Herons Formula Test

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Herons Formula Test - 27

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Question 1
1 / -0

Find the area of the parallelogram with sides $$9 cm$$ and $$11 cm$$ and one of the diagonals as $$14 cm. $$

A
$$22\sqrt{15}$$

B
$$24\sqrt{17}$$

C
$$21\sqrt{17}$$

D
None of these

Solution

Area of $$\parallel $$gm ABCD = 2 $$\times $$ area $$\triangle $$ABC
ar $$\triangle $$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 $$\times $$ $$12\sqrt { 17 } =24\sqrt { 17 } { cm }^{ 2 }$$
Question 2
1 / -0

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,

A
$$1430$$ sq m

B
$$1576$$ sq m

C
$$1500$$ sq m

D
None of these

Solution

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 $$=2\times 750\ m^2$$
$$=1500\ m^2$$
Question 3
1 / -0

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.

A
$$2100$$ sq m

B
$$2100\sqrt{15}$$ sq m

C
$$2000$$ sq m

D
$$2000\sqrt{15}$$ sq m

Solution

$$\textbf{Step - 1: Finding the sides}$$
$$\text{Given that the sides are in the ratio }6:7:8$$
$$\implies a=6x,b=7x,c=8x$$
$$\text{Perimeter}=a+b+c$$
$$\implies 420=6x+7x+8x=21x$$
$$\implies x=20$$
$$\implies a=120,b=140,c=160$$
$$\textbf{Step - 2: Calculating area}$$
$$s=\dfrac{a+b+c}2=210$$
$$\implies \text{Area}=\sqrt{s(s-a)(s-b)(s-c)}$$
$$\implies \text{Area} = \sqrt{210\times90\times70\times50}$$
$$\implies \text{Area}=2100\sqrt{15}$$
$$\textbf{Thus, the area of the given field is }\mathbf{2100\sqrt{15}}\textbf{ sq. m.}$$
Question 4
1 / -0

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

A
$$2\sqrt { 3 }$$

B
$$3\sqrt { 3 }$$

C
$$5\sqrt { 3 }$$

D
$$4\sqrt { 3 }$$

Solution

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
Question 5
1 / -0

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 $$\sqrt{2}=1.414$$)

A
Rs.$$4532.90$$

B
Rs.$$4242$$

C
Rs.$$1696.80$$

D
Rs.$$1000$$

Solution

$$Heron's\ Formula\ :$$
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 = $$4000\times0.4242=Rs.\ 1696.80$$. $$[C]$$
Question 6
1 / -0

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

A
$$8cm^2$$

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

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

D
$$24cm^2$$

Solution

Let third side be $$x$$
Perimeter $$=8+11+x$$
$$19+x=32$$
$$x=13$$
$$s=\cfrac { 32 }{ 2 } =16$$
Heron's formula
$$A=\sqrt { s\left( s-a \right) \left( s-b \right) \left( s-c \right) } $$
$$=\sqrt { 16\left( 16-8 \right) \left( 16-11 \right) \left( 16-13 \right) } $$
$$=\sqrt { 16\times 8\times 5\times 3 } $$
$$=8\sqrt { 30 }\ cm^2$$
Question 7
1 / -0

One side of an equilateral triangle measures $$8$$ cm. Find its area using Heron's formula. What is its altitude

A
$$12\sqrt 3\ \text{cm}^2, 3\sqrt 3$$ cm

B
$$16\sqrt 3\ \text{cm}^2, 4\sqrt 3$$ cm

C
$$14\sqrt 3\ \text{cm}^2, \dfrac {7}{2}\sqrt 3$$ cm

D
$$20\sqrt 3\ \text{cm}^2, 5\sqrt 3$$ cm

Solution

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 $$=\cfrac{16\sqrt{3}}{4}=4\sqrt{3}$$
Question 8
1 / -0

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

A
$$1500\sqrt{5}m^2$$

B
$$1500m^2$$

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

D
$$1100\sqrt{3}m^2$$

Solution

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

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

$$\implies1500\sqrt3\ m^2$$
Question 9
1 / -0

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

A
$$34cm^2$$

B
$$44cm^2$$

C
$$54cm^2$$

D
$$55cm^2$$

Solution

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.
Question 10
1 / -0

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.

A
$$2250m^2$$

B
$$6250m^2$$

C
$$5250m^2$$

D
$$6050m^2$$

Solution

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

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

$$\implies6250\ m^2$$

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Herons Formula Test - 27

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Herons Formula Test - 27

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Question 1

$$22\sqrt{15}$$

$$24\sqrt{17}$$

$$21\sqrt{17}$$

None of these

Solution

Question 2

$$1430$$ sq m

$$1576$$ sq m

$$1500$$ sq m

None of these

Solution

Question 3

$$2100$$ sq m

$$2100\sqrt{15}$$ sq m

$$2000$$ sq m

$$2000\sqrt{15}$$ sq m

Solution

Question 4

$$2\sqrt { 3 }$$

$$3\sqrt { 3 }$$

$$5\sqrt { 3 }$$

$$4\sqrt { 3 }$$

Solution

Question 5

Rs.$$4532.90$$

Rs.$$4242$$

Rs.$$1696.80$$

Rs.$$1000$$

Solution

Question 6

$$8cm^2$$

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

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

$$24cm^2$$

Solution

Question 7

$$12\sqrt 3\ \text{cm}^2, 3\sqrt 3$$ cm

$$16\sqrt 3\ \text{cm}^2, 4\sqrt 3$$ cm

$$14\sqrt 3\ \text{cm}^2, \dfrac {7}{2}\sqrt 3$$ cm

$$20\sqrt 3\ \text{cm}^2, 5\sqrt 3$$ cm

Solution

Question 8

$$1500\sqrt{5}m^2$$

$$1500m^2$$

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

$$1100\sqrt{3}m^2$$

Solution

Question 9

$$34cm^2$$

$$44cm^2$$

$$54cm^2$$

$$55cm^2$$

Solution

Question 10

$$2250m^2$$

$$6250m^2$$

$$5250m^2$$

$$6050m^2$$

Solution

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