MCQ Questions for Class 9 Maths Heron'S Formula Quiz 3

What is the area of a triangle whose sides are

$3\ \mathrm{cm}$ ,

$5\,\mathrm{ cm}$ and

$4\,\mathrm{ cm}$ ?

0%
$6 \ \displaystyle \mathrm{cm^{2}}$
0%
$7.5 \ \displaystyle \mathrm{cm^2}$
0%
$\displaystyle 5\sqrt{2} \,\mathrm{cm^{2}}$
0%
none

Explanation

Given

$a=3\,\mathrm{cm},b=5\,\mathrm{cm},c=4\,\mathrm{cm}$

Now

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

Hence

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

Now using Heron's Formula

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

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

$\Delta=6\,\mathrm{cm^2}$

$\therefore$ Area of the triangle

$= 6 \mathrm {cm^2}$

Find the area of the parallelogram with sides

$10 cm$ and

$12 cm$ and one of the diagonals as

$14 cm.$

0%
$48 \sqrt{6}$ $cm^2$
0%
$24 \sqrt{6}$ $cm^2$
0%
$40 \sqrt{6}$ $cm^2$
0%
$20 \sqrt{6}$ $cm^2$

Explanation

Let

$ABCD$ be the given parallelogram.

Area of parallelogram

$ABCD = 2$ x (area of triangle

$ABC$ )

Now

$a = 10$ cm,

$b = 12$ cm and

$c = 14$ cm

$S=\dfrac { 10+12+14 }{ 2 } =\dfrac { 36 }{ 2 } =18$ cm

$^2$

Area of triangle

$ABC$ :

$A=\sqrt { s(s-a)(s-b)(s-c) } =\sqrt { 18(18-10)(18-12)(18-14) } =\sqrt { 18\times 8\times 6\times 4 } =\sqrt { 3456 } =24\sqrt { 6 }$

Area of parallelogram

$ABCD$ :

$2\times 24\sqrt { 6 } =48\sqrt { 6 }$ cm

$^2$

Hence, area of parallelogram is

$48\sqrt { 6 }$ cm

$^2$ .

The perimeter of a triangular field is 144 m and the ratio of the sides is 3 : 4 : 5 The area of the field is

0%
$864\ \displaystyle m^{2}$
0%
$468\ \displaystyle m^{2}$
0%
$824\ \displaystyle m^{2}$
0%
none

Explanation

Let the side of the triangle

$a,b$ and

$c$ be

$3x,4x,5x$

Perimeter

$=144$ m

$\therefore 3x+4x+5x=144$

$\Rightarrow 12x=144$

$\Rightarrow x=12$

Then

$a=12\times 3=36m$

$b=4\times 12=48 m$

$c=5\times 12=60$

Then

$s=\dfrac{a+b+c}{2}=\dfrac{36+48+60}{2}=\dfrac{144}{2}=72$

$\therefore$ Area of the triangle

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

$\Rightarrow \sqrt{72(72-36)(72-48)(72-60)}$

$\Rightarrow \sqrt{72\times 36\times 24\times 12}=864 \ m^2$

Find the area of a triangle with sides having length

$40, 24$ and

$32 m$

0%
$384 m^2$
0%
$364 m^2$
0%
$374 m^2$
0%
None of the above

Explanation

We use heron's formula to find the area of the triangle.

Let us find the semi perimeter:

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

Now area of the triangle is:

$A=\sqrt { s(s-a)(s-b)(s-c) } =\sqrt { 48(48-40)(48-24)(48-32) } =\sqrt { 48\times 8\times 24\times 16 } =\sqrt { 147456 } =384$ m

$^2$

Hence, the area of the triangle is

$384$ m

$^2$ .

Find the area of a triangle with two equal sides of

$5 cm$ and remaining one of

$8 cm$ .

0%
$9cm^2$
0%
$8cm^2$
0%
$12cm^2$
0%
$16cm^2$

Explanation

We use heron's formula to find the area of the triangle.

Let us find the semi perimeter:

$S=\dfrac { 5+5+8 }{ 2 } =\dfrac { 18 }{ 2 } =9$ cm

Now area of the triangle is:

$A=\sqrt { s(s-a)(s-b)(s-c) } =\sqrt { 9(9-5)(9-5)(9-8) } =\sqrt { 9\times 4\times 4\times 1 } =\sqrt { 144 } =12$ cm

$^2$

Hence, the area of the triangle is

$12 cm ^2$

Find the area of the triangluar field whose sides are

$39\ m, 25\ m$ and

$56\ m$ respectively.

