Mensuration

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.

Find the area of quadrilateral whose one diagonal is 20 cm long and the perpendiculars to this diagonal from other vertices are of length 9 cm and 15 cm.

Answer

Consider ABCD as a quadrilateral in which BD = 20 cm, AY = 15 cm and CX = 9 cm.

Area of quadrilateral ABCD = Area of triangle ABD + Area of triangle BCD

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

∴ Area of quadrilateral ABCD = $\dfrac{1}{2}$ × BD × AY + $\dfrac{1}{2}$ × BD × CX

Substituting the values we get,

Area of quadrilateral ABCD = $\dfrac{1}{2}$ x BD x (AY + CX)

= $\dfrac{1}{2}$ x 20 x (15 + 9)

= 10 x 24

= 240 cm²

Hence, area of quadrilateral = 240 cm².

Find the area of a quadrilateral whose diagonals are of length 18 cm and 12 cm and they intersect each other at right angles.

Answer

Consider ABCD as a quadrilateral in which the diagonals AC and BD intersect each other at M at right angles.

From figure,

AC = 18 cm and BD = 12 cm

When diagonals of a quadrilateral intersect at right angles,

Area of quadrilateral = $\dfrac{1}{2}$ x d₁ x d₂, where d₁ and d₂ are diagonals.

Substituting the values we get,

Area of quadrilateral ABCD = $\dfrac{1}{2}$ x 12 x 18

= 6 x 18

= 108 cm²

Hence, area of quadrilateral = 108 cm².

Find the area of the quadrilateral field ABCD whose sides AB = 40 m, BC = 28 m, CD = 15 m, AD = 9 m and ∠A = 90°.

Answer

From figure,

ABCD is a quadrilateral field.

In triangle BAD,

∠A = 90°

Using the Pythagoras Theorem

⇒ BD² = AB² + AD²

Substituting the values we get,

⇒ BD² = 40² + 9²

⇒ BD² = 1600 + 81 = 1681

⇒ BD = $\sqrt{1681}$ = 41 m

We know that,

Area of quadrilateral ABCD = Area of △BAD + Area of △BDC

Calculating area of △BDC,

In △BDC,

Let a = BD = 41 m, b = BC = 28 m and c = CD = 15 m.

Semi-perimeter (s) = $\dfrac{a + b + c}{2} = \dfrac{41 + 28 + 15}{2} = \dfrac{84}{2}$ = 42 m.

By Heron's formula,

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

Substituting values we get,

$A = \sqrt{42(42 - 41)(42 - 28)(42 - 15)} \\[1em] = \sqrt{42 \times 1 \times 14 \times 27} \\[1em] = \sqrt{15876} \\[1em] = 126 \text{ m}^2.$

Calculating area of △BAD,

$\text{Area of △BAD } = \dfrac{1}{2} \times \text{ base × height} \\[1em] = \dfrac{1}{2} \times BA \times AD \\[1em] = \dfrac{1}{2} \times 40 \times 9 \\[1em] = 180 \text{ m}^2$

Area of quadrilateral ABCD = Area of △BAD + Area of △BDC

= 180 + 126

= 306 m².

Hence, area of quadrilateral ABCD = 306 m².

Find the area of the quadrilateral ABCD in which ∠BCA = 90°, AB = 13 cm and ACD is an equilateral triangle of side 12 cm.

Answer

In right-angled △ABC,

Using Pythagoras theorem,

⇒ AB² = AC² + BC²

Substituting the values we get,

⇒ 13² = 12² + BC²

⇒ BC² = 13² – 12²

⇒ BC² = 169 – 144 = 25

⇒ BC = $\sqrt{25}$ = 5 cm.

Calculating area of △BCA,

$\text{Area of △BCA } = \dfrac{1}{2} \times \text{ base × height} \\[1em] = \dfrac{1}{2} \times AC \times BC \\[1em] = \dfrac{1}{2} \times 12 \times 5 \\[1em] = 30 \text{ cm}^2.$

Calculating area of △ACD,

$\text{Area of △ACD } = \dfrac{\sqrt{3}}{4} \times \text{ (side)}^2 \\[1em] = \dfrac{\sqrt{3}}{4} \times (12)^2 \\[1em] = \dfrac{\sqrt{3}}{4} \times 144 \\[1em] = 36\sqrt{3} \\[1em] = 62.35 \text{ cm}^2 \\[1em]$

From figure,

Area of quadrilateral ABCD = Area of △BCA + Area of △ACD

= 30 cm² + 62.35 cm²

= 92.35 cm².

Hence, area of quadrilateral ABCD = 92.35 cm².

Find the area of quadrilateral ABCD in which ∠B = 90° , AB = 6 cm, BC = 8 cm and CD = AD = 13 cm.

Answer

In △ABC,

Using Pythagoras theorem,

AC² = AB² + BC²

Substituting the values we get,

⇒ AC² = 6² + 8²

⇒ AC² = 36 + 64 = 100

⇒ AC² = 10²

⇒ AC = 10 cm

Calculating area of △ADC,

In △ADC,

Let a = AD = 13 cm, b = DC = 13 cm and c = AC = 10 cm.

Semi-perimeter (s) = $\dfrac{a + b + c}{2} = \dfrac{13 + 13 + 10}{2} = \dfrac{36}{2}$ = 18 cm.

By Heron's formula,

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

Substituting values we get,

$\text{Area of △ADC} = \sqrt{18(18 - 13)(18 - 13)(18 - 10)} \\[1em] = \sqrt{18 \times 5 \times 5 \times 8} \\[1em] = \sqrt{3600} \\[1em] = 60 \text{ cm}^2.$

Calculating area of △ABC,

$\text{Area of △ABC } = \dfrac{1}{2} \times \text{ base × height} \\[1em] = \dfrac{1}{2} \times AB \times BC \\[1em] = \dfrac{1}{2} \times 6 \times 8 \\[1em] = 24 \text{ cm}^2$

From figure,

Area of quadrilateral ABCD = Area of △ADC + Area of △ABC

= 60 + 24

= 84 cm².

Hence, area of quadrilateral ABCD = 84 cm².

