MCQExams
0:0:1
CBSE
JEE
NTSE
NEET
Practice
Homework
×
CBSE Questions for Class 8 Maths Squares And Square Roots Quiz 8 - MCQExams.com
CBSE
Class 8 Maths
Squares And Square Roots
Quiz 8
The square root of $$\displaystyle\frac{36}{5}$$ correct to two decimal places is _____________.
Report Question
0%
$$2.68$$
0%
$$2.69$$
0%
$$2.67$$
0%
$$2.66$$
Explanation
We know $$\sqrt{\dfrac{36}{5}}=\dfrac {6}{\sqrt 5}$$
$$=\dfrac{6}{2.236}$$ (Taking approximate value of square root $$\sqrt{5}$$)
$$=2.6833$$
Upto two decimal places:
$$=2.68$$
The number that must be subtracted from $$16161$$ to get a perfect square is ________.
Report Question
0%
$$31$$
0%
$$32$$
0%
$$33$$
0%
$$34$$
Explanation
Let the number is $$x$$
Finding the square root of $$16161$$
$$\sqrt{16161}=127.1259$$
Finding the square of $$127$$
$$127^2=16129$$
$$x=16161-16129$$
$$x=32$$
Estiamate the square root of $$850$$
Report Question
0%
$$29.15$$
0%
$$30.21$$
0%
$$98.23$$
0%
$$23.11$$
Explanation
The square root of $$850$$ is $$\sqrt {850}=\sqrt {25 \times 34}=5\sqrt {34}$$
Square root of $$34$$ lie between $$5$$ and $$6$$.
square of $$5.5$$ is $$30.25$$
Now, we can say that square root of $$34$$ lie between $$5.5$$ and $$6$$.
Now, square of $$5.75$$ is $$33.06$$
So, square root of $$34$$ lie between $$5.75$$ and $$6$$.
Now, we have to choose the number $$5.85$$
$$(5.85)^2=34.225$$ which is greater than $$34$$ and close to $$34$$
So, assume a number $$5.84$$.
$$(5.84)^2=34.1056$$
$$(5.83)^2=33.9889$$
Hence, we can say that square root of 34 lie between $$5.83$$ and $$5.84$$.
So,
$$5\times 5.83=29.15$$.
What approximate value will come in place of the question mark $$(?)$$ in the following question? (you are not expected to calculate the exact value)
$$\sqrt{8000}=?$$
Report Question
0%
$$76$$
0%
$$89$$
0%
$$65$$
0%
$$97$$
0%
$$58$$
Find the square root of $$225$$ by division method.
Report Question
0%
$$5$$
0%
$$15$$
0%
$$25$$
0%
$$35$$
State true (T) or false (F):
The square of 86 will have 6 at the units place.
Report Question
0%
True
0%
False
Explanation
Digit at unit place of $$(86)^2 = $$ Digit at the unit place of $$6^2 = 36$$
Hence, $$6$$ will be correct answer.
Write a Pythagorean triplet whose one number is 8
Report Question
0%
$$8,10,6$$
0%
$$15,10,5$$
0%
$$63,27,3$$
0%
$$80,40,20$$
Explanation
$$8^2=64$$
$$\Rightarrow$$ $$100-64=36$$
$$\Rightarrow$$ $$\sqrt{36}=6$$
$$\Rightarrow$$ $$\sqrt{100}=10$$
Let
Hypotenuse $$=10$$ [ Since its greater than $$6$$ and $$8$$ ]
Base $$=8$$ and
Perpendicular $$=6$$
$$(Hypotenuse)^2=(Perpendicular)^2+(Base)^2$$ [ Pythagoras theorem ]
$$\Rightarrow$$ $$(10)^2=(6)^2+(8)^2$$
$$\Rightarrow$$ $$100=36+64$$
$$\Rightarrow$$ $$100=100$$
$$\therefore$$ Pythagorean triplet are $$6,8,10$$
What will be the unit digits of the square of the following numbers?
$$39$$, $$297$$, $$5125$$, $$7286$$
Report Question
0%
$$9,9,1,6$$
0%
$$1,9,5,6$$
0%
$$9,2,5,2$$
0%
$$3,2,5,7$$
Explanation
$$(a)$$The numbers with unit's digits $$1$$ or $$9$$ has its square as $$1$$
$$\therefore\,$$ the unit's digit of square of $$39$$ is $$1$$
$$(b)$$The numbers with unit's digits $$3$$ or $$7$$ has its square as $$9$$
$$\therefore\,$$ the unit's digit of square of $$297$$ is $$9$$
$$(c)$$The numbers with unit's digits $$5$$ has its square as $$5$$
$$\therefore\,$$ the unit's digit of square of $$5125$$ is $$5$$
$$(d)$$The numbers with unit's digits $$4$$ or $$6$$ has its square as $$6$$
$$\therefore\,$$ the unit's digit of square of $$7286$$ is $$6$$
Use Pythagoras theorem to check which of following triplets would make a right triangle.