0%
$\displaystyle420\ m^{2}$
0%
$\displaystyle520\ m^{2}$
0%
$\displaystyle820\ m^{2}$
0%
$\displaystyle320\ m^{2}$

Explanation

$\displaystyle s=\left ( 25+39+56 \right )\div 2$

$\displaystyle =60m$

Area

$\displaystyle =\sqrt{s\left ( s-a \right )\left ( s-b \right )\left ( s-c \right )}m^{2}$

$\displaystyle =\sqrt{60\left ( 60-39 \right )\left ( 60-25 \right )\left ( 60-56 \right )}$

$\displaystyle =\sqrt{60\times 35\times 21\times 4}$

$\displaystyle =\sqrt{\left ( 5\times 3\times 4 \right )\times \left ( 5\times 7 \right )\times \left ( 7\times 3 \right )\times 4}$

$\displaystyle =\left ( 4\times 3\times 5\times 7 \right )m^{2}=420\ m^{2}$

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

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

Explanation

Hypotenuse will be opposite to right angle which is

$13$ cm

The base and height will be

$5$ cm and

$12$ cm.

$Area\ of\ triangle=\dfrac { 1 }{ 2 } \times \dfrac { 5 }{ 100 } \times \dfrac { 12 }{ 100 }= \dfrac { 30 }{ 10000 } =0.003\ { m }^{ 2 }$ '

Which of the following is true?

0%
Area by heron's formula = Area general formula
0%
Area by heron's formula $\neq$ Area general formula
0%
$Area = 6cm^2$
0%
None of the above

Explanation

General formula of area of triangle is:

$A=\dfrac { 1 }{ 2 } \times base\times height=A=\dfrac { 1 }{ 2 } \times 3\times 4=6$ cm

$^2$

We will now find the area of triangle using herons formula:

$S=\dfrac { 3+4+5 }{ 2 } =\dfrac { 12 }{ 2 } =6$ cm

Now area is:

$A=\sqrt { s(s-a)(s-b)(s-c) } =\sqrt { 6(6-3)(6-4)(6-5) } =\sqrt { 6\times 3\times 2\times 1 } =\sqrt { 36 } =6$ cm

$^2$

Hence, area of triangle is

$6$ cm

$^2$ and

Area by Heron's formula

$=$ Area by general formula

$=6 cm ^2$

The sides of a triangular plot are in the ratio of

$3:5:7$ and its perimeter is

$300m$ . Find its area.

0%
$1500m^2$
0%
$1500\sqrt{3}m^2$
0%
$1400m^2$
0%
$1400\sqrt{3}m^2$

Explanation

The sides of a the triangular plot are in the ratio

$3:5:7$ . So, let the sides of the triangle be

$3x$ ,

$5x$ and

$7x$ .

Also it is given that the perimeter of the triangle is

$300$ m therefore,

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

$15x=300$

$x=20$

Therefore, the sides of the triangle are

$60,100$ and

$140$ .

Now using herons formula:

$S=\dfrac { 60+100+140 }{ 2 } =\dfrac { 300 }{ 2 } =150$ m

Area of the triangle is:

$A=\sqrt { s(s-a)(s-b)(s-c) } =\sqrt { 150(150-60)(150-100)(150-140) } =\sqrt { 150\times 90\times 50\times 10 } =\sqrt { 6750000 }$

$=1500\sqrt { 3 }$ m

$^2$

Hence, area of the triangular plot is

$1500\sqrt { 3 }$ m

$^2$ .

$s$ in Heron's formula is ___ .

0%
area of a triangle
0%
perimeter of a triangle
0%
Semi-perimeter of a triangle
0%
None of the above

Area of the quadrilateral ABCD is

0%
$307 m^2$
0%
$305 m^2$
0%
$306 m^2$
0%
$304 m^2$

Explanation

$ABCD$ is a quadrilateral,

$AB=9$ m,

$AD=28$ m,

$BC=40$ m and

$CD=15$ m and

$\angle ABC=90^{ 0 }$ .

