MCQ Questions for Class 9 Maths Heron'S Formula Quiz 4

A triangle and a parallelogram have the same base and the same area. If the sides of the triangle are

$15 cm, 14 cm, 13 cm$ , and the parallelogram stands on the base of

$15$ cm. Find the height of the parallelogram.

0%
$5.9 cm$
0%
$6.8 cm$
0%
$6.5 cm$
0%
$5.6 cm$

Explanation

Area of triangle =

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

$S=\dfrac { a+b+c }{ 2 } =\dfrac { 15+14+13 }{ 2 } =21cm$

$\therefore \quad$ area of

$\triangle =\sqrt { 21\left( 21-15 \right) \left( 21-14 \right) \left( 21-13 \right) } =\sqrt { 21\times 6\times 7\times 8 }$

$\sqrt { 3\times 7\times 3\times 2\times 7\times 4\times 2 } =2\times 2\times 3\times 7$

${ 84cm }^{ 2 }$

Now, area of

$\parallel$ gm = area of

$\triangle$ =

$b\times h=84{ cm }^{ 2 }$

$\Rightarrow \quad 15\times h=84\Rightarrow h=\dfrac { 84 }{ 15 } =5.6cm$

$\therefore$ Height of

$\parallel$ gm =

$5.6cm$

What is the area of a triangle whose sides are

$13\ cm, 14\ cm$ and

$15\ cm$ ?

0%
$84\ sq.cm$
0%
$64\ sq.cm$
0%
$825\ sq.cm$
0%
$105\ sq.cm$

Explanation

$S=\dfrac { a+b+c }{ 2 } =\dfrac { 13+14+15 }{ 2 } =21cm$

area of

$\triangle =\sqrt { s\left( s-a \right) \left( s-b \right) \left( s-c \right) } =\sqrt { 21\left( 21-13 \right) \left( 21-14 \right) \left( 21-15 \right) }$

$=\sqrt { 21\times 8\times 7\times 6 } =\sqrt { 3\times 7\times 2\times 2\times 2\times 7\times 2\times 3 } =2\times 2\times 3\times 7$

$={ 84cm }^{ 2 }$

Find the number of trees that can be planted in a triangular ground having sides

$51m, 37m$ and

$20m$ , when each tree occupies

$6m^2$ of space:

0%
$54$
0%
$51$
0%
$82$
0%
$43$

Explanation

Area of

$\triangle =\sqrt { s\left( s-a \right) \left( s-b \right) \left( s-c \right) } ,\quad s=\dfrac { a+b+c }{ 2 } =\dfrac { 51+37+20 }{ 2 } =54m$

$\therefore \quad$ Area of triangle ground

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

$=\sqrt { 54\times 3\times 17\times 34 } =306{ m }^{ 2 }$

Number of trees

$=\dfrac { 306 }{ 6 } =51$

The perimeter of a triangular field is

$420$ m and its sides are in the ratio

$6: 7: 8$ Find area of the triangular field.

0%
$2100$ sq m
0%
$2100\sqrt{15}$ sq m
0%
$2000$ sq m
0%
$2000\sqrt{15}$ sq m

Explanation

$\textbf{Step - 1: Finding the sides}$

$\text{Given that the sides are in the ratio }6:7:8$

$\implies a=6x,b=7x,c=8x$

$\text{Perimeter}=a+b+c$

$\implies 420=6x+7x+8x=21x$

$\implies x=20$

$\implies a=120,b=140,c=160$

$\textbf{Step - 2: Calculating area}$

$s=\dfrac{a+b+c}2=210$

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

$\implies \text{Area} = \sqrt{210\times90\times70\times50}$

$\implies \text{Area}=2100\sqrt{15}$

$\textbf{Thus, the area of the given field is }\mathbf{2100\sqrt{15}}\textbf{ sq. m.}$

$\triangle ABC, AB = 6\ cm, BC = 7\ cm$ and

$AC = 5\ cm$ . Find the area of

$\triangle ABC$

0%
$6\sqrt {6}\ cm^{2}$
0%
$6\sqrt {3}\ cm^{2}$
0%
$6\sqrt {2}\ cm^{2}$
0%
$9\sqrt {6}\,cm^{2}$