The perimeter of a rectangular cardboard is 96 cm; if its breadth is 18 cm, find the length and the area of the cardboard.

Answer

We know that,

Perimeter of rectangle = 2 × (l + b) = 96 cm

Substituting the values we get,

⇒ 2(l + 18) = 96

⇒ (l + 18) = 48

⇒ l = 48 - 18 = 30 cm.

Area of rectangular cardboard = l × b

Substituting the values we get,

Area = 30 × 18 = 540 cm².

Hence, length = 30 cm and area of rectanglular cardboard = 540 cm².

The length of a rectangular hall is 5 m more than its breadth. If the area of the hall is 594 m², find its perimeter.

Answer

Let breadth = x meters

So, length = (x + 5) meters

We know that,

Area of rectangular hall = length × breadth

Substituting the values we get,

⇒ 594 = x(x + 5)

⇒ 594 = x² + 5x

⇒ x² + 5x – 594 = 0

⇒ x² + 27x – 22x – 594 = 0

⇒ x(x + 27) – 22(x + 27) = 0

⇒ (x – 22)(x + 27) = 0

⇒ x – 22 = 0 or x + 27 = 0

⇒ x = 22 or x = -27.

Since, length of side cannot be negative so, x ≠ -27.

∴ Breadth = x = 22 m and Length = (x + 5) = 22 + 5 = 27 m.

Perimeter = 2(l + b)

Substituting the values we get,

Perimeter = 2(27 + 22) = 2 × 49 = 98 m.

Hence, perimeter of hall = 98 m.

The diagram (i) given below shows two paths drawn inside a rectangular field 50 m long and 35 m wide. The width of each path is 5 metres. Find the area of the shaded portion.

Answer

We know that,

Area of rectangle = length × breadth

Area of square = side × side.

From figure,

Area of shaded portion = Area of rectangle ABCD + Area of rectangle PQRS – Area of square LMNO

Substituting values we get,

Area of shaded portion = AB × AD + PR × RS - LM × MN

= 50 × 5 + 35 × 5 - 5 × 5

= 250 + 175 - 25

= 400 m².

Hence, area of shaded region = 400 m².

In the diagram (ii) given below, calculate the area of the shaded portion. All measurements are in centimetres.

Answer

We know that,

Area of rectangle = length × breadth

Area of square = side × side.

From figure,

Area of shaded portion = Area of large rectangle - 5 × Area of a small square.

Substituting values we get,

Area of shaded portion = (8 × 6) - (5 × 2 × 2)

= 48 - 20

= 28 cm².

Hence, area of shaded region = 28 cm².

A rectangular plot 20 m long and 14 m wide is to be covered with grass leaving 2 m all around. Find the area to be laid with grass.

Answer

Consider ABCD as a plot.

Length of plot = 20 m and breadth of plot = 14 m.

Let PQRS be the plot to be covered with grass.

From figure,

PQ = 20 - (2 × 2)

= 20 - 4

= 16 m

QR = 14 - (2 × 2)

= 14 - 4

= 10 m

Area of rectangular plot PQRS = length × breadth

Substituting the values we get,

Area = 16 × 10 = 160 m².

Hence, area to be laid with grass = 160 m².

The shaded region of the given diagram represents the lawn in front of a house. On three sides of the lawn there are flower beds of width 2 m.

(i) Find the length and the breadth of the lawn.

(ii) Hence, or otherwise, find the area of the flower–beds.

Answer

(i) Let PQRS be the lawn.

From figure,

QR = BC - BQ - RC = 30 - 2 - 2 = 26 m.

SR = CD - DS = 12 - 2 = 10 m.

Length of PQRS = QR = 26 m and,

Breadth of PQRS = SR = 10 m.

Hence, length and breadth of lawn are 26 m and 10 m respectively.

(ii) From figure,

Area of flower beds = Area of rectangle ABCD - Area of rectangle PQRS

= (AD × DC) - (QR × SR)

= (30 × 12) - (26 × 10)

= 360 - 260

= 100 m².

Hence, area of flower-beds = 100 m².

A footpath of uniform width runs all around the inside of a rectangular field 50 m long and 38 m wide. If the area of the path is 492 m², find its width.

Answer

Consider ABCD as a rectangular field having, length = 50 m and breadth = 38 m.

Let x meters be the width of foot path.

We know that,

Area of rectangular = l × b

From figure,

Area of path = Area of rectangle ABCD - Area of rectangle PQRS

Substituting the values we get,

Area of path = (AB × BC) - (PQ × QR) ........(1)

From figure,

PQ = AB - x - x = (50 - 2x) m,

QR = BC - x - x = (BC - 2x) m.

Substituting the values in equation 1 we get,

⇒ 492 = (50 × 38) - (50 - 2x) (38 - 2x)

⇒ 492 = 1900 - [50(38 - 2x) - 2x(38 - 2x)]

⇒ 492 = 1900 - (1900 - 100x - 76x + 4x²)

⇒ 492 = 1900 - 1900 + 100x + 76x - 4x²

⇒ 492 = 176x - 4x²

⇒ 492 = 4(44x - x²)

⇒ 123 = 44x - x²

⇒ x² - 44x + 123 = 0

⇒ x² - 41x - 3x + 123 = 0

⇒ x(x - 41) - 3(x - 41) = 0

⇒ (x - 3)(x - 41) = 0

⇒ x - 3 = 0 or x - 41 = 0

⇒ x = 3 m or x = 41 m.

Since, width of path cannot be greater than breadth of field,

So, x ≠ 41 m.

Hence, width of the footpath is 3 m.

The cost of enclosing a rectangular garden with a fence all around at the rate of ₹ 150 per metre is ₹ 54,000. If the length of the garden is 100 m, find the area of the garden.

Answer

Given,

Length = 100 m.

Let breadth = x meters.

By formula,

Perimeter of rectangle = 2(l + b)

Substituting the values we get,

Perimeter of rectangular garden = 2(100 + x) = (200 + 2x) m.

Given,

Cost of enclosing fence = ₹ 150 per meter.

