$$\Delta = \sqrt{s(s-a)(s-b)(s-c)}$$, $$2s = a+b+c$$

$$\Delta = \sqrt{(s-a)(s-b)(s-c)}$$

$$\Delta = \sqrt{s(s-a)(s-b)(s-c)}$$, $$s = a+b+c$$

$$\Delta =\sqrt{s(s-a)(s-b)(s-c)}, 2s=a+b+c$$

$$\Delta =\sqrt{s(s-a)(s-b)(s-c)}, s=a+b+c$$

MCQ Questions for Class 9 Maths Heron'S Formula Quiz 1

The sides of a triangle are

$3$ cm,

$4$ cm and

$5$ cm. Its area is .......

0%
$12 cm^2$
0%
$15 cm^2$
0%
$20 cm^2$
0%
$6 cm^2$

Explanation

By Heron's formula,
Area of the triangle

$= \sqrt{s(s - a)(s - b)(s -c )}$

We know,

$s = \displaystyle \frac{3 + 4 + 5}{2} = 6$

$\therefore$ Area of triangle

$= \sqrt{6(6 - 3)(6 - 4)(6 - 5)}$

$= \sqrt{6 \times 3 \times 2 \times1} = 6 cm^2$

State true or false:

$\triangle ABC$ ,

$AB= BC=CA=2a$ and AD is perpendicular to side BC, the area of

$\triangle ABC= a^2\sqrt2$ .

0%
True
0%
False

Explanation

Since all the sides of the triangle are equal, therefore it is an equilateral triangle.
Area of equilateral triangle

$=\cfrac{\sqrt{3}}{4} s^2$

$=\cfrac{\sqrt{3}}{4} (2a)^2$

$=a^2 \sqrt{3}$

Ans: False

Area of an equilateral triangle of side

$a$ units can be calculated by using the formula :

0%
$\sqrt{s^{2}(s-a)^{2}}$
0%
$(s-a)\sqrt{s^{2}(s-a)}$
0%
$\sqrt{s(s-a)^{2}}$
0%
$(s-a)\sqrt{s(s-a)}$

Explanation

We know that for an equilateral triangle,

$a=b=c$

By Heron's formula, we get

$\triangle=\sqrt{s(s-a)(s-b)(s-c)}$

So, area of given equilateral triangle

$=\sqrt { s\left( s-a \right) \left( s-a \right) \left( s-a \right) }$

$=\left( s-a \right) \sqrt { s\left( s-a \right) }$

The area of an equilateral triangle with side

$2 \sqrt{3}$ cm is

0%
$5.196 cm^2$
0%
$0.866 cm^2$
0%
$3.496 cm^2$
0%
$1.732 cm^2$

Explanation

Area of the triangle

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

$=3\sqrt{3}=3(1.732)=5.196cm^2$

The area of a triangle whose sides are 4 cm, 13 cm and 15 cm is

0%
$\displaystyle \sqrt{420}$ $\displaystyle cm^{2}$
0%
24 $\displaystyle cm^{2}$
0%
48 $\displaystyle cm^{2}$
0%
56 $\displaystyle cm^{2}$

Explanation

Given side of triangle is 4 cm , 13 cm and 15 cm

We know that area of triangle=

$\sqrt{s(s-a)(s-b)(s-c)}$

where $$s=\frac{a+b+c}{2} and a, b, c are three sides

$s=\frac{4+13+15}{2}= 16$

Then area of triangle=

$\sqrt{16(16-4)(16-13)(16-15)}=\sqrt{16\times 12\times 3\times 1}=\sqrt{576}=24 cm^{2}$

The sides of a triangle are

$4, 5$ and

$6$ cm. The area of the triangle is equal to

0%
$\displaystyle \frac{15}{4} cm^2$
0%
$\displaystyle \frac{15}{4}\sqrt 7 cm^2$
0%
$\displaystyle \frac{4}{15} \sqrt 7 cm^2$
0%
None of these

