Mensuration — Exercise 15.1

Find the area of a triangle whose base is 6 cm and corresponding height is 4 cm.

Answer

Given,

Base of triangle = 6 cm

Height of triangle = 4 cm

We know that,

Area of triangle = $\dfrac{1}{2}$ × base × height

Substituting the values we get,

Area of triangle = $\dfrac{1}{2}$ × 6 × 4

= 6 × 2

= 12 cm².

Hence, area of triangle = 12 cm².

Find the area of a triangle whose sides are 3 cm, 4 cm and 5 cm.

Answer

Consider a = 3 cm, b = 4 cm and c = 5 cm

We know that

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

Substituting the values we get,

s = $\dfrac{(3 + 4 + 5)}{2} = \dfrac{12}{2}$ = 6 cm.

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

Substituting values we get,

$A = \sqrt{6(6 - 3)(6 - 4)(6 - 5)} \\[1em] = \sqrt{6 \times 3 \times 2 \times 1} \\[1em] = \sqrt{36} \\[1em] = 6 \text{ cm}^2.$

Hence, area of triangle = 6 cm².

Find the area of a triangle whose sides are 29 cm, 20 cm and 21 cm.

Answer

Consider a = 29 cm, b = 20 cm and c = 21 cm

We know that,

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

Substituting the values we get,

s = $\dfrac{(29 + 20 + 21)}{2} = \dfrac{70}{2}$ = 35 cm.

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

Substituting values we get,

$A = \sqrt{35(35 - 29)(35 - 20)(35 - 21)} \\[1em] = \sqrt{35 \times 6 \times 15 \times 14} \\[1em] = \sqrt{44100} \\[1em] = 210 \text{ cm}^2.$

Hence, area of triangle = 210 cm².

Find the area of a triangle whose sides are 12 cm, 9.6 cm and 7.2 cm.

Answer

Consider a = 12 cm, b = 9.6 cm and c = 7.2 cm

We know that,

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

Substituting the values we get,

s = $\dfrac{(12 + 9.6 + 7.2)}{2} = \dfrac{28.8}{2}$ = 14.4 cm.

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

Substituting values we get,

$A = \sqrt{14.4(14.4 - 12)(14.4 - 9.6)(14.4 - 7.2)} \\[1em] = \sqrt{14.4 \times 2.4 \times 4.8 \times 7.2} \\[1em] = \sqrt{1194.39} \\[1em] = 34.56 \space \text{cm}^2.$

Hence, area of triangle = 34.56 cm².

Find the area of a triangle whose sides are 34 cm, 20 cm and 42 cm. Hence, find the length of the altitude corresponding to the shortest side.

Answer

Consider 34 cm, 20 cm and 42 cm as the sides of triangle.

a = 34 cm, b = 20 cm and c = 42 cm

We know that,

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

Substituting the values we get,

s = $\dfrac{(34 + 20 + 42)}{2} = \dfrac{96}{2}$ = 48 cm.

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

Substituting values we get,

$A = \sqrt{48(48 - 34)(48 - 20)(48 - 42)} \\[1em] = \sqrt{48 \times 14 \times 28 \times 6} \\[1em] = \sqrt{112896} \\[1em] = 336 \text{ cm}^2.$

Here the shortest side of the triangle is 20 cm. Let height = h cm be the corresponding altitude.

We know that,

Area of triangle = $\dfrac{1}{2}$ × base × height

Substituting the values we get,

⇒ 336 = $\dfrac{1}{2}$ × 20 × h

⇒ h = $\dfrac{336 \times 2}{20}$

⇒ h = $\dfrac{336}{10}$

⇒ h = 33.6 cm.

Hence, area of triangle = 336 cm² and length of the altitude corresponding to the shortest side = 33.6 cm.

The sides of a triangular field are 975 m, 1050 m and 1125 m. If this field is sold at the rate of ₹ 10 lakh per hectare, find its selling price. [1 hectare = 10000 m²]

Answer

Consider, a = 975 m, b = 1050 m and c = 1125 m.