∴ Cost of enclosing fence all round the rectangular garden = ₹150(200 + 2x) = ₹(30,000 + 300x).

Given, total cost of fencing = ₹ 54,000

∴ 30,000 + 300x = 54,000

⇒ 300x = 54,000 – 30,000

⇒ 300x = 24,000

⇒ x = $\dfrac{24,000}{300}$

⇒ x = 80 m.

∴ Breadth of garden = 80 m.

So, the area of rectangular garden = length × breadth

= 100 × 80

= 8000 m².

Hence, the area of rectangular garden = 8000 m².

A rectangular floor which measures 15 m × 8 m is to be laid with tiles measuring 50 cm × 25 cm. Find the number of tiles required. Further, if a carpet is laid on the floor so that a space of 1 m exists between its edges and the edges of the floor, what fraction of the floor is uncovered?

Answer

Let ABCD be the rectangular floor and PQRS be the carpet.

Area of floor = l × b = 15 × 8 = 120 m² = 120 × (100 cm)² = 1200000 cm²

Area of a tile = 50 cm × 25 cm = 1250 cm²

No. of required tiles = $\dfrac{\text{Area of rect. floor}}{\text{Area of a tile}}$

Substituting the values we get,

No. of required tiles = $\dfrac{1200000}{1250}$ = 960.

From figure,

Length of carpet (PQ) = 15 – 1 – 1

= 15 – 2

= 13 m

Breadth of carpet (QR) = 8 – 1 – 1

= 8 – 2

= 6 m

Area of carpet = l × b

= 13 × 6

= 78 m².

Area of floor which is uncovered by carpet = Area of floor – Area of carpet

= 120 – 78

= 42 m²

Fraction of floor uncovered = $\dfrac{\text{Area of floor uncovered}}{\text{Area of floor}}$

= $\dfrac{42}{120} = \dfrac{7}{20}$.

Hence, number of tiles required to cover the floor = 960 tiles and $\dfrac{7}{20}$ is the fraction of floor uncovered.

The width of a rectangular room is $\dfrac{3}{5}$ of its length $x$ metres. If its perimeter is $y$ metres, write an equation connecting $x$ and $y$. Find the floor area of the room if its perimeter is 32 m.

Answer

Given,

Length of rectangular room = x meters

Width of rectangular room = $\dfrac{3}{5}x$ meters.

Perimeter = y meters.

We know that,

Perimeter = 2(l + b)

Substituting the values we get,

$\Rightarrow y = 2\Big[x + \dfrac{3}{5}x\Big] \\[1em] \Rightarrow y = 2\Big[\dfrac{5x + 3x}{5}\Big] \\[1em] \Rightarrow 5y = 2 \times 8x \\[1em] \Rightarrow 5y = 16x \text{ ........ (1)}$

The above equation is the required relation between x and y.

Given, perimeter = y = 32 m.

Now substituting the value of y in equation (1)

⇒ 16x = 5 × 32

⇒ x = $\dfrac{160}{16}$ = 10 m,

⇒ Breadth = $\dfrac{3}{5} \times x = \dfrac{3}{5} \times 10$ = 6 m.

Floor area of the room = l × b

= 10 × 6

= 60 m².

Hence, 16x = 5y is the equation connecting x and y and the floor area of room = 60 m².

A rectangular garden 10 m by 16 m is to be surrounded by a concrete walk of uniform width. Given that the area of the walk is 120 square metres, assuming the width of the walk to be x, form an equation in x and solve it to find the value of x.

Answer

Let ABCD be a rectangular garden.

Length = 10 m and Breadth = 16 m.

Area of garden ABCD = l × b

= 10 × 16 = 160 m²

Given, width of the walk = x meters.

From figure,

Length of rectangular garden PQRS = 10 + x + x = (10 + 2x) m

Breadth of rectangular garden PQRS = 16 + x + x = (16 + 2x) m

From figure,

⇒ Area of walk = Area of rectangle PQRS - Area of rectangle ABCD

⇒ 120 = (10 + 2x)(16 + 2x) - 160

⇒ 120 = 160 + 20x + 32x + 4x² - 160

⇒ 120 = 4x² + 52x

⇒ 4x² + 52x - 120 = 0

⇒ 4(x² + 13x - 30) = 0

⇒ x² + 13x - 30 = 0

The above equation is the equation in x.

Solving further,

⇒ x² + 15x - 2x - 30 = 0

⇒ x(x + 15) - 2(x + 15) = 0

⇒ (x - 2)(x + 15) = 0

⇒ x = 2 or x = -15.

Since, length cannot be negative.

∴ x = 2.

Hence, equation is x² + 13x - 30 = 0 and x = 2 metres.

A rectangular room is 6 m long, 4.8 m wide and 3.5 m high. Find the inner surface area of the four walls.

Answer

It is given that

Length of rectangular room = 6 m

Breadth of rectangular room = 4.8 m

Height of rectangular room = 3.5 m

By formula,

Inner surface area of four walls = 2(l + b) × h

= 2(6 + 4.8) × 3.5

= 2 × 10.8 × 3.5

= 21.6 × 3.5

= 75.6 m².

Hence, inner surface area of four walls = 75.6 m².

A rectangular plot of land measures 41 metres in length and 22.5 metres in width. A boundary wall 2 metres high is built all around the plot at a distance of 1.5 m from the plot. Find the inner surface area of the boundary wall.

Answer

Let ABCD be the rectangular plot.

Given,

Length of rectangular plot = 41 metres,

Breadth of rectangular plot = 22.5 metres.

Height of boundary wall = 2 metres.

Boundary wall is built at a distance of 1.5 m. It means wall is built on base PQRS.

From figure,

Length of plot PQRS = 41 + 1.5 + 1.5 = 44 m.

Breadth of plot PQRS = 22.5 + 1.5 + 1.5 = 25.5 m.

By formula,

Inner surface area of the boundary wall = 2(l + b) × h

= 2 (44 + 25.5) × 2

= 2 × 69.5 × 2

= 278 m².

Hence, inner surface area of boundary wall = 278 m².

Find the perimeter and area of the figure (i) given below in which all corners are right angles.