Explanation

Given that, a = 4 cm, b = 5 cm, and c = 6 cm
We know

$\displaystyle s = \frac{a + b + c}{2} = \frac{4+ 5 + 6}{2} = \frac{15}{2}$

$\therefore$ Area of the triangle

$= \sqrt{s(s - a) (s - b) (s - c)}$

$\displaystyle = \sqrt{\left ( \frac{15}{2} \right ) \left ( \frac{15}{2}-4 \right ) \left ( \frac{15}{2} - 5 \right ) \left ( \frac{15}{2} - 6 \right )}$

$= \displaystyle \sqrt{\frac{15}{2} \times \frac{7}{2} \times \frac{5}{2} \times \frac{3}{2}} = \frac{15}{4} \sqrt 7 cm^2$

If the sides of triangle are doubled then is area

0%
remains the same
0%
becomes doubled
0%
becomes three times
0%
become four times

Explanation

Let the sides of triangle is a,b,c

Then area of triangle =

$\sqrt{s(s-a)(s-b)(s-c)}$ where s=

$\frac{1}{2}(a+b+c)$

If sides are doubled the new sides 2a,2b,2c

Then S=

$\frac{1}{2}(2a+2b+2c)\Rightarrow2\times \frac{1}{2} (a+b+c)=2s$

Then area of new triangle =$$\sqrt{S(S-2a)(S-2b)(S-2c)}= \sqrt{2s(2s-2a)(2s-2b)(2s-2c)}=\sqrt{16}s(s-a)(s-b)(s-c)= 4\sqrt{s(s-a)(s-b)(s-c)}

Then area of new triangle =4 times of old triangle

Heron's formula is :

0%
$\Delta = \sqrt{s(s+a)(s+b)(s+c)}$
0%
$\Delta = \sqrt{(s-a)(s-b)(s-c)}$
0%
$\Delta = \sqrt{s(s-a)(s-b)(s-c)}$ , $s = a+b+c$
0%
$\Delta = \sqrt{s(s-a)(s-b)(s-c)}$ , $2s = a+b+c$

Explanation

${\textbf{Step -1: A triangle having side lengths as }}\mathbf{a,}{\textbf{ }}\mathbf b{\textbf{ and }}\mathbf c.{\text{ }}$

${\text{Area of triangle }} = \sqrt {s\left( {s - a} \right)\left( {s - b} \right)\left( {s - c} \right)}$

${\text{Where, s is the semi perimeter of the triangle;}}$

${\text{that is, }}s = \dfrac{{a + b + c}}{2}$

${\text{Thus, the Heron's formula is}}$ $\vartriangle = \sqrt {s\left( {s - a} \right)\left( {s - b} \right)\left( {s - c} \right)}$ ${\text{Where, }}2s = a + b + c$

${\textbf{Hence, option D is correct. }}$

Area of traingle ABC whose sides are 24m, 40m and 32m is-

0%
$96{m}^{2}$
0%
$384{m}^{2}$
0%
$43{m}^{2}$
0%
$192{m}^{2}$

Explanation

Area of triangle with sides $a$ , $b$ and $s$ $= \dfrac { a+b+c }{ 2 }$ is $\sqrt {s(s-a)(s-b)(s-c) } \$ .
For triangle with sides 24 m, 40 m and 32 m, $s = \dfrac { 24+40+32 }{ 2 }=48 m\$

Area of the triangle with sides 24 m, 40 m and 32 m $=\sqrt {48(48-24)(48-40)(48-32) } =\sqrt { 48\times 24\times 8\times 16 } =\sqrt { 24\times 2\times 24\times 8\times 8\times 2 } = 24\times 8\times 2 = 384 {m}^{2}$

Hence, option 'B' is correct.

The sides of a triangle are 5 cm, 12 cm and 13 cm. Then its area is

0%
$0.0024 m^2$
0%
$0.0026 m^2$
0%
$0.003 m^2$
0%
$0.0015 m^2$

Explanation

Using Heron's formula i.e.