Explanation

Using Heron's formula,

Area of a triangle

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

where

$s=$ semiperimeterof the triangle

$a=$ length of side

$BC$

$b=$ length of side

$AC$

$c=$ length of side

$AB$

$\triangle ABC,\ a=7,\ b=5,\ c=6$

Semiperemeter of

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

$=\dfrac{7+5+6}{2}$

$=9$

Hence, the area of

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

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

$={6\sqrt{6}\ cm^{2}}$

Find the area of the parallelogram with sides

$11 cm$ and

$13 cm$ and one of the diagonals as

$16 cm.$

0%
$23\sqrt{32}$ sq. cm
0%
$25\sqrt{35}$ sq. cm
0%
$24\sqrt{35}$ sq. cm
0%
None of these

Explanation

$\parallel$ gm ABCD =

$2\times ar\triangle ABC$

$\triangle$ ABC =

$\sqrt { s\left( s-a \right) \left( s-b \right) \left( s-c \right) } ,\quad s=\dfrac { a+b+c }{ 2 } =\dfrac { 16+13+11 }{ 2 } =20cm$

$\therefore$ ar

$\triangle$ ABC =

$\sqrt { 20\left( 20-16 \right) \left( 20-13 \right) \left( 20-11 \right) } =\sqrt { 20\times 4\times 7\times 9 } =\sqrt { 5\times 4\times 4\times 7\times 9 } =4\times 3\sqrt { 35 } =12\sqrt { 35 } { cm }^{ 2 }$

$\therefore$ ar

$\triangle$ ABCD =

$2\times 12\sqrt { 35 } =24\sqrt { 35 } { cm }^{ 2 }$

Find the area of the quadrilateral whose sides are

$9 m ,40 m, 28 m, 15 m$ . The angle between first two sides is

$90^o$ .

0%
$348$ sq. m
0%
$315$ sq. m
0%
$306$ sq. m
0%
None of these

Explanation

Area of quadrilateral

$ABCD$ = ar

$[\triangle ABC]$ + ar

$[\triangle ACD]$

$\triangle ABC$ :

$ar\ [\triangle ABC] = \dfrac { 1 }{ 2 } \times BC\times AB\\=\dfrac { 1 }{ 2 } \times 40m\times 9m=180{ m }^{ 2 }$

Using pythagoras theorem,

$AC= \sqrt { { AB }^{ 2 }+{ BC }^{ 2 } } \\ =\sqrt { { \left( 9 \right) }^{ 2 }+{ \left( 40 \right) }^{ 2 } }\\ =\sqrt { 1681 } =41m$

Using Heron's formula,

$ar [\triangle ACD] = \sqrt { s\left( s-a \right) \left( s-b \right) \left( s-c \right) } \\\quad s=\dfrac { a+b+c }{ 2 } \\\ \ \ \quad =\dfrac { 41+28+15 }{ 2 } \\\quad \ \ \ =\dfrac { 84 }{ 2 } =42m$

$ar [\triangle ACD ]= \sqrt { 42\left( 42-41 \right) \left( 42-28 \right) \left( 42-15 \right) }\\ \quad \quad \quad \ \ \ \quad =\sqrt { 42\times 1\times 14\times 27 } =126{ m }^{ 2 }$

Therefore, area of quadrilateral

$ABCD$

$=180 + 126$

$=306\ m^{2}$

Find the area of the parallelogram with sides

$9 cm$ and

$11 cm$ and one of the diagonals as

$14 cm.$

0%
$22\sqrt{15}$
0%
$24\sqrt{17}$
0%
$21\sqrt{17}$
0%
None of these

Explanation

Area of

$\parallel$ gm ABCD = 2

$\times$ area

$\triangle$ ABC

$\triangle$ ABC =

$\sqrt { s\left( s-a \right) \left( s-b \right) \left( s-c \right) } ,\quad \& \quad s=\dfrac { a+b+c }{ 2 } =\dfrac { 14+11+19 }{ 2 } =17cm$

$\therefore$ ar

$\triangle$ ABC =

$\sqrt { 17\left( 17-14 \right) \left( 17-11 \right) \left( 17-9 \right) } =\sqrt { 17\times 3\times 6\times 8 } =12\sqrt { 17 } { cm }^{ 2 }$

