$ ii) $144$

A society collected $\text{Rs. }9,216$ wherein, each member contributed as many rupees as there were members in the society. Find the number of members.

Find the smallest whole number by which $1458$ should be multiplied, so as to get a perfect square number.

For the following number, find the smallest whole number by which it should be multiplied so as to get a perfect square number.
252

Find the least number that must be added to $1300$ so as to get a perfect square. Also find the square root of the perfect square.

Find the square root of each of the following by Long division method:
(i) $2304$ (ii) $4489$
(iii) $3481$ (iv) $529$
(v) $3249$ (vi) $1369$
(vii) $5776$ (viii) $7921$
(ix) $576$ (x) $3136$

Find the square root of the following decimal numbers:
(i) $2.56$ (ii) $7.29$
(iii) $51.84$ (iv) $42.25$
(v) $31.36$ (vi) $0.2916$
(vii) $11.56$ (viii) $0.001849$

Find the square root of the following numbers by long division method:
531441
4401604

Find the square root of each of the following numbers by using the method of prime factorization:
$729$

Find the square root of each of the following numbers by using the method of prime factorization:
$4096$

Find the square root of each of the following numbers by using the method of prime factorization:
$7056$

Find the square root of each of the following numbers by using the method of prime factorization:
$2025$

Find the square root of each of the following numbers by using the method of prime factorization:
$8100$

Write a Pythagorean triplet whose smallest number is
$20$

Write a Pythagorean triplet whose smallest number is
$14$

Find the least square number which exactly divisible by each of the numbers $8,12,15$ and $20$

Find the largest number of $3$ digits which is perfect square.

Evaluate:
$\surd 576$

Write a Pythagorean triplet whose number is
$63$

Find the square root of the following decimal numbers by division method:
$42.25$

Find the square root of each of the following by division method:
$106929$

Write a Pythagorean triplet whose number is
$8$

Evaluate :
$\sqrt { 92416 }$

Find the unit digit in the square of the numbers $81$ and $79$ .

$22453$

The square of which of the following numbers would be an odd number?
$731$

Give reason.
$8948$

Show that the following number is not perfect square:
$4058$

. Give reason.
$333333$

What will be the units digit of the squares of the following numbers?
$977$

What will be the units digit of the squares of the following number?
$53924$

What will be the units digit of the squares of the following numbers?
$4583$

What will be the units digit of the squares of the following numbers?
$52698$

What will be the units digit of the squares of the following numbers?
$12796$

What will be the units digit of the squares of the following numbers?
$99880$