Answer

From figure,

Area of rectangle ABFG = l × b

= BF × AB

= (BC + CF) × AB

= (4 + 1.5) × 2

= 5.5 × 2

= 11 m².

Area of rectangle CDEF = l × b

= CD × DE

= 4 × 1.5

= 6 m².

Total Area = Area of rectangle ABFG + Area of rectangle CDEF

= 11 + 6 = 17 m².

From figure,

AG = BF, GF = AB and FE = CD (As opposite sides of rectangle are equal.)

Perimeter of figure = AB + BC + CD + DE + EF + FG + AG

= 2 + 4 + 4 + 1.5 + 4 + 2 + 5.5

= 23 m.

Hence, perimeter = 23 m and area = 17 m².

Find the perimeter and area of the figure (ii) given below in which all corners are right angles.

Answer

The points are labelled on the figure as shown below:

Area of rectangle ABIJ = l × b

= AB × BI

= AB × (BC + CI)

= 3 × (5 + 2)

= 3 × 7 = 21 m².

Area of rectangle EFGH = l × b

= EF × FG

= 2 × 7 = 14 m².

Area of rectangle CDHI = l × b

= CD × DH

= 8 × 2 = 16 m².

Total area = Area of rectangle ABIJ + Area of rectangle EFGH + Area of rectangle CDHI

= 21 + 14 + 16

= 51 m².

Perimeter of figure = AB + BC + CD + DE + EF + FG + GH + HI + IJ + JA

= 3 + 5 + 8 + 5 + 2 + 7 + 2 + 8 + 3 + 7

= 50 m.

Hence, perimeter = 50 m and area = 51 m².

Find the area and perimeter of the figure (iii) given below in which all corners are right angles and all measurements are in cm.

Answer

The points are labelled on the figure as shown below:

Area of rectangle BCDE = l × b

= ED × CD

= 5 × 2

= 10 cm².

Area of rectangle FGHI = l × b

= FG × GH

= 3 × 2

= 6 cm².

Area of rectangle JKLM = l × b

= JK × KL

= 5 × 2

= 10 cm².

Area of rectangle ABMN = l × b

= AB × AN

(From figure AB = AC - BC = 7 - 5 = 2 cm.)

= 2 × 12

= 24 cm².

From figure,

Total area = Area of rectangle BCDE + Area of rectangle FGHI + Area of rectangle JKLM + Area of rectangle ABMN

= 10 + 6 + 10 + 24

= 50 cm².

From figure,

The vertical distance between C and L will be equal to vertical distance between A and N,

So ignoring the vertical sides in right side and replacing it will CL.

Perimeter = AC + DE + FG + HI + JK + LN + NA + CL

= 7 + 5 + 3 + 3 + 5 + 7 + 12 + 12

= 54 cm.

Hence, area = 50 cm² and perimeter = 54 cm.

The length and the breadth of a rectangle are 12 cm and 9 cm respectively. Find the height of a triangle whose base is 9 cm and whose area is one-third that of rectangle.

Answer

Area of rectangle = l × b = 12 × 9

= 108 cm².

Given,

Area of triangle = one-third the area of rectangle.

Substituting the values we get,

Area of triangle = $\dfrac{1}{3}$ × 108 = 36 cm².

Consider h cm as the height of triangle.

By formula,

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

Substituting the values we get,

⇒ 36 = $\dfrac{1}{2}$ × 9 × h

⇒ 36 × 2 = 9 × h

⇒ h = $\dfrac{72}{9}$

⇒ h = 8 cm.

Hence, height of triangle is 8 cm.

The area of a square plot is 484 m². Find the length of its one side and the length of its one diagonal.

Answer

Let ABCD be the square plot having area 484 m².

Let length of each side of the plot be x meters.

We know that,

Area of square = side × side

Substituting the values we get,

⇒ 484 = (x)²

⇒ x = $\sqrt{484}$ = 22 m.

Since, each angle = 90° in a square.

In right angle triangle ABC,

Using Pythagoras Theorem,

⇒ AC² = AB² + BC²

⇒ AC² = 22² + 22²

⇒ AC² = 484 + 484 = 968

⇒ AC = $\sqrt{968} = 22\sqrt{2}$

⇒ AC = 22 × 1.414 = 31.11 m.

Hence, length of side = 22 m and length of diagonal = 31.11 m.

A square has the perimeter 56 m. Find its area and the length of one diagonal correct up to two decimal places.

Answer

Let ABCD be a square with side x metres.

Perimeter of square = 4 × side

Substituting the values we get,

⇒ 56 = 4x

⇒ x = $\dfrac{56}{4}$ = 14 m.

Since, each angle = 90° in a square.

In right angle triangle ABC

Using Pythagoras theorem,

⇒ AC² = AB² + BC²

⇒ AC² = 14² + 14²

⇒ AC² = 196 + 196 = 392

⇒ AC = $\sqrt{392}$

⇒ AC = $14\sqrt{2}$ = 14 × 1.414 = 19.80 m.

Area of square = (side)²

= 14² = 196 m².

Hence, the area of square = 196 m² and length of diagonal = 19.80 m.

A wire when bent in the form of an equilateral triangle encloses an area of $36\sqrt{3}$ cm². Find the area enclosed by the same wire when bent to form:

(i) a square, and

(ii) a rectangle whose length is 2 cm more than its width.

Answer

Given,

Area of equilateral triangle = $36\sqrt{3}$ cm²

Consider x cm as the side of equilateral triangle

We know that,

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

Substituting the values we get,

$\Rightarrow \dfrac{\sqrt{3}}{4}(\text{ side})^2 = 36\sqrt{3} \\[1em] \Rightarrow \text{side}^2 = \dfrac{36 \sqrt{3} \times 4}{\sqrt{3}} \\[1em] \Rightarrow \text{side}^2 = 144 \\[1em] \Rightarrow \text{side} = \sqrt{144} = 12 \text{ cm}.$

By formula,

Perimeter of equilateral triangle = 3 × side = 3 × 12 = 36 cm.

(i) As the same wire is now bent to form a square.