We know that,

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

Substituting the values we get,

$s = \dfrac{(975 + 1050 + 1125)}{2}\\[1em] = \dfrac{3150}{2}\\[1em] = 1575 \text{ m}.$

By formula,

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

Substituting values we get :

$A = \sqrt{1575(1575 - 975)(1575 - 1050)(1575 - 1125)}\\[1em] = \sqrt{1575 \times 600 \times 525 \times 450}\\[1em] = \sqrt{(525 \times 3) \times (150 \times 2 \times 2) \times (525) \times (150 \times 3)}\\[1em] = \sqrt{525^2 \times 150^2 \times 2^2 \times 3^2}\\[1em] = 525 \times 150 \times 2 \times 3\\[1em] = 472500 \text{ m}^2 \\[1em] = \dfrac{472500}{10000} \\[1em] = 47.25 \text{ hectares}.$

We know that,

Selling price of 1 hectare field = ₹ 10 lakh.

∴ Selling price of 47.25 hectare field = ₹ 10,00,000 × 47.25 = ₹ 4,72,50,000.

Hence, selling price of the triangular field = ₹ 4,72,50,000.

The base of a right angled triangle is 12 cm and its hypotenuse is 13 cm long. Find its area and the perimeter.

Answer

It is given that,

ABC is a right angled triangle.

From figure,

BC = 12 cm and AC = 13 cm

Using the Pythagoras theorem,

AC² = AB² + BC²

Substituting the values we get,

⇒ 13² = AB² + 12²

⇒ AB² = 13² – 12²

⇒ AB² = 169 – 144 = 25

⇒ AB = $\sqrt{25}$ = 5 cm.

We know that,

Area of triangle ABC = $\dfrac{1}{2}$ × base × height.

Substituting the values we get,

A = $\dfrac{1}{2} \times 12 \times 5$ = 30 cm².

Perimeter of triangle ABC (P) = AB + BC + CA

Substituting the values we get,

= 5 + 12 + 13

= 30 cm.

Hence, area of triangle = 30 cm² and perimeter = 30 cm.

Find the area of an equilateral triangle whose side is 8 m. Give your answer correct to two decimal places.

Answer

Given,

Side of equilateral triangle = 8 m.

We know that,

Area of equilateral triangle = $\dfrac{\sqrt{3}}{4}$(side)²

Substituting the values we get,

$A = \dfrac{\sqrt{3}}{4} \times (8)^2 \\[1em] = \dfrac{\sqrt{3}}{4} \times 64 \\[1em] = 16\sqrt{3} \\[1em] = 16 \times 1.732 \\[1em] = 27.71$

Hence, area of equilateral triangle = 27.71 m².

If the area of an equilateral triangle is $81\sqrt{3}$ cm², find its perimeter.

Answer

We know that,

Area of equilateral triangle = $\dfrac{\sqrt{3}}{4}$(side)²

Substituting the values,

$\Rightarrow 81\sqrt{3} = \dfrac{\sqrt{3}}{4}(side)^2 \\[1em] \Rightarrow (side)^2 = \dfrac{81\sqrt{3} \times 4}{\sqrt{3}} \\[1em] \Rightarrow (side)^2 = 81 \times 4 \\[1em] \Rightarrow (side)^2 = 324 \\[1em] \Rightarrow \text{side} = \sqrt{324} = 18 \text{ cm}.$

So the perimeter of equilateral triangle = 3 × side

= 3 × 18 = 54 cm.

Hence, perimeter of equilateral triangle = 54 cm.

If the perimeter of an equilateral triangle is 36 cm, calculate its area and height.

Answer

We know that,

Perimeter of an equilateral triangle = 3 × side.

Substituting the values,

⇒ 36 = 3 × side

⇒ side = $\dfrac{36}{3}$ = 12 cm.

Area of equilateral triangle = $\dfrac{\sqrt{3}}{4}(side)^2$

Substituting the values we get,

$A = \dfrac{\sqrt{3}}{4} \times (12)^2 \\[1em] = \dfrac{\sqrt{3}}{4} \times 144 \\[1em] = 36\sqrt{3} \\[1em] = 36 \times 1.732 \\[1em] = 62.4 \text{ cm}^2$

From figure,

In triangle ABD,

Using Pythagoras Theorem,

AB² = AD² + BD² .......(1)

The perpendicular from a vertex of an equilateral triangle to the opposite side, bisects it.

So, BD = $\dfrac{12}{2}$ = 6 cm.