$A=\sqrt{s(s-a)(s-b)(s-c)}$
where,

$a=5$ cm

$b=12$ cm

$c=13$ cm

$s=\displaystyle \frac{a+b+c}{2}=\frac {5+12+13}{2}=15$ cm

$A=\sqrt {15(15-5)(15-12)(15-13)}$

$A=\sqrt{15\times 10\times 3\times 2}$

$\therefore A=30$

$cm^2$

$\therefore A=0.003$

$m^2$

If s is the semi-perimeter of a

$\Delta A B C$ whose sides are a,b,c then s=.............?

0%
$a + b + c$
0%
$\dfrac { a + b + c } { 2 }$
0%
$\dfrac { a + b + c } { 3 }$
0%
$\dfrac { a + b + c } { 4 }$

Explanation

semiperimeter of a triangle is given as half of the perimeter which means half of the sum of the lengths of all three sides

Hence, in given question

$\triangle ABC$

semiperimeter

$=s=\cfrac{a+b+c}{2}$

Find the area of a triangle whose two sides are

$18$ cm and

$10$ cm and the perimeter is

$42$ cm.

0%
$83.67$ cm $^2$
0%
$46.12$ cm $^2$
0%
$77.15$ cm $^2$
0%
$69.69$ cm $^2$

Explanation

Let

$a$ ,

$b$ and

$c$ be the sides of triangle.

Perimeter

$=a+b+c$

$=42 \text{ cm}$

$a=18$ ,

$b=10$

$a+b+c=42$

$\therefore$

$c=42-b-a = 42 - 18-10$

$\therefore c=14$

Area of triangle

$= \sqrt{s (s -a) (s -b) ( s -c)}$

where,

$s=\dfrac{a+b+c}{2}=21 \text{ cm}$

$\therefore$ Area of triangle

$=\sqrt {21(21-18)(21-10)(21-14)}$

$\therefore$ Area of triangle

$=69.69 \text{ cm}^2$

The sides of a triangle are 56 cm, 60 cm and 52 cm long. Then the area of the triangle is

0%
$1322 cm^2$
0%
$1311 cm^2$
0%
$1344 cm^2$
0%
$1392 cm^2$

Explanation

$s=\dfrac{a+b+c}{2}$
By Heron's formula,Area of the triangle=

$\sqrt{s(s-a)(s-b)(s-c)}$
Where a,b,c are sides of the triangle.

$s=\dfrac{56+60+52}{2}=84$
Area of the triangle

$=\sqrt{84(84-56)(84-60)(84-52)}=\sqrt{1806336}=1344cm^2$

Find the cost of laying grass in a triangular field of sides

$50\>m, 65\>m$ and

$65\> m$ at the rate of Rs

$7$ per

$m^2$ .

0%
Rs $10500$
0%
Rs $11375$
0%
Rs $10000$
0%
Rs $21000$

Explanation

Sides of the triangle are

$a=50\> m, b=65\> m, c=65\> m$
Area of triangle, by Heron's formula

$=\sqrt{s(s-a)(s-b)(s-c)}$
where,

$s= \dfrac{a+b+c}{2}$

$s = \dfrac{50 + 65 + 65}{2}$

$s = 90$
Area of triangle =

$\sqrt{90(40) (25)(25)}$
Area of triangle =

$1500\>m^2$

Cost of laying grass

$=$ Area

$\times 7$
Cost of laying grass

$=1500 \times 7$
Cost of laying grass

$=$ Rs

$10500$

The area of an isosceles triangle having base 2 cm and the length of one of the equal sides 4 cm, is

0%
$\sqrt{15} cm^2$
0%
$\sqrt{\dfrac{15}{2}}cm^2$
0%
$2 \sqrt{15} cm^2$
0%
$4 \sqrt{15}cm^2$

Explanation

$s=\dfrac{4+4+2}{2}=5$

Area of the triangle

$\Delta =\sqrt{s(s-a)(s-b)(s-c)}\\=\sqrt{5(5-4)(5-4)(5-2)}=\sqrt{15}cm^2$

In a triangle, the sides are given as 11 cm, 12 cm and 13 cm. The length of the altitude corresponding to the side having length 12 cm is