∴ Perimeter of equilateral triangle = Perimeter of square

36 = 4 × side

Side = $\dfrac{36}{4}$ = 9 cm.

Area of square = side × side = 9 × 9 = 81 cm².

Hence, area enclosed by wire in form of square = 81 cm².

(ii) As the same wire is now bent to form a rectangle.

∴ Perimeter of triangle = Perimeter of rectangle ........(1)

According to the condition given for rectangle,

Length is 2 cm more than its width

Let width of rectangle = x cm

∴ Length of rectangle = (x + 2) cm

Perimeter of rectangle = 2(l + b)

Substituting the values in equation 1 we get,

⇒ 36 = 2[(x + 2) + x]

⇒ 36 = 2[2x + 2]

⇒ 4x + 4 = 36

⇒ 4x = 32

⇒ x = 8 cm.

∴ Length of rectangle = x + 2 = 8 + 2 = 10 cm and Breadth of rectangle = x = 8 cm.

By formula,

Area of rectangle = length × breadth

= 10 × 8

= 80 cm².

Hence, area enclosed by wire in form of square = 80 cm².

Two adjacent sides of a parallelogram are 15 cm and 10 cm. If the distance between the longer sides is 8 cm, find the area of the parallelogram. Also find the distance between shorter sides.

Answer

Let ABCD be a parallelogram with side AB = 15 cm and side BC = 10 cm.

Distance between longer side DM = 8 cm

Consider DN as the distance between the shorter side

Area of parallelogram ABCD = base × height

= AB × DM = 15 × 8 = 120 cm².

Considering base BC and height DN

Area of parallelogram = BC × DN

⇒ 120 = 10 × DN

⇒ DN = $\dfrac{120}{10}$ = 12 cm.

Hence, the area of parallelogram = 120 cm² and the distance between shorter side = 12 cm.

ABCD is a parallelogram with sides AB = 12 cm, BC = 10 cm and diagonal AC = 16 cm. Find the area of the parallelogram. Also find the distance between its shorter sides.

Answer

In triangle ABC,

Let,

BC = a = 10 cm, AC = b = 16 cm and AB = c = 12 cm.

We know that,

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

= $\dfrac{10 + 16 + 12}{2} = \dfrac{38}{2}$ = 19 cm.

By Heron's formula,

$\text{Area of triangle } = \sqrt{s(s - a)(s - b)(s - c)} \\[1em] = \sqrt{19(19 - 10)(19 - 16)(19 - 12)} \\[1em] = \sqrt{19 \times 9 \times 3 \times 7} \\[1em] = \sqrt{3591} \\[1em] = 59.9 \text{ cm}^2.$

We know that,

Diagonal of a parallelogram divides it into two triangles of equal area.

∴ Area of triangle ABC = Area of triangle ADC

∴ Area of parallelogram = 2 × Area of triangle ABC.

= 2 × 59.9

= 119.8 cm².

Let DM be the distance between the shorter sides of the parallelogram.

By formula,

Area of parallelogram = base × height = BC × DM

Substituting the values we get,

⇒ 119.8 = 10 × DM

⇒ DM = $\dfrac{119.8}{10}$

⇒ DM = 11.98 cm.

Hence, the distance between shorter sides = 11.98 cm and area of parallelogram = 119.8 cm².

Diagonals AC and BD of a parallelogram ABCD intersect at O. Given that AB = 12 cm and perpendicular distance between AB and DC is 6 cm. Calculate the area of the triangle AOD.

Answer

Let ABCD be a parallelogram with AC and BD the diagonals intersecting at O.

From figure,

AB = 12 cm and DM = 6 cm.

By formula,

Area of parallelogram ABCD = base × height = AB × DM

= 12 × 6

= 72 cm².

Since, diagonals of parallelogram intersect each other so O is the mid-point of BD.

∴ AO is the median of the △ABD.

Since, median divides the triangle into two triangles of equal area,

∴ Area of △AOD = $\dfrac{1}{2}$ × Area of △ABD ......(1)

Since, diagonal of a parallelogram divides it into two triangles of equal area.

∴ Area of △ABD = $\dfrac{1}{2}$ × Area of || gm ABCD.

Substituting above value of △ABD in equation 1 we get,

Area of △AOD = $\dfrac{1}{2} \times \dfrac{1}{2}$ Area of || gm ABCD

= $\dfrac{1}{4} \times 72$ = 18 cm².

Hence, area of △AOD = 18 cm².

ABCD is a parallelogram with side AB = 10 cm. Its diagonals AC and BD are of length 12 cm and 16 cm respectively. Find the area of the parallelogram ABCD.

Answer

Let ABCD be a parallelogram with diagonals intersecting at O.

Since, diagonals of a parallelogram bisect each other.

∴ AO = $\dfrac{12}{2}$ = 6 cm and OB = $\dfrac{16}{2}$ = 8 cm.

In triangle AOB,

Let AB = a = 10 cm, BO = b = 8 cm and OA = c = 6 cm.

We know that,

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

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

By Heron's formula,

$\text{Area of triangle } = \sqrt{s(s - a)(s - b)(s - c)} \\[1em] = \sqrt{12(12 - 10)(12 - 8)(12 - 6)} \\[1em] = \sqrt{12 \times 2 \times 4 \times 6} \\[1em] = \sqrt{576} \\[1em] = 24 \text{ cm}^2.$

Since, diagonals of parallelogram intersect each other so O is the mid-point of BD.

∴ AO is the median of the △ABD.

Since, median divides the triangle into two triangles of equal area.

∴ Area of △AOB = $\dfrac{1}{2}$ × Area of △ABD ......(1)

Since, diagonal of a parallelogram divides it into two triangles of equal area.

∴ Area of △ABD = $\dfrac{1}{2}$ × Area of || gm ABCD.

Substituting above value of △ABD in equation 1 we get,

Area of △AOB = $\dfrac{1}{2} \times \dfrac{1}{2}$ Area of || gm ABCD

Substituting values in above equation we get,

24 = $\dfrac{1}{4}$ Area of || gm ABCD

⇒ Area of || gm ABCD = 24 × 4 = 96 cm².