Report Question
0%
$$5, 20, 25$$
0%
$$7, 24, 25$$
0%
$$7, 23, 25$$
0%
$$15, 20, 25$$
Explanation
The pythagoras theorem is
$$a^2=b^2+c^2$$
$$20^2+5^2=425$$ and $$25^2=625$$
so $$25^2\neq20^2+5^2$$
now check for b
$$25^2=24^2+7^2$$
$$625=576+49$$
$$625=625$$
so,$$25^2=24^2+7^2$$
now check for c
$$25^2=23^2+7^2$$
$$625=529+49$$
$$625\neq578$$
Now ckeck for d
$$25^2=20^2+15^2$$
$$625=400+225$$
$$625=625$$
so only option $$B$$ and $$D$$ is correct.
Find the square root of $$126736$$ by long division method.
Report Question
0%
$$355$$
0%
$$356$$
0%
$$366$$
0%
$$365$$
Find the square root of which of the following numbers will be the least :
Report Question
0%
$$7\dfrac{58}{81}$$
0%
$$11\dfrac{14}{25}$$
0%
$$10\dfrac{1}{36}$$
0%
$$0.3481$$
Explanation
$$A.$$
$$7\dfrac{58}{81}=\dfrac{625}{81}$$
$$\Rightarrow$$ $$\sqrt{\dfrac{625}{81}}=\dfrac{25}{9}=2.77$$
$$B.$$
$$11\dfrac{14}{25}=\dfrac{289}{25}$$
$$\Rightarrow$$
$$\sqrt{\dfrac{289}{25}}=\dfrac{17}{5}=3.4$$
$$C.$$
$$10\dfrac{1}{36}=\dfrac{361}{36}$$
$$\Rightarrow$$
$$\sqrt{\dfrac{361}{36}}=\dfrac{19}{6}=3.16$$
$$D.$$
$$0.3481=\dfrac{3481}{10000}$$
$$\Rightarrow$$
$$\sqrt{\dfrac{3481}{10000}}=\dfrac{59}{100}=0.59$$
$$\therefore$$ We can see, $$0.3481$$ has least square root.
Find value of $$\sqrt{5184}$$.
Report Question
0%
$$72$$
0%
$$52$$
0%
$$82$$
0%
$$62$$
Explanation
$$\sqrt{5184}\\=\sqrt{2\times 2\times 2\times 2\times 2\times 2\times 3\times 3\times 3\times 3}\\=\sqrt{2^6\times 3^4}\\=2^3\times3^2\\=8\times9=72$$
Find the square root by prime factorisation:
i) $$196$$
ii) $$225$$
Report Question
0%
$$14, 25$$
0%
$$14, 15$$
0%
$$24, 15$$
0%
$$24, 25$$
Explanation
(1)
$$196=2\times 2\times 7\times 7$$
$$\sqrt{196}=\sqrt{2\times 2\times 7\times 7}$$
$$=\sqrt{2^{2}\times 7^{2}}$$
$$=2\times 7$$
$$=14$$
(2)
$$225=3\times 3\times 5\times 5$$
$$\sqrt{225}=\sqrt{3\times 3\times 5\times 5}$$
$$=\sqrt{3^{2}\times 5^{2}}$$
$$=3\times 5$$
$$=15$$
Find the value of $$\sqrt{66049}$$.
Report Question
0%
$$253$$
0%
$$257$$
0%
$$347$$
0%
$$343$$
Explanation
The number $$66049$$ can only be divided by $$257$$
$$66049=257\times 257$$
$$66049=(257)^2$$
$$\sqrt{66049}=\sqrt{257^2}$$
$$\therefore$$ $$\sqrt{66049}=257$$
What are the number which cannot come to the unit place of perfect square?