0%
10.25 cm
0%
11.25 cm
0%
12.25 cm
0%
9.25 cm

Explanation

Sides of triangle are

$11,12,13$
Area of triangle =

$\sqrt{s(s-a)(s-b)(s-c)}$

$s = \dfrac{a + b+ c}{2}$

$s = \dfrac{11 + 12 + 13}{2}$

$s = 18$
Area of triangle =

$\sqrt{18(7) (6) (5)}$
Area of triangle =

$\sqrt{18(7) (6) (5)} = 61.5$

Area of triangle =

$\dfrac{1}{2} (base) (altitude)$
61.5 =

$\dfrac{1}{2} \times 12 \times (altitude)$
Altitude =

$10.25$

The sides of a

$\Delta$ are

$7 cm, 24 cm$ and

$25 cm$ . Find its area.

0%
$168 cm^2$
0%
$84 cm^2$
0%
$87.5 cm^2$
0%
$300 cm^2$

Explanation

Let

$a=7$ cm

$b=24$ cm

$c=25$ cm

$S=\cfrac { a+b+c }{ 2 } \\ =\cfrac { 7+24+25 }{ 2 } \\ =28\ cm$

Now, area of

$\triangle =\sqrt { S(S-a)(S-b)(S-c) }$

$=\sqrt { 28(28-7)(28-24)(28-25) } \\ =\sqrt { 24\times 24\times 4\times 3 } \\ =\sqrt { 4\times 7\times 7\times 3\times 4\times 3 } \\ =4\times 7\times 3\\ =84\ { cm }^{ 2 }$

The cost of levelling the ground in the form of a triangle having the sides 51 m, 37 m and 20 m at the rate of Rs

$3$ per

$m^2$ is

0%
Rs. 918
0%
Rs. 818
0%
Rs. 718
0%
Rs. 618

Explanation

$s=\dfrac{51+37+20}{2}=\dfrac{108}{2}=54$

Area of the triangle

$=\sqrt{54(54-51)(54-37)(54-20)}$

$=\sqrt{54(3)(17)(34)}=\sqrt{93636}$

Area

$=$

$306 m^2$ .

Cost of levelling at

$Rs. 3$ per

$m^2=306(3)=$ Rs.

$918$

Area of triangle ABC whose sides are 24 m, 40 m and 32 m, is -

0%
$90 m^2$
0%
$384 m^2$
0%
$43 m^2$
0%
$192 m^2$

Explanation

$S=\dfrac {24+40+32}{2}=\dfrac {96}{2}=48 m$

Area

$=\sqrt {48\times (48-24)(48-40)(48-32)}$

$=\sqrt {48\times 24\times 8\times 16}$

$=\sqrt {24\times 2\times 24\times 8\times 16}$

$=24\times 16=384 m^2$

The area of an isosceles triangle whose base is

$'b'$ and equal sides are of length

$'a'$ is:

0%
$\dfrac{a}{4}\sqrt{4b^2-a^2}$
0%
$\dfrac{b}{4}\sqrt{4a^2-b^2}$
0%
$\dfrac{b}{2}\sqrt{4b^2-a^2}$
0%
$\dfrac{a}{2}\sqrt{4a^2-b^2}$

Explanation

In an isosceles triangle, the altitude bisects the base.

$Altitude=\sqrt { (side\quad length\quad of\quad equal\quad sides)^{ 2 }+\left( \cfrac { base }{ 2 } \right) ^{ 2 } } \\ =\sqrt { { a }^{ 2 }-\cfrac { { b }^{ 2 } }{ 4 } }$

Area of isosceles triangle :

$=\cfrac { 1 }{ 2 } \times altitude\times base\\ =\cfrac { 1 }{ 2 } \times \sqrt { { a }^{ 2 }-\cfrac { { b }^{ 2 } }{ 4 } } \times b\\ =\cfrac { 1 }{ 4 } \times b\sqrt { 4a^{ 2 }-b^{ 2 } }$

The sides of a triangular plot are in the ratio

$4 : 5 : 6$ and its perimeter is

$150\ \mathrm{cm}$ . Then the sides are:

0%
$4\ \mathrm{cm}, 5\ \mathrm{cm}, 6\ \mathrm{cm}$
0%
$40\ \mathrm{ cm}, 50\ \mathrm{ cm}, 60\ \mathrm{ cm}$
0%
$8\ \mathrm{cm}, 10\ \mathrm{cm}, 12\ \mathrm{cm}$
0%
$120\ \mathrm{cm}, 150\ \mathrm{cm}, 180\ \mathrm{cm}$

Explanation

Let the sides of the triangle be

$4x,5x,6x$ , respectively.