Hence, area of || gm ABCD = 96 cm².

The area of a parallelogram is p cm² and its height is q cm. A second parallelogram has equal area but its base is r cm more than that of the first. Obtain an expression in terms of p, q and r for the height h of the second parallelogram.

Answer

By formula,

Area of a parallelogram = base × height ........(1)

Substituting values of 1st || gm in above equation,

⇒ p = base × q

⇒ base = $\dfrac{p}{q}$ cm.

Given,

Base of 2nd || gm is r cm more than the base of 1st || gm.

∴ Base of 2nd || gm = $\Big(\dfrac{p}{q} + r\Big)$ cm.

Given,

Height of second parallelogram = h cm

Substituting the values of 2nd || gm in equation 1,

$\Rightarrow p = \Big(\dfrac{p}{q} + r\Big) \times h \\[1em] \Rightarrow p = \Big(\dfrac{p + qr}{q}\Big) \times h \\[1em] \Rightarrow h = \Big(\dfrac{pq}{p + qr}\Big).$

Hence, h = $\Big(\dfrac{pq}{p + qr}\Big)$ cm.

What is the area of a rhombus whose diagonals are 12 cm and 16 cm?

Answer

By formula,

Area of rhombus = $\dfrac{1}{2}$ × d₁ × d₂, where d₁ and d₂ are diagonals.

Substituting the values we get,

Area of rhombus = $\dfrac{1}{2}$ × 16 × 12

= 8 × 12 = 96 cm².

Hence, area of rhombus = 96 cm².

The area of a rhombus is 98 cm². If one of its diagonal is 14 cm, what is the length of the other diagonal?

Answer

By formula,

Area of rhombus = $\dfrac{1}{2}$ × d₁ × d₂, where d₁ and d₂ are diagonals.

Substituting the values we get,

⇒ 98 = $\dfrac{1}{2}$ × 14 × d₂

⇒ d₂ = $\dfrac{98 \times 2}{14}$ = 14 cm.

Hence, the length of other diagonal = 14 cm.

The perimeter of a rhombus is 45 cm. If its height is 8 cm, calculate its area.

Answer

Let length of each side of rhombus = x cm.

Given,

Perimeter = 45 cm

⇒ x + x + x + x = 45

⇒ 4x = 45

⇒ x = $\dfrac{45}{4}$ cm

By formula,

Area of rhombus = base × height

Substituting the values we get,

Area of rhombus = $\dfrac{45}{4}$ × 8 = 90 cm².

Hence, area of rhombus = 90 cm².

PQRS is a rhombus. If it is given that PQ = 3 cm and the height of the rhombus is 2.5 cm, calculate its area.

Answer

From figure,

PQ is the base of rhombus PQRS and SM is the height of rhombus.

By formula,

Area of rhombus PQRS = base × height

= 3 × 2.5

= 7.5 cm².

Hence, area of rhombus PQRS = 7.5 cm².

If the diagonals of a rhombus are 8 cm and 6 cm, find its perimeter.

Answer

Let ABCD be a rhombus with AC and BD as two diagonals.

Let AC = 8 cm and BD = 6 cm.

Since, diagonals of a rhombus bisect each other at right angles.

∴ AO = 4 cm and BO = 3 cm.

In right angle triangle AOB,

Using Pythagoras theorem

⇒ AB² = AO² + BO²

⇒ AB² = 4² + 3²

⇒ AB² = 16 + 9 = 25

⇒ AB = $\sqrt{25}$ = 5 cm.

Side of rhombus ABCD = 5 cm

By formula,

Perimeter of rhombus = 4 × side = 4 × 5 = 20 cm.

Hence, perimeter of rhombus = 20 cm.

If the sides of a rhombus are 5 cm each and one diagonal is 8 cm, calculate

(i) the length of the other diagonal, and

(ii) the area of the rhombus.

Answer

(i) Let ABCD be a rhombus with AC and BD diagonals and each side = 5 cm.

Let AC = 8 cm.

Since, diagonals of rhombus bisect each other at right angles.

∴ AO = 4 cm.

In right angle triangle AOB

Using Pythagoras theorem,

⇒ AB² = AO² + BO²

⇒ 5² = 4² + BO²

⇒ 25 = 16 + BO²

⇒ BO² = 25 – 16 = 9

⇒ BO = $\sqrt{9}$ = 3 cm.

⇒ BD = 2 × BO = 2 × 3 = 6 cm.

Hence, length of other diagonal = 6 cm.

(ii) Area of rhombus = $\dfrac{1}{2}$ × product of diagonals

= $\dfrac{1}{2}$ × 8 × 6

= 4 × 6

= 24 cm².

Hence, area of rhombus = 24 cm².

The figure (i) given below is a trapezium. Find the length of BC and the area of the trapezium. Assume AB = 5 cm, AD = 4 cm, CD = 8 cm.

Answer

(a) Construct BN perpendicular to CD.

So, BADN is a rectangle.

As opposite sides of rectangle are equal.

∴ BN = AD = 4 cm and ND = BA = 5 cm.

From figure,

CN = CD – ND = 8 - 5 = 3 cm.

In right angle triangle BCN,

Using Pythagoras theorem,

⇒ BC² = BN² + CN²

⇒ BC² = 4² + 3²

⇒ BC² = 16 + 9 = 25

⇒ BC = $\sqrt{25}$ = 5 cm.

By formula,

Area of trapezium = $\dfrac{1}{2}$ × sum of parallel sides × height

= $\dfrac{1}{2}$ × (AB + CD) × AD

= $\dfrac{1}{2}$ × (5 + 8) × 4

= 13 × 2 = 26 cm².

Hence, BC = 5 cm and area of trapezium = 26 cm².

The figure (ii) given below is a trapezium. Find

(i) AB

(ii) area of trapezium ABCD.

Answer

(i) Construct a perpendicular from C to AD parallel to AB.

So, ABCM is a rectangle. Since, opposite sides of a rectangle are equal.

∴ AM = CB = 2 units.