Substituting the values in (1) we get,

⇒ 12² = AD² + 6²

⇒ 144 = AD² + 36

⇒ AD² = 144 – 36 = 108

⇒ AD = $\sqrt{108}$ = 10.4 cm.

Hence, area of triangle = 62.4 cm² and height = 10.4 cm.

If the lengths of the sides of a triangle are in the ratio 3: 4 : 5 and its perimeter is 48 cm, find its area.

Answer

Let a, b and c be the sides of the triangle.

Given,

Ratio of the sides are 3 : 4 : 5.

Let a = 3x cm, b = 4x cm and c = 5x cm.

Given,

⇒ Perimeter = 48 cm

⇒ a + b + c = 48

⇒ 3x + 4x + 5x = 48

⇒ 12x = 48

⇒ x = $\dfrac{48}{12}$ = 4.

Substituting value of x,

a = 3x = 3 × 4 = 12 cm,

b = 4x = 4 × 4 = 16 cm,

c = 5x = 5 × 4 = 20 cm.

We know that,

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

$s = \dfrac{12 + 16 + 20}{2} \\[1em] = \dfrac{48}{2} \\[1em] = 24 \text{ cm}.$

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

Substituting values we get,

$A = \sqrt{24(24 - 12)(24 - 16)(24 - 20)} \\[1em] = \sqrt{24 \times 12 \times 8 \times 4} \\[1em] = \sqrt{9216} \\[1em] = 96 \text{ cm}^2.$

Hence, area of triangle = 96 cm².

The sides of a triangular plot are in the ratio 3 : 5 : 7 and its perimeter is 300 m. Find its area. Take $\sqrt{3} = 1.732$.

Answer

Given,

Sides of a triangle are in the ratio = 3 : 5 : 7

Perimeter = 300 m

Let a = 3x cm, b = 5x cm and c = 7x cm.

Given,

⇒ Perimeter = 300 m

⇒ a + b + c = 300

⇒ 3x + 5x + 7x = 300

⇒ 15x = 300

⇒ x = $\dfrac{300}{15}$ = 20.

Substituting value of x,

a = 3x = 3 × 20 = 60 m,

b = 5x = 5 × 20 = 100 m,

c = 7x = 7 × 20 = 140 m.

We know that,

Semi-perimeter (s) = $\dfrac{\text{Perimeter}}{2} = \dfrac{300}{2}$ = 150 m.

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

Substituting values we get,

$A = \sqrt{150(150 - 60)(150 - 100)(150 - 140)} \\[1em] = \sqrt{150 \times 90 \times 50 \times 10} \\[1em] = \sqrt{6750000} \\[1em] = 2598.07 ≈ 2598 m^2.$

Hence, area of triangle = 2598 m².

ABC is a triangle in which AB = AC = 4 cm and ∠A = 90°. Calculate the area of △ABC. Also find the length of perpendicular from A to BC.

Answer

It is given that

AB = AC = 4 cm

From figure,

Using the Pythagoras theorem,

BC² = AB² + AC²

Substituting the values we get,

⇒ BC² = 4² + 4²

⇒ BC² = 16 + 16 = 32

⇒ BC = $\sqrt{32} = 4\sqrt{2}$ cm.

Let perpendicular from A to BC be h cm.

Area of △ABC = $\dfrac{1}{2}$ × base × height

= $\dfrac{1}{2}$ × AC × AB

= $\dfrac{1}{2} \times 4 \times 4$

= 8 cm.

From figure,

Area of △ABC = $\dfrac{1}{2}$ × BC × h.

$\therefore 8 = \dfrac{1}{2} \times BC \times h \\[1em] \Rightarrow 8 = \dfrac{1}{2} \times 4\sqrt{2} \times h \\[1em] \Rightarrow 8 = 2\sqrt{2}h \\[1em] \Rightarrow h = \dfrac{8}{2\sqrt{2}} \\[1em] \Rightarrow h = 2\sqrt{2} = 2.83$

Hence, area of △ABC = 8 cm² and length of perpendicular from A to BC = 2.83 cm.

Find the area of an isosceles triangle whose equal sides are 12 cm each and the perimeter is 30 cm.

Answer

Consider △ABC as the isosceles triangle.

Here, AB = AC = 12 cm.

Perimeter = 30 cm

⇒ AB + AC + BC = 30

⇒ 12 + 12 + BC = 30

⇒ BC = 30 - 24 = 6 cm.