Perimeter of triangle

$=4x+5x+6x$

$4x+5x+6x = 150\ \mathrm{cm}$

$15x = 150\ \mathrm{cm}$

$\Rightarrow x=10\ \mathrm{cm}$

$4x = 4\times 10 \ \mathrm{cm} = 40 \ \mathrm{cm}$

$5x = 5\times 10 \ \mathrm{cm} = 50 \ \mathrm{cm}$

$6x = 6\times 10 \ \mathrm{cm} = 60 \ \mathrm{cm}$

Therefore, the sides are

$40\ \mathrm{cm}, \ 50 \ \mathrm{cm},$ and

$60 \ \mathrm{cm}$ .

Two sides of a triangle are

$13 cm$ and

$14 cm$ and its semi perimeter is

$18 cm$ . Then third side of the triangle is :

0%
$12 cm$
0%
$11 cm$
0%
$10 cm$
0%
$9 cm$

In figure,

$Ar(||gm ABCD)$ is:

0%
$10 cm$
0%
$20 cm$
0%
$10\sqrt{3}cm^2$
0%
$20\sqrt{3}cm^2$

Explanation

Area of parallelogram ABCD

$=2 \times area \ of \triangle ABC$

Now,

$S=\cfrac { a+b+c }{ 2 } =\cfrac { 5+7+8 }{ 2 } =10\ cm$

Area of

$\triangle ABC=\sqrt { s(s-a)(s-b)(s-c) }$

$=\sqrt { 10(10-5)(10-7)(10-8) } \\ =\sqrt { 10\times 5\times 3\times 2 } \\ =\sqrt { 300 } \\ =\sqrt { 3\times 100 } \\ =10\sqrt { 3 } \quad { cm }^{ 2 }$

Area of parallelogram ABCD

$=2\times \triangle ABC\\ =2\times 10\sqrt { 3 } \\ =20\sqrt { 3 } \ { cm }^{ 2 }$

Two sides of a triangle are

$16 cm$ and

$14 cm$ and its semi-perimeter is

$18 cm$ . Then the third side of the triangle is:

0%
$12 cm$
0%
$11 cm$
0%
$10 cm$
0%
$6 cm$

Explanation

Semiperimeter,

$S=18 \ cm$

Let the side of the triangle be

$a,b,c$ such that

$a=16 \ cm ,\ b=14 \ cm \\$

$S=\cfrac { a+b+c }{ 2 } \\ =\cfrac { 16+14+c }{ 2 } \\ 18=\cfrac { 30+c }{ 2 } \\ c=36-30=6 \ cm$

Therefore, third side of triangle is

$6 \ cm$

Area of an equilateral triangle of side 'a' units can be calculated by using the formula :

0%
$\sqrt{s^2(s-a)^2}$
0%
$(s-a)\sqrt{s^2(s-a)}$
0%
$\sqrt{s(s-a)^2}$
0%
$(s-a)\sqrt{s(s-a)}$

Explanation

When the sides of triangle are given we solve the area of the triangle using Heron's Formula which is

area of a triangle

$=\sqrt { s(s−a)(s−b)(s−c) }$

where

$s$ is the semi-perimeter and

$s=\dfrac{(a+b+c)}{2}$

Here

$a=b=c$ ,

Therefore, the formula becomes :

$Area\,=\,\sqrt { s(s−a)\times(s-a)\times(s-a) }$

$\Rightarrow Area\,=\,\sqrt{s(s-a)\times(s-a)^2}$

$\Rightarrow Area\,=\,(s-a)\sqrt{s(s-a)}$

Hence option D is correct.