Area of

$ABCD$ = Area of triangle

$ABC+$ Area of triangle

$ACD$

$\triangle ABC$ ,

$\angle B=90^{ 0 }$ , therefore,

$AC^2=AB^2+BC^2=9^2+40^2=81+1600=1681$

$\Rightarrow AC=\sqrt { 1681 } =41$ m

Area of

$\triangle ABC$ is:

$A=\dfrac { 1 }{ 2 } \times AB\times AC==\dfrac { 1 }{ 2 } \times 9\times 40=180$ m

$^2$

$\triangle ACD$ ,

$S=\dfrac { 28+15+41 }{ 2 } =\dfrac { 84 }{ 2 } =42$ m

Now area of

$\triangle ACD$ is:

$A=\sqrt { s(s-a)(s-b)(s-c) } =\sqrt { 42(42-28)(42-15)(42-41) } =\sqrt { 42\times 14\times 27\times 1 } =\sqrt { 15876 } =126$ m

$^2$

Hence, area of quadrilateral

$ABCD$ is

$180+126=306$ m

$^2$ .

Which of the following is / are true?

0%
Attitude is required for calculating area by heron's formula
0%
Attitude is not required for calculating area by heron's formula
0%
Semi-perimeter is required for calculating area by heron's formula
0%
Semi-perimeter is not required for calculating area by heron's formula

Explanation

By Heron's Formula,

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

where

$s=semi-perimeter$ and

$a,b,c$ are

$side\ lengths$

Semi-perimeter

$=\dfrac{sum\ of\ side\ lengths}{2}$

So only the

$semi-perimeter$ and the

$side\ lengths$ are required.

Find the area of a triangle, two of its sides are

$8 cm$ and

$11 cm$ and the perimeter is

$32 cm$ .

0%
$8\sqrt{30}cm^2$
0%
$15\sqrt{2}cm^2$
0%
$4\sqrt{30}cm^2$
0%
$5\sqrt{31}cm^2$

Explanation

Since the perimeter of the triangle with sides

$a$ ,

$b$ and

$c$ is

$a+b+c$ , therefore, perimeter of triangle with sides

$8$ cm,

$11$ cm and

$c$ cm with perimeter

$32$ cm is:

$32=8+11+c$

$c=32-19=13$

We use heron's formula to find the area of the triangle.

Let us find the semi perimeter:

$S=\dfrac { 8+11+13 }{ 2 } =\dfrac { 32 }{ 2 } =16$ cm

Now area of the triangle is:

$A=\sqrt { s(s-a)(s-b)(s-c) } =\sqrt { 16(16-8)(16-11)(16-13) } =\sqrt { 16\times 8\times 5\times 3 } =\sqrt { 1920 } =8\sqrt { 30 }$ cm

$^2$

Hence, the area of the triangle is

$8\sqrt { 30 }$ cm

$^2$ .

Find the area of the triangle by Heron's formula.

0%
$12m^2$
0%
$24m^2$
0%
$40m^2$
0%
$6m^2$

Explanation

We use heron's formula to find the area of the triangle.

Let us find the semi perimeter:

$S=\dfrac { 5+3+4 }{ 2 } =\dfrac { 12 }{ 2 } =6$ cm

Now area of the triangle is:

$A=\sqrt { s(s-a)(s-b)(s-c) } =\sqrt { 6(6-5)(6-3)(6-4) } =\sqrt { 6\times 1\times 3\times 2 } =\sqrt { 36 } =6$ m

$^2$

Hence, the area of the triangle is

$6$ m

$^2$ .

The area of a triangle by Heron's formula is

0%
$\sqrt{(s-a)(s-b)(s-c)}$
0%
$\sqrt{s(s-a)(s-b)(s-c)}$
0%
$\sqrt{(s+a)(s-b)(s-c)}$
0%
$\sqrt{(s+a)(s+c)(s+c)}$

Explanation

Using Herons formula, we first have to find the semi perimeter with sides

$a,b$ and

$c$ :

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

And then area of the triangle is:

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

Hence, area of a triangle by heron's formula is

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

Find the area of a triangle using Heron's formula

0%
$170 cm^2$
0%
$160 cm^2$
0%
$180 cm^2$
0%
$200 cm^2$

Explanation

We find the area of triangle using herons formula:

$S=\dfrac { 9+40+41 }{ 2 } =\dfrac { 90 }{ 2 } =45$ cm

Now area is:

$A=\sqrt { s(s-a)(s-b)(s-c) } =\sqrt { 45(45-9)(45-40)(45-41) } =\sqrt { 45\times 36\times 5\times 4 } =\sqrt { 32400 } =180$ cm

$^2$

Hence, area of triangle is

$180$ cm

$^2$ .