From figure,

⇒ AD = AM + MD

⇒ MD = AD - AM = 8 - 2 = 6 units.

In right angle triangle MDC,

⇒ CD² = MD² + CM²

⇒ 10² = 6² + CM²

⇒ 100 = 36 + CM²

⇒ CM² = 64

⇒ CM = $\sqrt{64}$ = 8 units.

Since, ABCM is a rectangle.

∴ AB = CM = 8 units.

Hence, AB = 8 units.

(ii) Area of trapezium ABCD = $\dfrac{1}{2}$ × (sum of || sides) × distance between them

= $\dfrac{1}{2}$ × (AD + BC) × AB

= $\dfrac{1}{2}$ × (8 + 2) × 8

= 40 sq. units.

Hence, area of trapezium ABCD = 40 sq. units.

The cross-section of a canal is shown in figure (iii) given below. If the canal is 8 m wide at the top and 6 m wide at the bottom and the area of the cross-section is 16.8 m², calculate its depth.

Answer

Consider ABCD as the cross section of canal in the shape of trapezium.

Draw a perpendicular AM from A to CD.

So, AM is the depth of canal.

Given, the area of cross-section of canal = 16.8 m².

∴ $\dfrac{1}{2}$ × sum of parallel sides × depth = 16.8

⇒ $\dfrac{1}{2}$ × (AB + DC) × AM = 16.8

⇒ $\dfrac{1}{2}$ × (6 + 8) × AM = 16.8

⇒ $\dfrac{1}{2}$ × 14 × AM = 16.8

⇒ AM = $\dfrac{16.8 \times 2}{14}$

⇒ AM = $\dfrac{33.6}{14}$ = 2.4 m

Hence, depth of canal = 2.4 meters.

The distance between parallel sides of a trapezium is 12 cm and the distance between mid-points of other sides is 18 cm. Find the area of the trapezium.

Answer

Let ABCD be the trapezium in which AB || DC. Let E and F be mid-points of sides AD and BC respectively, then EF = 18 cm.

Given,

Distance between parallel sides of a trapezium is 12 cm.

∴ Height = 12 cm.

By formula,

Sum of the lengths of two parallel sides = 2 × Distance between mid-points of two non-parallel sides

⇒ AB + CD = 2 × EF = 2 × 18 = 36 cm.

Area of trapezium = $\dfrac{1}{2}$ × (Sum of parallel sides) × height

= $\dfrac{1}{2}$ × (AB + CD) × 12

= $\dfrac{1}{2}$ × 36 × 12

= 18 × 12

= 216 cm².

Hence, area of trapezium = 216 cm².

The area of a trapezium is 540 cm². If the ratio of parallel sides is 7 : 5 and the distance between them is 18 cm, find the length of parallel sides.

Answer

Given,

Area of trapezium = 540 cm²

Ratio of parallel sides = 7 : 5

Let the sides be 7x and 5x cm.

Distance between the parallel sides = height = 18 cm

By formula,

Area of trapezium = $\dfrac{1}{2}$ × sum of parallel sides × height

⇒ 540 = $\dfrac{1}{2}$ × (7x + 5x) × 18

⇒ 540 = $\dfrac{1}{2}$ × 12x × 18

⇒ 540 = 6x × 18

⇒ 540 = 108x

⇒ x = $\dfrac{540}{108}$ = 5 cm.

⇒ 7x = 7 × 5 = 35 cm and 5x = 5 × 5 = 25 cm.

Hence, the length of parallel sides are 25 cm and 35 cm.

The parallel sides of an isosceles trapezium are in the ratio 2 : 3. If its height is 4 cm and area is 60 cm², find the perimeter.

Answer

Since, ABCD is an isosceles trapezium so, BC = AD.

Since, parallel sides of an isosceles trapezium are in the ratio 2 : 3.

∴ CD = 2a and AB = 3a.

Construct perpendicular DN from D to AB and perpendicular CM from C to AB.

Given,

Area = 60 cm²

By formula,

Area of trapezium = $\dfrac{1}{2}$ × sum of parallel sides × height

⇒ 60 = $\dfrac{1}{2}$ × (AB + DC) × DN

⇒ 60 = $\dfrac{1}{2}$ × (3a + 2a) × 4

⇒ 60 = 2 × 5a

⇒ 10a = 60

⇒ a = 6 cm.

⇒ AB = 3a = 3 × 6 = 18 cm and CD = 2a = 2 × 6 = 12 cm.

In △ADN and △BCM,

⇒ ∠AND = ∠CMB = 90°

⇒ DN = CM = 4 cm

⇒ AD = CB = x cm (let) (As ABCD is an isosceles trapezium).

∴ △ADN ≅ △BCM by RHS axiom.

∴ AN = MB ........(1)

Since, DNMC is a rectangle.

∴ NM = DC = 12 cm. (As opposite sides of a rectangle are equal.)

From figure,

⇒ AN + NM + MB = 18

⇒ AN + 12 + MB = 18

⇒ AN + MB = 6

⇒ 2AN = 6 (As AN = MB)

⇒ AN = $\dfrac{6}{2}$ = 3 cm.

⇒ MB = 3 cm.

In right angle triangle AND,

⇒ AD² = AN² + DN²

⇒ x² = 4² + 3²

⇒ x² = 16 + 9

⇒ x² = 25

⇒ x = $\sqrt{25}$ = 5 cm.

From figure,

Perimeter = AB + BC + CD + DA

= 18 + 5 + 12 + 5

= 40 cm.

Hence, perimeter of trapezium = 40 cm.

The area of a parallelogram is 98 cm². If one altitude is half the corresponding base, determine the base and the altitude of the parallelogram.

Answer

Let base = x cm

Corresponding altitude = $\dfrac{x}{2}$ cm

By formula,

Area of parallelogram = base × altitude

Substituting the values we get,

⇒ 98 = $x \times \dfrac{x}{2}$

⇒ 98 = $\dfrac{x^2}{2}$

⇒ x² = 98 × 2 = 196

⇒ x = $\sqrt{196}$ = 14 cm

⇒ Base = x = 14 cm

⇒ Altitude = $\dfrac{x}{2}$ = 7 cm.