We know that,

Semi-perimeter (s) = $\dfrac{\text{Perimeter}}{2} = \dfrac{30}{2}$ = 15 cm.

Area of an isosceles triangle = $\dfrac{1}{4}b\sqrt{4a^2 - b^2}$, where a is length of equal sides and b is the length of other side.

Substituting values we get,

$A = \dfrac{1}{4} \times 6 \times \sqrt{4(12)^2 - 6^2} \\[1em] = \dfrac{1}{4} \times 6 \times \sqrt{4 \times 144 - 36} \\[1em] = \dfrac{1}{4} \times 6 \times \sqrt{540} \\[1em] = \dfrac{1}{4} \times 6 \times 23.24 \\[1em] = \dfrac{139.44}{4} \\[1em] = 34.86 \text{ cm}^2.$

Hence, area of isosceles triangle = 34.86 cm².

Find the area of an isosceles triangle whose base is 6 cm and perimeter is 16 cm.

Answer

Given,

base = 6 cm and perimeter = 16 cm

Consider △ABC as an isosceles triangle in which,

Let, AB = AC = x cm.

So, BC = 6 cm.

We know that,

Perimeter of △ABC = AB + BC + AC

Substituting the values we get,

⇒ 16 = x + 6 + x

⇒ 16 = 2x + 6

⇒ 16 – 6 = 2x

⇒ 10 = 2x

⇒ x = $\dfrac{10}{2}$ = 5 cm.

Area of an isosceles triangle = $\dfrac{1}{4}b\sqrt{4a^2 - b^2}$, where a is length of equal sides and b is the length of other side.

Substituting values we get,

$A = \dfrac{1}{4} \times 6 \times \sqrt{4(5)^2 - 6^2} \\[1em] = \dfrac{1}{4} \times 6 \times \sqrt{4 \times 25 - 36} \\[1em] = \dfrac{1}{4} \times 6 \times \sqrt{100 - 36} \\[1em] = \dfrac{1}{4} \times 6 \times \sqrt{64} \\[1em] = \dfrac{1}{4} \times 6 \times 8 \\[1em] = 12 \text{ cm}^2.$

Hence, area of isosceles triangle = 12 cm².

The sides of a right-angled triangle containing the right angle are 5x cm and (3x – 1) cm. Calculate the length of the hypotenuse of the triangle if its area is 60 cm².

Answer

Consider △ABC as a right angled triangle.

AB = 5x cm and BC = (3x – 1) cm

We know that,

Area of △ABC = $\dfrac{1}{2}$ × base × height = $\dfrac{1}{2}$ × BC × AB

Substituting the values we get,

⇒ 60 = $\dfrac{1}{2}$ × (3x – 1) × 5x

⇒ 120 = 5x(3x – 1)

⇒ 120 = 15x² – 5x

⇒ 15x² – 5x – 120 = 0

⇒ 5(3x² – x – 24) = 0

⇒ 3x² – x – 24 = 0

⇒ 3x² – 9x + 8x – 24 = 0

⇒ 3x(x – 3) + 8(x - 3) = 0

⇒ (3x + 8)(x - 3) = 0

⇒ 3x + 8 = 0 or x - 3 = 0

⇒ 3x = -8 or x = 3

⇒ x = $-\dfrac{8}{3}$ or x = 3

Since, x cannot be negative as length of a side cannot be negative. So, x = 3.

AB = 5 × 3 = 15 cm

BC = (3 × 3 – 1) = 9 – 1 = 8 cm

In right angled △ABC,

Using Pythagoras theorem,

⇒ AC² = AB² + BC²

Substituting the values we get,

⇒ AC² = 15² + 8²

⇒ AC² = 225 + 64 = 289

⇒ AC² = 17²

So, AC = 17 cm.

Hence, the hypotenuse of the right angled triangle is 17 cm.

In △ABC, ∠B = 90°, AB = (2x + 1) cm and BC = (x + 1) cm. If the area of the △ABC is 60 cm², find its perimeter.