Report Question
0%
$$2$$
0%
$$3$$
0%
$$7$$
0%
$$8$$
Explanation
Perfect square $$\rightarrow$$ Unit place digit
$$1^2=1$$
$$2^2=4$$
$$3^2=9$$
$$4^2=16$$
$$5^2=25$$
$$6^2=36$$
$$7^2=14$$
$$8^2=64$$
$$9^2=81$$
$$10^2=100$$
We can see number which can comes to unit place of perfect squares are $$1,4,5,6,9$$ and $$0$$.
$$\therefore$$ The numbers which can not come to the unit place of perfect square are $$2,3,7$$ and $$8.$$
A pythagorean triplet whose two numbers are 8 and 10, then third number is
Report Question
0%
$$6$$
0%
$$14$$
0%
$$16$$
0%
$$18$$
Explanation
Require pythagorean triplet is (6,8,10) as $$6^2+8^2 = 36+64= 100 = 10^2 $$
Which of the following is a perfect square number?
Report Question
0%
257
0%
256
0%
300
0%
None of these
Which of the following is pythagorean triplet?
Report Question
0%
3 ,4 ,5
0%
5 ,12 ,13
0%
7 ,24 ,25
0%
All of these
Explanation
$$\textbf{Step-Check option one by one using condition of pythagorian triplet}$$
$$\text{$$a,b$$ and $$c$$ are called pythagorean triplet if}$$ $$a^2+b^2=c^2$$
$$\text{Clearly,}$$ $$3^2+4^2=9+16=25=5^2$$
$$5^2+12^2=25+144=169=13^2$$
$$7^2+24^2=49+576=625=25^2$$
$$\text{Here all three of the given options are satisfy the pythagorean triplet.}$$
$$\textbf{Hence option D is correct}$$
Find the atleast number which must be added to each of the following numbers to get a perfect square. Also find the square root of the perfect square numbers.
$$a)525$$
Report Question
0%
4
0%
3
0%
1
0%
6
0%
12
Explanation
$$i)\ 23^2=529$$
$$\therefore \ $$ it will add $$4$$ to $$525$$ we get $$529$$
which is payout square
$$\therefore \ 525+4=529=23^2$$
$$\therefore 4$$ should be added
State whether the statements are true (T) or false (F).
For every natural number $$m, (2m - 1, 2m^2 - 2m, 2m^2 - 2m^2 + 1)$$ is a Pythagorean triplet.
Report Question
0%
True
0%
False
Explanation
$$(2m- 1)^2 \neq (2m^3 - 2m )^2 + (2m^2 - 2m + 1)^2$$
Or $$(2m^3 - 2m)^2 \neq (2m - 1)^2 + (2m^2 - 2m + 1)^2$$
Or $$ (2m^2 - 2m + 1)^2 \neq (2m -1 )^2 + (2m^3 - 2m )^2$$
Which of the following is a Pythagorean triplet?
Report Question
0%
$$8, 6, 10$$
0%
$$8, 9, 10$$
0%
$$10, 12, 13$$
0%
None of these
Explanation
"Pythagorean triples" are integer solutions to the Pythagorean Theorem, $$a^2 + b^2 = c^2.$$
$$A.$$ $$8,6,10$$
$$(Hypotenuse)^2=(Perpendicular)^2+(Base)^2$$
$$\Rightarrow$$ $$(10)^2=(8)^2+(6)^2$$
$$\Rightarrow$$
$$100=64+36$$
$$\Rightarrow$$
$$100=100$$
$$\therefore$$ $$8,6,10$$ is a Pythagorean triplet
$$B.$$ $$8,9,10$$
$$(Hypotenuse)^2=(Perpendicular)^2+(Base)^2$$
$$\Rightarrow$$ $$(10)^2=(8)^2+(9)^2$$
$$\Rightarrow$$
$$100=64+81$$
$$\Rightarrow$$
$$100\ne 146$$
$$\therefore$$ $$8,9,10$$ is not a Pythagorean triplet
$$C.$$ $$10,12,13$$
$$(Hypotenuse)^2=(Perpendicular)^2+(Base)^2$$
$$\Rightarrow$$ $$(13)^2=(10)^2+(12)^2$$
$$\Rightarrow$$
$$169=100+144$$
$$\Rightarrow$$
$$100\ne 244$$
$$\therefore$$ $$10,12,13$$ is not a Pythagorean triplet
Find $$\sqrt {125}$$ given $$\sqrt {5}=2.23607$$
Report Question
0%
$$11.21$$
0%
$$11.38$$
0%
$$11.18$$
0%
$$11.24$$
Explanation
$$\sqrt{125}$$
$$=\sqrt{5\times 5\times 5}$$
$$=5\sqrt{5}$$
$$=5\times 2.23607$$
$$=11.180$$.