Therefore, ar

$\parallel$ gm ABCD = 2

$\times$

$12\sqrt { 17 } =24\sqrt { 17 } { cm }^{ 2 }$

The parallel sides of a parallelogram are

$60$ m and

$25$ m and one of the diagonals is

$65$ m. Find the area of the parallelogram,

0%
$1430$ sq m
0%
$1576$ sq m
0%
$1500$ sq m
0%
None of these

Explanation

The area of parallelogram

$ABCD = 2 \times Ar \triangle ABC$

Now,

$Ar\ \triangle ABC =\sqrt { s\left( s-a \right) \left( s-b \right) \left( s-c \right) } \quad \\ s=\dfrac { a+b+c }{ 2 } \\=\dfrac { 65+60+25 }{ 2 } \\=75m$

Thus,

$\therefore Ar \triangle ABC = \sqrt { 75\left( 75-65 \right) \left( 75-60 \right) \left( 75-25 \right) } \\=\sqrt { 75\times 10\times 15\times 50 } \\=750{ m }^{ 2 }$

Thus, area of parallelogram

$=2\times 750\ m^2$

$=1500\ m^2$

Find the area of an equilateral triangle each of whose side is 4 cm long?

0%
$2\sqrt { 3 }$
0%
$3\sqrt { 3 }$
0%
$5\sqrt { 3 }$
0%
$4\sqrt { 3 }$

Explanation

solution:

$Area = \dfrac{\sqrt{3}}{4} r^2$

$Area = \dfrac{\sqrt{3}}{4} 4^2$

$Area = \dfrac{\sqrt{3}}{1} 4$

hence the correct opt: D

In the given figure, the area of the

$\triangle {ABC}$ is

0%
$13.24{cm}^{2}$
0%
$12.28{cm}^{2}$
0%
$11.32{cm}^{2}$
0%
$15.37{cm}^{2}$

Explanation

We know that the area of the triangle using heron's formula is given by

Area

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

So,

$S = \dfrac{8+4+11}{2} = \dfrac{23}{2} = 11.5$

Area

$= \sqrt{11.5(11.5-11)(11.5-8)(11.5-4)}$

$=\sqrt{11.5\times 0.5\times 3.5\times 7.5}$

$= 12.28\text{ cm}^{2}$

If the sides of a triangle are

$\dfrac{y}{z} + \dfrac{z}{x},\dfrac{z}{x} + \dfrac{x}{y}$ and

$\dfrac{x}{y} + \dfrac{y}{z}$ then its area is

0%
xyz
0%
$\sqrt {\dfrac{x}{y} + \dfrac{y}{z} + \dfrac{z}{x}}$
0%
$\sqrt {xyz}$
0%
$\dfrac{{{x^2}{y^2}{z^2}}}{2}$

A triangular park in a city has dimensions 100m X 90m X110m.A contract is given to a company for planting grass in the park at the rate of Rs.4000 per hectare.Find the amount to be paid to the company.(Take

$\sqrt{2}=1.414$ )

0%
Rs. $4532.90$
0%
Rs. $4242$
0%
Rs. $1696.80$
0%
Rs. $1000$

Explanation

$Heron's\ Formula\ :$

Area of a triangle =

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

where

$s$ is the semi-perimeter and

$a,\ b$ and

$c$ are the sides of the triangle.

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

Here, for the triangular park,

$a=100\ m,\ \ b=90\ m,\ \ c=110\ m$ and

$s=\dfrac{100+90+110}{2}=150\ m$

Area of the triangular park =

$\Delta=\sqrt{150\times50\times60\times40}=3000\sqrt{2}\ m^2=0.3\sqrt{2}\ ha=0.4242\ ha$

Rate of planting grass =

$Rs.\ 4000\ /ha$

So, the amount to be paid =

$4000\times0.4242=Rs.\ 1696.80$ .

$[C]$

Find the area of a triangle whose two sides are

$8$ cm and

$11$ cm and the perimeter is

$32$ cm.