Answer

Given,

AB = (2x + 1) cm

BC = (x + 1) cm

We know that,

Area of △ABC = $\dfrac{1}{2}$ × base × height

= $\dfrac{1}{2}$ × BC × AB

Substituting the values we get,

⇒ 60 = $\dfrac{1}{2}$ × (x + 1) × (2x + 1)

⇒ 60 × 2 = (2x + 1)(x + 1)

⇒ 120 = 2x² + 3x + 1

⇒ 2x² + 3x + 1 – 120 = 0

⇒ 2x² + 3x – 119 = 0

⇒ 2x² + 17x – 14x – 119 = 0

⇒ x(2x + 17) – 7(2x + 17) = 0

⇒ (x – 7)(2x + 17) = 0

⇒ x – 7 = 0 or 2x + 17 = 0

⇒ x = 7 or 2x = -17

⇒ x = 7 or x = $-\dfrac{17}{2}$

Since, x cannot be negative as length of a side cannot be negative. So, x = 7.

⇒ AB = (2x + 1) = 2 × 7 + 1 = 15 cm

⇒ BC = (x + 1) = 7 + 1 = 8 cm.

In right angled △ABC,

Using Pythagoras Theorem,

⇒ AC² = AB² + BC²

Substituting the values we get,

⇒ AC² = 15² + 8²

⇒ AC² = 225 + 64

⇒ AC² = 289

⇒ AC² = 17²

⇒ AC = 17 cm

Perimeter of △ABC = AB + BC + AC = 15 + 8 + 17 = 40 cm.

Hence, perimeter of △ABC = 40 cm.

If the perimeter of a right angled triangle is 60 cm and its hypotenuse is 25 cm, find its area.

Answer

Let △ABC be the right angle triangle.

We know that,

Perimeter of a right-angled triangle = 60 cm

Hypotenuse = 25 cm

So, the sum of other two sides of triangle = 60 – 25 = 35 cm

Let base (BC) = x cm

So, AB = (35 - x) cm

Using the Pythagoras theorem,

⇒ AC² = AB² + BC²

⇒ 25² = (35 - x)² + x²

⇒ 625 = 1225 + x² - 70x + x²

⇒ 2x² - 70x + 600 = 0

Dividing by 2 on both sides,

⇒ x² - 35x + 300 = 0

⇒ x² - 15x - 20x + 300 = 0

⇒ x(x – 15) - 20(x - 15) = 0

⇒ (x - 15)(x - 20) = 0

⇒ x - 15 = 0 or x - 20 = 0

⇒ x = 15 or x = 20.

If x = 15, then 35 - x = 35 - 15 = 20 cm.

If x = 20, then 35 - x = 35 - 20 = 15 cm.

So, length of other two sides apart from hypotenuse are 15 cm and 20 cm.

Area = $\dfrac{1}{2}$ × base × height

Substituting the values we get,

A = $\dfrac{1}{2}$ × 15 × 20 = 150 cm².

Hence, area of triangle = 150 cm².

The perimeter of an isosceles triangle is 40 cm. The base is two third of the sum of equal sides. Find the length of each side.

Answer

Let the length of equal sides be x cm.

Length of base = $\dfrac{2}{3}(x + x) = \dfrac{2}{3} \times 2x = \dfrac{4x}{3}$ cm.

Given,

Perimeter = 40 cm

$\therefore x + x + \dfrac{4x}{3} = 40 \\[1em] \Rightarrow 2x + \dfrac{4x}{3} = 40 \\[1em] \Rightarrow \dfrac{6x + 4x}{3} = 40 \\[1em] \Rightarrow 10x = 120 \\[1em] \Rightarrow x = 12 \text{ cm}.$

Length of Base = $\dfrac{4x}{3}$ = $\dfrac{4}{3} \times 12$ = 16 cm

Hence, length of equal sides = 12 cm and length of base = 16 cm.

If the area of an isosceles triangle is 60 cm² and the length of each of its equal sides is 13 cm, find its base.

Answer

Let the length of equal sides be a cm and length of base be b cm.

Area of isosceles △ABC = $\dfrac{1}{4}b\sqrt{4a^2 - b^2}$, where a is length of equal sides and b is the length of other side.