Find the area of an equilateral triangle with side

$10cm$ .

0%
$25\sqrt{3}cm^2$
0%
$24\sqrt{3}cm^2$
0%
$26\sqrt{3}cm^2$
0%
None of the above

Explanation

Formula: Are of an equilateral triangle having side length

$a$ is given by,

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

Here it is given that, side length

$, a=10$ cm

Thus its area

$=\dfrac{\sqrt 3}{4}\times 10^2$ cm

$^2=25\sqrt 3$ cm

$^2$

If the edges of a triangular board are

$15$ cm,

$36$ cm and

$39$ cm, then the cost of painting its one of the faces at the rate of

$5$ paise per cm

$^2$ is

0%
Rs. $1212$
0%
Rs. $13.5$
0%
Rs. $1350$
0%
Rs $11.6$

Explanation

Semi-perimeter,

$s=\dfrac{15+36+39}{2}$

$=45$

From Heron's formula

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

$=\sqrt{45\times 30\times 9\times 6}$

$=9\times 5\times 6=30\times 9=270$ cm

given that cost is 5 paisa per

$cm^2$

so, for 270

$cm^2$ cost in rs will be:

Cost

$=\dfrac{270\times 5}{100}$ Rs.

$=\dfrac{1350}{100}$

$\Rightarrow (B)$ .

Which of the following is true about the above quadrilateral?

0%
Area of quadrilateral can be calculated by one general formula of area of triangle
0%
Area of quadrilateral cannot be calculate by the general formula of area of triangle because of the absence of height
0%
Height is AD and BC in $\triangle ADC$ and $\triangle ABC$ respectively.
0%
Area of Quadrilateral can be easily calculated by the heron's formula.

Explanation

$ABCD$ is a quadrilateral,

$AB=3$ cm,

$AD=5$ cm,

$BC=4$ cm and

$CD=4$ cm and

$AC=5$ cm.

Area of

$ABCD$ = Area of triangle

$ACD+$ Area of triangle

$ACB$

$\triangle ACD$ ,

$S=\dfrac { 5+5+4 }{ 2 } =\dfrac { 14 }{ 2 } =7$ cm

Now area of

$\triangle ACD$ is:

$A=\sqrt { s(s-a)(s-b)(s-c) } =\sqrt { 7(7-5)(7-5)(7-4) } =\sqrt { 7\times 2\times 2\times 3 } =\sqrt { 84 } =2\sqrt { 21 }$ cm

$^2$

$\triangle ACB$ ,

$S=\dfrac { 3+4+5 }{ 2 } =\dfrac { 12 }{ 2 } =6$ cm

Now area of

$\triangle ACB$ is:

$A=\sqrt { s(s-a)(s-b)(s-c) } =\sqrt { 6(6-3)(6-4)(6-5) } =\sqrt { 6\times 3\times 2\times 1 } =\sqrt { 36 } =6$ cm

$^2$

Hence, area of quadrilateral

$ABCD$ is

$(6+2\sqrt { 21 })$ cm

$^2$ .

Area of quadrilateral

$ABCD$ cannot be found by the general formula of area of triangle because the height is not given.

Use Heron's formula to find the area of a triangle of lengths 3,3 and 4

0%
$5\sqrt{2}$
0%
$2\sqrt{5}$
0%
$2\sqrt{7}$
0%
$7\sqrt{2}$

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

$5 cm, 9 cm$ and

$12 cm$ .

0%
$\sqrt{26}\ cm^2$
0%
$2\sqrt{26}\ cm^2$
0%
$3\sqrt{26}\ cm^2$
0%
$4\sqrt{26}\ cm^2$

Explanation

We will find the area of triangle with lengths

$5$ cm,

$9$ cm and

$12$ cm using heron's formula:

$S=\dfrac { 5+9+12 }{ 2 } =\dfrac { 26 }{ 2 } =13$ units

Now area is:

$A=\sqrt { s(s-a)(s-b)(s-c) } =\sqrt { 13(13-5)(13-9)(13-12) } =\sqrt { 13\times 8\times 4\times 1 } =\sqrt { 416 } =4\sqrt { 26 }$ sq.units

Hence, area of triangle is

$4\sqrt { 26 } cm^2$

A land ABCD is in the shape of a Quadrilateral has

$\angle C = 90, AB = 9m, BC = 12m, CD = 5m$ and

$AD = 8m$ . How much area does it occupy?