0%
$8cm^2$
0%
$8\sqrt{30}cm^2$
0%
$\dfrac{8}{\sqrt{30}}cm^2$
0%
$24cm^2$

Explanation

Let third side be

$x$

Perimeter

$=8+11+x$

$19+x=32$

$x=13$

$s=\cfrac { 32 }{ 2 } =16$

Heron's formula

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

$=\sqrt { 16\left( 16-8 \right) \left( 16-11 \right) \left( 16-13 \right) }$

$=\sqrt { 16\times 8\times 5\times 3 }$

$=8\sqrt { 30 }\ cm^2$

One side of an equilateral triangle measures

$8$ cm. Find its area using Heron's formula. What is its altitude

0%
$12\sqrt 3\ \text{cm}^2, 3\sqrt 3$ cm
0%
$16\sqrt 3\ \text{cm}^2, 4\sqrt 3$ cm
0%
$14\sqrt 3\ \text{cm}^2, \dfrac {7}{2}\sqrt 3$ cm
0%
$20\sqrt 3\ \text{cm}^2, 5\sqrt 3$ cm

Explanation

According to the Heron's formula, Area (A) of the triangle having sides

$a,b,c$ units is

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

where

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

For the given triangle,

$a=b=c=8cm$

$s=\cfrac{8+8+8}{2}=12$

$A=\sqrt{12(12-8)(12-8)(12-8)}$

$A=\sqrt{3\times 4\times 4^{3}}$

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

$=16\sqrt{3}$

Area of the triangle =

$\cfrac{1}{2}\times base \times altitude=16\sqrt{3}$

$\cfrac{1}{2}\times 8 \times$ altitude

$=16\sqrt{3}$

Altitude

$=\cfrac{16\sqrt{3}}{4}=4\sqrt{3}$

The perimeter of a triangle is

$300m$ . If its sides are in the ratio

$3:5:7$ . Find the area of the triangle.

0%
$1500\sqrt{5}m^2$
0%
$1500m^2$
0%
$1500\sqrt{3}m^2$
0%
$1100\sqrt{3}m^2$

Explanation

Let, 1 side be =

$3x$

2 side be =

$5x$

3 side be =

$7x$

According to question,

$3x+5x+7x=300$

$\implies x= 20$

So, sides are

$60 ,100, 140$

Area =

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

$s=semi-perimeter$

$s=150$ ,

$a = 60$ ,

$b = 100$ ,

$c = 140$

Area=

$\sqrt {150(150-60)(150-100)(150-140)}$

$\implies \sqrt{150(90)(50)(10)}$

$\implies1500\sqrt3\ m^2$

Find the area of a triangle whose sides are 9 cm, 12 cm, and 15 cm.

0%
$34cm^2$
0%
$44cm^2$
0%
$54cm^2$
0%
$55cm^2$

Explanation

Given sides of the triangle

$a=9$ cm

$b=12$ cm

$c=15$ cm

The given triangle is a right angle triangle

because

$9^2+12^2=15^2$

$\therefore$ Area

$=\dfrac{1}{2}\times 9\times 12=54$ sq cm

Area

$=54$ sq cm.

The perimeter of a triangle field is

$450\ m$ and its sides are in the ratio

$25:17:12$ . Find the area of the triangle.