Heron's formunla is :

0%
$\Delta = \sqrt{s(s+a)(s+b)(s+c)}$
0%
$\Delta = \sqrt{(s-a)(s-b)(s-c)}$
0%
$\Delta = \sqrt{s(s-a)(s-b)(s-c)} , s = a+b+c$
0%
$\Delta = \sqrt{s(s-a)(s-b)(s-c)} , 2s = a+b+c$

Heron's formula is :

0%
$\Delta =\sqrt{s(s+a)(s+b)(s+c)}$
0%
$\Delta =\sqrt{s(s-a)(s-b)(s-c)}$
0%
$\Delta =\sqrt{s(s-a)(s-b)(s-c)}, s=a+b+c$
0%
$\Delta =\sqrt{s(s-a)(s-b)(s-c)}, 2s=a+b+c$

Explanation

${\textbf{Step -1: A triangle having side lengths as }}\mathbf{a,}{\textbf{ }}\mathbf b{\textbf{ and }}\mathbf c.{\text{ }}$

${\text{Area of triangle }} = \sqrt {s\left( {s - a} \right)\left( {s - b} \right)\left( {s - c} \right)}$

${\text{Where, s is the semi perimeter of the triangle;}}$

${\text{that is, }}s = \dfrac{{a + b + c}}{2}$

${\text{Thus, the Heron's formula is}}$ $\vartriangle = \sqrt {s\left( {s - a} \right)\left( {s - b} \right)\left( {s - c} \right)}$ ${\text{Where, }}2s = a + b + c$

${\textbf{Hence, option D is correct. }}$

The area of a triangle whose sides are

$13$ cm,

$14$ cm and

$15$ cm is :

0%
$42 ~cm^{2}$
0%
$86~cm^{2}$
0%
$84~cm^{2}$
0%
$100~cm^{2}$

Area of an equilateral triangle of side a units can be calculated by using the formula :

0%
$\sqrt{s^2(s-a)^2}$
0%
$(s-a)\sqrt{s^2(s-a)}$
0%
$\sqrt{s(s-a)^3}$
0%
$(s-a)\sqrt{s(s-a)}$

The lengths of four sides and a diagonal of the given quadrilateral are indicated in the diagram. If

$A$ denotes the area and

$l$ the length of the other diagonal, then

$A$ and

$l$ are respectively

0%
$12\sqrt{6},4\sqrt{6}$
0%
$12\sqrt{6},5\sqrt{6}$
0%
$6\sqrt{6},4\sqrt{6}$
0%
$6\sqrt{6},5\sqrt{6}$

Explanation

For $\Delta ABC$ , a $=$ 6 cm, b $=$ 5, c $=$ 7 cm

$\therefore s = \dfrac{6+5+7}{2} = 9$ cm

$\therefore$ Area of $\Delta ABC = \sqrt{s(s-a)(s-b)(s-c)}$

$= \sqrt{9\times (9-6)(9-5)(9-7)}$

$= \sqrt{9\times 3\times 4\times 2}$

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

Thus, area of quadrilateral $= 2\times$ Area of $\Delta ABC$
$= 12\sqrt{6}$ sq cm

From congruency of triangles, it can be proved that $AE=ED= \dfrac{1}{2}AD$ , and $AE \perp BC$

$AE = \dfrac{1}{2}AD = \dfrac{l}{2}$

Area of $\Delta ABC = \dfrac{1}{2}\times BC\times AE$

or $6\sqrt{6} = \dfrac{1}{2}\times 6\times \dfrac{l}{2}$

or $l = 4\sqrt{6}$

CBSE Questions for Class 9 Maths Heron'S Formula Quiz 1 - MCQExams.com

0:0:3

Practice Class 9 Maths Quiz Questions and Answers

CBSE Questions for Class 9 Maths Heron'S Formula Quiz 1 - MCQExams.com

0:0:3

Practice Class 9 Maths Quiz Questions and Answers

Support mcqexams.com by disabling your adblocker.