0%
$30+6\sqrt{5}m^2$
0%
$30+6\sqrt{35}m^2$
0%
$30+6\sqrt{125}m^2$
0%
None of the above

Explanation

$BD = \sqrt{12^2+5^2} = \sqrt{144+25} = \sqrt{169}=13m$
Area of

$\triangle BDC = \dfrac{1}{2} \times 12\times 5m^2 = 6\times 5=30m^2$
Area of \triangle ABD

$S=\dfrac{9+8+13}{2} = \dfrac{17+13}{2} = \dfrac{30}{2} = 15$

$=\sqrt{15\times (15-9)(15-8)(15-13)}$

$=\sqrt{15\times 6\times 7\times 2}$

$=\sqrt {3\times 5\times 3\times 2\times 7\times 2}$

$=3\times 2\sqrt{35}=6\sqrt{35}$
Total area

$= 30+6\sqrt{35}m^2$

The sides of a triangle are

$5y, 6y$ and

$9y$ . find an expression for its area

$A$ .

0%
$10y^2\sqrt{2}$
0%
$5y^2\sqrt{2}$
0%
$20y^2\sqrt{2}$
0%
$100y^2\sqrt{2}$

Explanation

We will find the area of triangle with sides

$5y$ ,

$6y$ and

$9y$ using heron's formula:

Let

$S$ be the semi perimeter of the given triangle

$S=\dfrac { 5y+6y+9y }{ 2 } \\=\dfrac { 20y }{ 2 }\\=10y$

Now area is:

$A=\sqrt { s(s-a)(s-b)(s-c) } \\=\sqrt { 10y\left( 10y-5y \right) \left( 10y-6y \right) \left( 10y-9y \right) } \\=\sqrt { 10y\times 5y\times 4y\times y } \\=\sqrt { 200y^{ 4 } } \\=10y^{ 2 }\sqrt { 2 }$

Hence, area of triangle is

$10y^{ 2 }\sqrt { 2 }$ sq.units.

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

$7, 8$ and

$9$ .

0%
${12\sqrt{5}}$
0%
$11\sqrt{5}$
0%
$6\sqrt{5}$
0%
None of the above

Explanation

We will find the area of triangle with lengths

$7,8$ and

$9$ using herons formula:

$S=\dfrac { 7+8+9 }{ 2 } =\dfrac { 24 }{ 2 } =12$ units

Now area is:

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

$=\sqrt { 12(12-7)(12-8)(12-9) }$

$=\sqrt { 12\times 5\times 4\times 3 }$

$=\sqrt { 720 } =12\sqrt { 5 }$ sq.units

Hence, area of triangle is

$12\sqrt { 5 }$ sq.units.

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

$4, 5$ and

$6.$

0%
$\dfrac{15}{4}\sqrt{7}$
0%
$\dfrac{13}{4}\sqrt{7}$
0%
$4\sqrt{7}$
0%
$3\sqrt{7}$

Explanation

Given: Lengths of a triangle

$=4,$

$5,$ and

$6$

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

Using heron's formula to find the area of the triangle:

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

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

$=\sqrt { \dfrac { 15 }{ 2 } \times \dfrac { 7 }{ 2 } \times \dfrac { 5 }{ 2 } \times \dfrac { 3 }{ 2 } }$

$=\sqrt { \dfrac { 1575 }{ 16 } }$

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

Hence, area of triangle is

$\dfrac { 15 }{ 4 } \sqrt { 7 }$ sq. units.

Calculate the area of quadrilateral

$ABCD$ .

0%
$6+\sqrt{21}cm^2$
0%
$6+2\sqrt{21}cm^2$
0%
$6+2\sqrt{7}cm^2$
0%
None of the above

Explanation

Area of

$\triangle ABC$

$S=\dfrac{5+4+3}{2}=\dfrac{9+3}{2} = \dfrac{12}{2} = 6cm^2$

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

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

$=\sqrt{6\times 1\times 2\times 3}$

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

$=6cm^2$
Area of

$\triangle ABC$

$s=\dfrac{5+5+4}{2} = \dfrac{14}{2} = 7cm^2$

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

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

$=2\sqrt{21}$
Total Area

$=(6+2\sqrt{21}) cm^2$

Which of the following are true?

