Herons Formula Test

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Herons Formula Test - 24

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Question 1
1 / -0

Find the area of the parallelogram with sides $$10 cm$$ and $$12 cm$$ and one of the diagonals as $$14 cm. $$

A
$$48 \sqrt{6}$$ $$cm^2$$

B
$$24 \sqrt{6}$$ $$cm^2$$

C
$$40 \sqrt{6}$$ $$cm^2$$

D
$$20 \sqrt{6}$$ $$cm^2$$

Solution

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$$^2$$.
Question 2
1 / -0

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

A
$$864\ \displaystyle m^{2}$$

B
$$468\ \displaystyle m^{2}$$

C
$$824\ \displaystyle m^{2}$$

D
none

Solution

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

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

$$\Rightarrow \sqrt{72\times 36\times 24\times 12}=864 \ m^2$$
Question 3
1 / -0

Find the area of a triangle with sides having length $$40, 24$$ and $$32 m$$

A
$$384 m^2$$

B
$$364 m^2$$

C
$$374 m^2$$

D
None of the above

Solution

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$$^2$$.
Question 4
1 / -0

Find the area of a triangle with two equal sides of $$5 cm$$ and remaining one of $$8 cm$$.

A
$$9cm^2$$

B
$$8cm^2$$

C
$$12cm^2$$

D
$$16cm^2$$

Solution

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 $$12 cm ^2$$
Question 5
1 / -0

Find the area of the triangluar field whose sides are $$39\ m, 25\ m$$ and $$56\ m$$ respectively.

A
$$\displaystyle420\ m^{2}$$

B
$$\displaystyle520\ m^{2}$$

C
$$\displaystyle820\ m^{2}$$

D
$$\displaystyle320\ m^{2}$$

Solution

$$\displaystyle s=\left ( 25+39+56 \right )\div 2$$
$$\displaystyle =60m$$

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

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

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

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

$$\displaystyle =\left ( 4\times 3\times 5\times 7 \right )m^{2}=420\ m^{2}$$
Question 6
1 / -0

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

A
$$0.003{ m }^{ 2 }$$

B
$$0.0024{ m }^{ 2 }$$

C
$$0.0026{ m }^{ 2 }$$

D
$$0.0015{ m }^{ 2 }$$

Solution

Hypotenuse will be opposite to right angle which is $$13$$ cm

The base and height will be $$5$$ cm and $$12$$ cm.

$$Area\ of\ triangle=\dfrac { 1 }{ 2 } \times \dfrac { 5 }{ 100 } \times \dfrac { 12 }{ 100 }= \dfrac { 30 }{ 10000 } =0.003\ { m }^{ 2 }$$'
Question 7
1 / -0

The sides of a triangular plot are in the ratio of $$3:5:7$$ and its perimeter is $$300m$$. Find its area.

A
$$1500m^2$$

B
$$1500\sqrt{3}m^2$$

C
$$1400m^2$$

D
$$1400\sqrt{3}m^2$$

Solution

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$$^2$$.
Question 8
1 / -0

$$s$$ in Heron's formula is ___ .

A
area of a triangle

B
perimeter of a triangle

C
Semi-perimeter of a triangle

D
None of the above
Question 9
1 / -0

Area of the quadrilateral ABCD is

A
$$307 m^2$$

B
$$305 m^2$$

C
$$306 m^2$$

D
$$304 m^2$$

Solution

If $$ABCD$$ 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 $$ACD$$

In $$\triangle ABC$$, $$\angle B=90^{ 0 }$$, 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$$^2$$

In $$\triangle ACD$$,

$$S=\dfrac { 28+15+41 }{ 2 } =\dfrac { 84 }{ 2 } =42$$ 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$$^2$$.
Question 10
1 / -0

Find the area of a triangle, two of its sides are $$8 cm$$ and $$11 cm$$ and the perimeter is $$32 cm$$.

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

B
$$15\sqrt{2}cm^2$$

C
$$4\sqrt{30}cm^2$$

D
$$5\sqrt{31}cm^2$$

Solution

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$$^2$$.

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Herons Formula Test - 24

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Herons Formula Test - 24

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SHARING IS CARING

Question 1

$$48 \sqrt{6}$$ $$cm^2$$

$$24 \sqrt{6}$$ $$cm^2$$

$$40 \sqrt{6}$$ $$cm^2$$

$$20 \sqrt{6}$$ $$cm^2$$

Solution

Question 2

$$864\ \displaystyle m^{2}$$

$$468\ \displaystyle m^{2}$$

$$824\ \displaystyle m^{2}$$

none

Solution

Question 3

$$384 m^2$$

$$364 m^2$$

$$374 m^2$$

None of the above

Solution

Question 4

$$9cm^2$$

$$8cm^2$$

$$12cm^2$$

$$16cm^2$$

Solution

Question 5

$$\displaystyle420\ m^{2}$$

$$\displaystyle520\ m^{2}$$

$$\displaystyle820\ m^{2}$$

$$\displaystyle320\ m^{2}$$

Solution

Question 6

$$0.003{ m }^{ 2 }$$

$$0.0024{ m }^{ 2 }$$

$$0.0026{ m }^{ 2 }$$

$$0.0015{ m }^{ 2 }$$

Solution

Question 7

$$1500m^2$$

$$1500\sqrt{3}m^2$$

$$1400m^2$$

$$1400\sqrt{3}m^2$$

Solution

Question 8

area of a triangle

perimeter of a triangle

Semi-perimeter of a triangle

None of the above

Question 9

$$307 m^2$$

$$305 m^2$$

$$306 m^2$$

$$304 m^2$$

Solution

Question 10

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

$$15\sqrt{2}cm^2$$

$$4\sqrt{30}cm^2$$

$$5\sqrt{31}cm^2$$

Solution

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