Substituting values in above equation we get,

$\Rightarrow 60 = \dfrac{1}{4}b\sqrt{4(13)^2 - b^2} \\[1em] \Rightarrow 240 = b\sqrt{4 \times 169 - b^2} \\[1em] \Rightarrow 240 = b\sqrt{676 - b^2}$

Squaring both sides we get,

$\Rightarrow 57600 = b^2(676 - b^2) \\[1em] \Rightarrow 676b^2 - b^4 - 57600 = 0 \\[1em] \Rightarrow b^4 - 676b^2 + 57600 = 0 \\[1em] \Rightarrow b^4 - 576b^2 - 100b^2 + 57600 = 0 \\[1em] \Rightarrow b^2(b^2 - 576) - 100(b^2 - 576) = 0 \\[1em] \Rightarrow (b^2 - 100)(b^2 - 576) = 0 \\[1em] \Rightarrow (b^2 - 100) = 0 \text{ or } (b^2 - 576) = 0 \\[1em] \Rightarrow b^2 = 100 \text{ or } b^2 = 576 \\[1em] \Rightarrow b = 10 \text{ or } b = 24.$

Hence, the length of base = 10 cm or 24 cm.

The base of a triangular field is 3 times its height. If the cost of cultivating the field at the rate of ₹25 per 100 m² is ₹60000, find its base and height.

Answer

Given,

Cost of cultivating the field at the rate of ₹25 per 100 m² = ₹ 60000

In ₹25, area of field cultivated = 100 m²

∴ In ₹60000, area of field cultivated = $\dfrac{100 \times 60000}{25}$ = 240000 m².

∴ Area of field = 240000 m².

Let base of field = b meters and height = 3b meters.

Area of triangle = $\dfrac{1}{2} \times$ base × height

Substituting values we get,

$\Rightarrow 240000 = \dfrac{1}{2} \times b \times 3b \\[1em] \Rightarrow 240000 = \dfrac{3b^2}{2} \\[1em] \Rightarrow b^2 = \dfrac{240000 \times 2}{3} \\[1em] \Rightarrow b^2 = 160000 \\[1em] \Rightarrow b = \sqrt{160000} = 400 \text{ m} \\[1em] \Rightarrow \text{height } = 3b = 3 \times 400 = 1200 \text{ m}.$

Hence, base = 400 m and height = 1200 m.

A triangular park ABC has sides 120 m, 80 m and 50 m (as shown in the adjoining figure). A gardner Dhania has to put a fence around it and also plant grass inside. How much area does she need to plant? Find the cost of fencing it with barbed wire at the rate of ₹ 200 per metre leaving a space 3 m wide for a gate on one side.

Answer

It is given that,

ABC is a triangular park with sides 120 m, 80 m and 50 m.

Here, the perimeter of △ABC = 120 + 80 + 50 = 250 m

Portion at which a gate is built = 3 m

Remaining perimeter = 250 – 3 = 247 m.

So, the length of the fence required around the park = 247 m.

Rate of fencing = ₹ 200 per metre

Total cost of fencing = 200 × 247 = ₹ 49,400.

We know that,

Semi perimeter (s) = $\dfrac{\text{Perimeter}}{2} = \dfrac{250}{2}$ = 125 m.

By formula,

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

Substituting values we get,

$A = \sqrt{125(125 - 120)(125 - 80)(125 - 50)}\\[1em] = \sqrt{125 \times 5 \times 45 \times 75}\\[1em] = \sqrt{(25 \times 5) \times 5 \times (5 \times 3 \times 3) \times (25 \times 3)} \\[1em] = \sqrt{25^2 \times 5 \times 5^2 \times 3^2 \times 3}\\[1em] = 25 \times 5 \times 3 \times \sqrt{5 \times 3}\\[1em] = 375 \sqrt{15} \text{ m}^2.$

Hence, area needed for plantation = $375 \sqrt{15}$ m² and cost of fencing = ₹49,400.

An umbrella is made by stitching 10 triangular pieces of cloth of two different colors (shown in the adjoining figure), each piece measuring 20 cm, 50 cm and 50 cm. How much cloth of each colour is required for the umbrella?

Answer

Semi-perimeter (s) = $\dfrac{a + b + c}{2} = \dfrac{20 + 50 + 50}{2} = \dfrac{120}{2}$ = 60 cm.

Total 10 triangular pieces of cloth are required. So, 5 of each colour.