Which of the following numbers are not perfect square?
Report Question
0%
2397
0%
121000
0%
25008
0%
19640
0%
All of above
Explanation
These all are $$not$$ perfect square numbers because perfect square numbers always ends with $$(0,1,4,5,6,9)$$ and if if last digit is $$0$$ it should be even number of zeros.
Therefore all the given numbers are not perfect square.
Simplify : $$(64)^{1/2}$$
Report Question
0%
$$12$$
0%
$$2$$
0%
$$8$$
0%
$$14$$
Explanation
$$(64)^{1/2}$$
$$=(8\times 8)^{1/2}$$
$$=(8^2)^{1/2}$$
$$=8^1$$
$$=8$$.
If $$\sqrt{2}=1.414$$ then the value of $$\sqrt{8}$$ is
Report Question
0%
$$2.828$$
0%
$$1.828$$
0%
$$2.282$$
0%
$$2.288$$
Explanation
$$\sqrt{2}=1.414$$(given)
$$\sqrt{8}=\sqrt{2\times 2\times 2}=2\sqrt{2}=2\times 1.414=2.828$$
$$\therefore \sqrt{8}=2.828$$
Square root of 169 is
Report Question
0%
13
0%
14
0%
15
0%
None of these
Find which of the square of the following numbers are even ?
Report Question
0%
$$431$$
0%
$$2826$$
0%
$$7779$$
0%
$$82004$$
Explanation
We know that square of an odd number is odd and square of an even number is even
$$(i)$$The square of $$431$$ is an odd number $$\because 431$$ is an odd number.
$$(ii)$$The square of $$2826$$ is an even number $$\because 2826$$ is an even number.
$$(iii)$$The square of $$7779$$ is an odd number $$\because 7779$$ is an odd number.
$$(iv)$$The square of $$82004$$ is an even number $$\because 82004$$ is an even number.
Mark the correct alternative of the following.
Which of the following is/are not Pythagorean triplet(s)?
Report Question
0%
$$3, 4, 5$$
0%
$$8, 15, 17$$
0%
$$7, 24, 25$$
0%
$$13, 26, 29$$
Explanation
We have to check for pythagorean triplets.
Option (A) $$3, 4, 5$$
It is easy to check that $$3^2 + 4^2 = 5^2$$
$$25 = 25$$
$$3, 4, 5$$ is pythagorean triplet.
Option (B) $$8 , 15 , 17$$
Observe that $$8^2 + 15^2 = 64 + 225$$
$$= 289 $$
$$= 17^2$$
Hence $$8, 15, 17$$ is pythagorean triplet
Option (C) $$7, 24, 25$$
observe that $$7^2 + 24^2 = 49 + 576 $$
$$= 625$$
$$= 25^2$$
So $$7, 24, 25$$ is pythagorean triplet.
Option (D) $$13, 26, 29$$
check that $$13^2 + 26^2 = 169 + 676$$
$$= 845$$
and $$29^2 = 841$$
So $$13^2 + 26^2 \neq 29^2$$
Hence option D is correct.
Which of the following number is not a perfect square?
Report Question
0%
$$1444$$
0%
$$3136$$
0%
$$961$$
0%
$$2222$$
Explanation
Explanation: According to the property of square, a number ending with $$2,3,7$$ and $$8$$ is not a perfect square.
Hence, $$2222$$ is not a perfect square.
Which of the following cannot be the unit digit of a perfect square number?
Report Question
0%
$$6$$
0%
$$1$$
0%
$$9$$
0%
$$8$$
Explanation
As we know that all the perfect square number ends with $$1,4,9,6,5,00$$
Hence, D will be correct answer.
0:0:1
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
0
Answered
0
Not Answered
0
Not Visited
Correct : 0
Incorrect : 0
Report Question
×
What's an issue?
Question is wrong
Answer is wrong
Other Reason
Want to elaborate a bit more? (optional)
Practice Class 8 Maths Quiz Questions and Answers
<
>
Support mcqexams.com by disabling your adblocker.
×
Please disable the adBlock and continue.
Thank you.
Reload page