0%
$2250m^2$
0%
$6250m^2$
0%
$5250m^2$
0%
$6050m^2$

Explanation

Let, 1 side be =

$25x$

2 side be =

$17x$

3 side be =

$12x$

According to question,

$25x+17x+12x=450$

$\implies x= \dfrac{450}{54}$

$\implies x=\dfrac{25}{3}$

So, sides are

$\dfrac{625}{3} , \dfrac{425}{3}, \dfrac{100}{1}$

Area =

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

$s=semi-perimeter$

$s= 225$ ,

$a = \dfrac{625}{3}$ ,

$b = \dfrac{425}{3}$ ,

$c = \dfrac{100}{1}$

Area=

$\sqrt {225(225-\dfrac{625}{3})(225-\dfrac{425}{3})(225-100)}$

$\implies \sqrt{225(\dfrac{50}{3})(\dfrac {250}{3})(125)}$

$\implies6250\ m^2$

Find the area of triangle of sides

$6cm ,8cm ,10cm$

0%
$24cm^2$
0%
$12cm^2$
0%
$30cm^2$
0%
$none$

Explanation

The sides are

$6cm,8cm,10cm$

$s=\dfrac {6+8+10}2=\dfrac {24}2=12$

The area is given as

$\sqrt {s(s-a)(s-b)(s-c)}\\\sqrt {12(6)(4)(2)}\\\sqrt {576}=24cm^2$

Find the area of a triangle whose sides are respectively

$150\ cm$ ,

$120\ cm$ , and

$200\ cm.$

0%
$8545.77\text{ sq. cm}$
0%
$8966.57\text{ sq. cm}$
0%
$9257.66\text{ sq. cm}$
0%
$5627.66 \text{ sq. cm}$

Explanation

Let

$a=150\ cm,\ b=120\ cm$ and

$c=200\ cm.$

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

$\Rightarrow s=\dfrac{150+120+200}{2}$

$\Rightarrow s=235\ cm$

Now, by using the heron's formula,

Area

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

$=\sqrt{235(235-150)(235-120)(235-200)}$

$=\sqrt{235(85)(115)(35)}$

$=\sqrt{80399375}$

$=8966.57\text{ sq. cm}$

If every side of a triangle is doubled, the area of the new triangle is

$K$ times the area of the old one. Find the value of

$K$ .

0%
$\sqrt{2}$
0%
$2$
0%
$3$
0%
$4$

Explanation

Let, the sides of the original triangle be

$a,b$ and

$c$

so, the semi-perimeter,

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

Original area

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

Now every side is doubled

So, the new sides are

$2a, 2b$ and

$2c$

new semi-perimeter,

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

New area

$=\sqrt{s'(s'-2a)(s'-2b)(s'-2c)}$

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

$=\sqrt{2.2.2.2s(s-a)(s-b)(s-c)}$

$=4\sqrt{s(s-a)(s-b)(s-c)}$

$=4\times Original\ Area$

Hence, the value of

$K$ is

$4$ .

Use Heron's formula to find the area of a triangle of lengths

$5, 7$ and

$8$ .

0%
$10\sqrt{3}$
0%
$20\sqrt{3}$
0%
$2\sqrt{6}$
0%
$20\sqrt{6}$

Three sides of a triangular field are 20m, 21m, and 29m long, respectively. The area of the field is :

0%
$215{ m }^{ 2 }$
0%
$230{ m }^{ 2 }$
0%
$210{ m }^{ 2 }$
0%
None of these

Explanation

Given: Sides of the triangular field;

$a=20m, b=21m, c=39m$

Now, Semiperimeter

$S = \dfrac{a+b+c}{2} = \dfrac{20+21+29}{2} = 35m$

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

$= \sqrt{35\times (35-20)\times (35-21)\times (35-29)}$

$= \sqrt{35\times 15\times 14\times 6}$

$=210 sq.m$

The area of a triangle whose sides are 15 m, 16 m and 17 m is :

0%
$24\sqrt { 4 } { m }^{ 2 }$
0%
$24\sqrt { 3 } { m }^{ 2 }$
0%
$24\sqrt { 21 } { m }^{ 2 }$
0%
None of these

Explanation

The sides of the triangle are

$a = 15 \text{ m }, b = 16 \text{ m }, c = 17 \text{ m }$

Area of the triangle is given by

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

$= \cfrac {15 + 16 + 17}{2}$

$= \cfrac {48}{2}$

$= 24 \text{ m}$

Area

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

$= \sqrt {24 (25 - 15) (24 - 16) (24 - 17)}$

$= \sqrt {24\times 9\times 8\times 7}$

$= \sqrt {\underline {8}\times \underline {3\times 3}\times 3\times \underline {8}\times 7}$

$= 24\sqrt {21} \text{ sq. m.}$

The sides of a triangle are

$7$ cm,

$9$ cm and

$14$ cm. Its area is

0%
$12\sqrt 5 \ \text{cm}^2$
0%
$12\sqrt 3 \ \text{cm}^2$
0%
$24\sqrt 5 \ \text{cm}^2$
0%
$63 \ \text{cm}^2$

Explanation

Area of triangle =

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

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

Given

$a= 7$ cm

$b= 9$ cm

$c= 14$ cm

$S= \dfrac{7+9+14}{2}$

$=\dfrac{30}{2}$

$S= 15 cm$

$\therefore$ Area of triangle =

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

$=\sqrt{15(15-7)(15-9)(15-14)}\\$

$=\sqrt{15.8.6.1}\\$

$=\sqrt{720}\\$

$=\sqrt{5\times 144}\\$

$=\sqrt{12\times 12\times 5}\\$

$=12\sqrt{5}cm^{2}\\$

$\therefore$ Area of triangle =

$12.\sqrt{5}\ cm^{2}$

So, the answer is A.