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

Substituting values we get,

$A = \sqrt{60(60 - 20)(60 - 50)(60 - 50)} \\[1em] = \sqrt{60 \times 40 \times 10 \times 10} \\[1em] = \sqrt{240000} \\[1em] = 200\sqrt{6} \text{ cm}^2.$

There are 5 triangular pieces. So area of 5 pieces = $5 \times 200\sqrt{6} = 1000\sqrt{6}$ cm².

Hence, $1000\sqrt{6}$ cm² of each colour cloth is required.

In the figure (i) given below, ABC is an equilateral triangle with each side of length 10 cm. In △BCD, ∠D = 90° and CD = 6 cm. Find the area of the shaded region. Give your answer correct to one decimal place.

Answer

Given,

ABC is an equilateral triangle of side = 10 cm

We know that,

Area of equilateral triangle ABC = $\dfrac{\sqrt{3}}{4}\text{ (side)}^2$

Substituting the values we get,

$A = \dfrac{\sqrt{3}}{4} \times (10)^2 \\[1em] = \dfrac{\sqrt{3}}{4} \times 100 \\[1em] = 25\sqrt{3} \\[1em] = 43.3 \text{ cm}^2.$

In right angled triangle BCD,

⇒ BC² = BD² + CD²

⇒ 10² = BD² + 6²

⇒ BD² = 100 - 36

⇒ BD² = 64

⇒ BD = 8 cm.

We know that,

Area of right angled triangle = $\dfrac{1}{2}$ × base × height.

Area of △BCD = $\dfrac{1}{2} \times CD \times BD = \dfrac{1}{2} \times 6 \times 8$ = 24 cm²

From figure,

Area of shaded portion = Area of triangle ABC - Area of triangle BCD

Substituting the values we get,

Area of shaded portion = 43.3 - 24 = 19.3 cm².

Hence, area of shaded region = 19.3 cm².

In the figure (ii) given, ABC is an isosceles right-angled triangle and DEFG is a rectangle. If AD = AE = 3 cm and DB = EC = 4 cm, find the area of the shaded region.

Answer

From figure,

In right angle triangle ADE,

Using pythagoras theorem,

⇒ DE² = AD² + AE²

⇒ DE² = 3² + 3²

⇒ DE² = 9 + 9

⇒ DE² = 18

⇒ DE = $\sqrt{18} = 3\sqrt{2}$ cm

Since, DEFG is a rectangle.

∴ GF = DE = $3\sqrt{2}$ cm.

In △DBG and △ECF,

DB = EC = 4 cm

DG = EF (Opposite sides of rectangle are equal)

∠DGB = ∠EFC = 90° (∵ DEFG is a rectangle)

Hence, by RHS axiom △DBG ≅ △ECF.

So, BG = FC (By C.P.C.T.)

Let BG = FC = x.

In right angle triangle ABC,

⇒ BC² = AB² + AC²

⇒ BC² = 7² + 7²

⇒ BC² = 49 + 49

⇒ BC² = 98

⇒ BC = $\sqrt{98} = 7\sqrt{2}$ cm

From figure,

BG + GF + FC = BC

⇒ BG + GF + FC = $7\sqrt{2}$

⇒ x + $3\sqrt{2}$ + x = $7\sqrt{2}$

⇒ 2x = $7\sqrt{2} - 3\sqrt{2}$

⇒ 2x = $4\sqrt{2}$

⇒ x = $2\sqrt{2}$ cm.

In right angle triangle DBG,

⇒ DB² = BG² + DG²

⇒ 4² = $(2\sqrt{2})$² + DG²

⇒ 16 = 8 + DG²

⇒ DG² = 16 - 8 = 8

⇒ DG = $\sqrt{8} = 2\sqrt{2}$ cm.

Area of right angle triangle DBG = $\dfrac{1}{2}$ x BG x DG

= $\dfrac{1}{2}$ x $2\sqrt{2}$ x $2\sqrt{2}$

= $\dfrac{1}{2}$ x 8 = 4 cm².

Since, △DBG ≅ △ECF.

∴ Areas of both triangle are equal.

Area of right angle triangle ADE = $\dfrac{1}{2}$ x AD x AE

= $\dfrac{1}{2}$ x 3 x 3

= $\dfrac{9}{2}$ = 4.5 cm².

Area of shaded region = Area of (△ADE + △DBG + △ECF)

= 4.5 + 4 + 4 = 12.5 cm².

Hence, area of shaded region = 12.5 cm².