0%
Area of $\triangle ABC$ cannot be calculated by Heron's formula.
0%
Area of $\triangle ABC$ can be calculated by Heron's formula.
0%
Area of $\triangle ABC$ cannot be calculated by the general formula of area.
0%
Area of $\triangle ABC$ can be calculated by the general formula of area.

Explanation

$\Delta ABC$ is right-angled,

$AB^2 + BC^2 = AC^2$ (hypotenuse theorem)

$AC^2 = 20^2+ 50^2$

$=400+2500$

$AC^2 = 2900$

$AC = \sqrt{2900}cm$ .

a) Area of

$\Delta ABC$ using general formula

$= \dfrac{1}{2}\times b \times h$

$=\dfrac{1}{2} \times 50cm \times 20cm$

$=500cm^2$ .

$\therefore$ Area of

$\Delta ABC$ can be calculated using general formula of area.

b) Area of

$\Delta ABC$ using Heron's formula

Area

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

$s\to$ semiperimeter)

Area

$=499.9999 cm^2$

$s\to \dfrac{a+b+c}{2} = 61.928$

$\therefore$ Area of

$\Delta ABC$ can be calculated by Herou's formula

$\therefore$ Hence, option (B) and option (D) are correct.

2063855_443591_ans_dbb81df91e84426ebb0096cf012d58bc.png

The parallel sides of a parallelogram are

$40$ and

$15$ m and one of the diagonals is

$35$ m. Find the area of the parallelogram, if the distance between the parallel sides is

$50$ m. (Use Heron's formula)

0%
$120$ $\sqrt{3}$ $m^{2}$
0%
$150$ $\sqrt{3}$ $m^{2}$
0%
$200$ $\sqrt{3}$ $m^{2}$
0%
$300$ $\sqrt{3}$ $m^{2}$

Explanation

$\parallel$ gm

$ABCD, AB = CD = 40\ m$ and

$AD = BC = 15\ m$ , Diagonal,

$AC = 35\ m$
In

$\Delta ABC$ ,

$s= \dfrac{40 + 15 + 35}{2}$

$= 45\ m$
Area of

$\parallel$ gm =

$2$ x Area of

$\Delta ABC$
=

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

$2\times \sqrt{45(45-40)(45-15)(45-35)}$
=

$2\times \sqrt{45(5)(30)(10)}$
=

$300\sqrt{3}$

$m^{2}$

The parallel sides of a parallelogram are

$20$ and

$10$ m and one of the diagonals is

$12$ m. Find the area of the parallelogram. (Use Heron's formula)

0%
6 $\sqrt{31}$ $m^{2}$
0%
6 $\sqrt{231}$ $m^{2}$
0%
6 $\sqrt{21}$ $m^{2}$
0%
3 $\sqrt{231}$ $m^{2}$

Explanation

$\parallel$ gm ABCD, AB = CD = 20 m and AD = BC = 10m, Diagonal, AC = 12 m
In

$\Delta$ ABC,

$s= \dfrac{20 + 10 + 12}{2}$ = 21 m
Area of

$\parallel$ gm = 2 x Area

$\Delta$ ABC
= 2

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

$\sqrt{21(21-20)(21-10)(21-12)}$
= 2

$\sqrt{21(1)(11)(9)}$
= 6

$\sqrt{231}$

$m^{2}$

The sides of a triangle are

$5\ cm, 12\ cm$ and

$13\ cm$ , what is its area?

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

Explanation

Area of

$\triangle =\sqrt { s\left( s-a \right) \left( s-b \right) \left( s-c \right) } ,\quad \quad s=\dfrac { a+b+c }{ 2 } =\dfrac { 13+12+5 }{ 2 } =15cm$

$\therefore$ ar of

$\triangle =\sqrt { 15\left( 15-13 \right) \left( 15-12 \right) \left( 15-5 \right) } =\sqrt { 15\times 2\times 3\times 10 } =30{ cm }^{ 2 }$

Therefore area of triangle

$={ 30cm }^{ 2 }=0.003{ m }^{ 2 }\left[ \because \quad 1{ m }^{ 2 }=10000{ cm }^{ 2 } \right]$

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CBSE Questions for Class 9 Maths Heron'S Formula Quiz 3 - MCQExams.com

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Practice Class 9 Maths Quiz Questions and Answers

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