$12.\sqrt{5}\ cm^{2}$

The sides of a triangle are in the ratio

$3:5:7.$ If the perimeter of the triangle is

$60$ cm, then its area

0%
$60\sqrt 3 \ \text{cm}^2$
0%
$30\sqrt 3\ \text{cm}^2$
0%
$15\sqrt 3 \ \text{cm}^2$
0%
$120\ \text{cm}^2$

Explanation

Given sides in the ratio

$3:5:7$

perimeter of the triangle

$=60cm$

$3x+5x+7x=60$

$\Rightarrow 15x=60$

$\Rightarrow x=60/15$

$\Rightarrow x=4$

$3x=3\times 4=12$ cm

$5x=5\times 4=20$ cm

$7x=7\times 4=28$ cm

Let

$S$ be the semiperimeter of given triangle.

$S=\dfrac{1}{2}(a+b+c)$

$=\dfrac{1}{2}\cdot 60$

$=30$

Now use formula foe area of triangle

Area

$=\sqrt{30(30-12)(30-20)(30-28)}$

$=\sqrt{30\cdot18\cdot10\cdot2}$

$=\sqrt{60\cdot180}$

$=\sqrt{10800}$

$=\sqrt{3600\cdot3}$

$=60\sqrt{3}cm^2$

$\therefore$ Area of triangle

$=60\sqrt{3}cm^2$ .

The percentage increase in the area of a triangle if its each side is doubled, is

0%
$200\%$
0%
$400\%$
0%
$300\%$
0%
$500\%$

If the sides of triangle are 4, 5 and 6 cm. Then the area (in sq cm) of triangle is

0%
$\dfrac{\pi}{4}$
0%
$\dfrac{\pi}{4} \sqrt{7}$
0%
$\dfrac{4}{15}$
0%
$\dfrac{4}{15} \sqrt{7}$
0%
$\dfrac{15}{4} \sqrt{7}$

Explanation

Given, triangle of sides

$a,b,c=4,5,6cm$

$\therefore S=\dfrac{a+b+c}{2}$

$=\dfrac{4+5+6}{2}=\dfrac{15}{2}$

Then, area of triangle

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

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

$=\sqrt{\dfrac{15}{2}.\left(\dfrac{7}{2}\right)\left(\dfrac{5}{2}\right)\left(\dfrac{3}{2}\right)}$

$=\dfrac{15}{4}\sqrt{7}$

Find the area of equilateral triangle each of whose sides measures 20 cm.

0%
173 ${cm^2}$
0%
150 ${cm^2}$
0%
120 ${cm^2}$
0%
153 ${cm^2}$

Explanation

Area of Equilateral Triangle =

$\dfrac{\sqrt{3}{b^2}}{4}$

where b is the Side of the Triangle.

Area=

$\dfrac{\sqrt{3}\times {20^2}}{4}$

Area=

$173.2{cm^2}$

Find the area of equilateral triangle (in

${cm^2}$ )each of whose sides measure 18 cm.

0%
150
0%
140
0%
180
0%
190

Explanation

Area of Equilateral Triangle =

$\dfrac{\sqrt{3}{a^2}}{4}$

where

$a$ is the Side of the Triangle.

Given

$a= 18\ cm$

Area

$=\dfrac{\sqrt{3}\times {18^2}}{4}$

Area

$=140.29\ {cm^2}$

CBSE Questions for Class 9 Maths Heron'S Formula Quiz 4 - MCQExams.com

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Practice Class 9 Maths Quiz Questions and Answers

CBSE Questions for Class 9 Maths Heron'S Formula Quiz 4 - MCQExams.com

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Practice Class 9 Maths Quiz Questions and Answers

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