Hence, base = 14 cm and altitude = 7 cm.

The length of a rectangular garden is 12 m more than its breadth. The numerical value of its area is equal to 4 times the numerical value of its perimeter. Find the dimensions of the garden.

Answer

Let breadth of rectangular garden = x meters,

∴ Length = (x + 12) meters.

Area of garden = length × breadth = x(x + 12) m²

Perimeter of garden = 2(l + b)

= 2[(x + 12) + x]

= 2[2x + 12] = (4x + 24) meters.

According to question,

⇒ Area of garden = 4 × Perimeter of garden

⇒ x(x + 12) = 4 × (4x + 24)

⇒ x² + 12x = 16x + 96

⇒ x² + 12x - 16x - 96 = 0

⇒ x² - 4x - 96 = 0

⇒ x² - 12x + 8x - 96 = 0

⇒ x(x - 12) + 8(x - 12) = 0

⇒ (x + 8)(x - 12) = 0

⇒ x + 8 = 0 or x - 12 = 0

⇒ x = -8 or x = 12.

Since, breadth cannot be negative.

∴ x ≠ -8.

Breadth = x = 12 m and Length = (x + 12) = (12 + 12) = 24 m.

Hence, length and breadth of garden are 24 m and 12 m respectively.

If the perimeter of a rectangular plot is 68 m and length of its diagonal is 26 m, find its area.

Answer

Let ABCD be a rectangular plot of length x m and breadth y m.

By formula,

Perimeter = 2(length + breadth)

Substituting the values we get,

⇒ 68 = 2(x + y)

⇒ 34 = x + y

⇒ x = 34 - y ......... (1)

In right angle triangle ABC

⇒ AC² = AB² + BC² (By pythagoras theorem)

⇒ 26² = x² + y²

⇒ x² + y² = 676

Substituting the value of x from equation (1),

⇒ (34 – y)² + y² = 676

⇒ 1156 + y² – 68y + y² = 676

⇒ 2y² – 68y + 1156 – 676 = 0

⇒ 2y² – 68y + 480 = 0

⇒ 2(y² – 34y + 240) = 0

⇒ y² – 34y + 240 = 0

⇒ y² – 24y – 10y + 240 = 0

⇒ y(y – 24) – 10(y – 24) = 0

⇒ (y – 10)(y – 24) = 0

⇒ y – 10 = 0 or y – 24 = 0

⇒ y = 10 m or y = 24 m.

Now substituting the value of y in equation (1)

⇒ y = 10 m, x = 34 – 10 = 24 m

⇒ y = 24 m, x = 34 – 24 = 10 m

Area in both cases = xy

= 24 × 10 or 10 × 24

= 240 m².

Hence, the area of the rectangular block is 240 m².

A rectangle has twice the area of a square. The length of the rectangle is 12 cm greater and the width is 8 cm greater than a side of a square. Find the perimeter of the square.

Answer

Let length of a side of a square = x cm.

According to question,

Length of rectangle = (x + 12) cm

Breadth of rectangle = (x + 8) cm

Given,

⇒ Area of rectangle = 2 × area of square

⇒ (x + 12)(x + 8) = 2 × (x × x)

⇒ x(x + 8) + 12(x + 8) = 2x²

⇒ x² + 8x + 12x + 96 = 2x²

⇒ x² – 2x² + 8x + 12x + 96 = 0

⇒ -x² + 20x + 96 = 0

⇒ x² – 20x – 96 = 0

⇒ x² – 24x + 4x – 96 = 0

⇒ x(x - 24) + 4(x - 24) = 0

⇒ (x + 4)(x – 24) = 0

⇒ x + 4 = 0 or x - 24 = 0

⇒ x = -4 or x = 24 cm

Since, side of a square cannot be negative.

∴ x ≠ -4.

Side of square = 24 cm

Perimeter of square = 4 × side = 4 × 24

= 96 cm.

Hence, perimeter of square = 96 cm.

The perimeter of a square is 48 cm. The area of a rectangle is 4 cm² less than the area of the square. If the length of the rectangle is 4 cm greater than its breadth, find the perimeter of the rectangle.

Answer

Perimeter of a square = 48 cm

Length of side of square = $\dfrac{\text{Perimeter}}{4} = \dfrac{48}{4}$ = 12 cm.

By formula,

Area = (side)² = 12² = 144 cm².

∴ Area of rectangle = 144 – 4 = 140 cm²

Let breadth of rectangle = x cm

∴ Length of rectangle = (x + 4) cm

Area of rectangle = l × b = x(x + 4) cm²

Substituting the values we get,

⇒ x(x + 4) = 140

⇒ x² + 4x – 140 = 0

⇒ x² + 14x – 10x – 140 = 0

⇒ x(x + 14) – 10(x + 14) = 0

⇒ (x + 14)(x – 10) = 0

⇒ x + 14 = 0 or x - 10 = 0

⇒ x = -14 or x = 10

Since, breadth cannot be negative.

∴ x ≠ -14.

Breadth = x = 10 cm and Length = x + 4 = 10 + 4 = 14 cm

Perimeter of rectangle = 2(l + b)

= 2(14 + 10)

= 2 × 24 = 48 cm.

Hence, perimeter of rectangle = 48 cm.

In the adjoining figure, ABCD is a rectangle with sides AB = 10 cm and BC = 8 cm. HAD and BFC are equilateral triangles; AEB and DCG are right angled isosceles triangles. Find the area of the shaded region and the perimeter of the figure.

Answer

In △AEB,

Let AE = BE = x cm, then from right angled triangle AEB,

⇒ AB² = AE² + EB²

⇒ 10² = x² + x²

⇒ 2x² = 100

⇒ x² = 50

⇒ x = $\sqrt{50} = 5\sqrt{2}$ cm.

Area of right angled △AEB = $\dfrac{1}{2}$ × base × height

$= \dfrac{1}{2} \times x \times x = \dfrac{1}{2}x^2 \\[1em] = \dfrac{1}{2} \times 50 \\[1em] = 25 \text{ cm}^2$

In △DGC,

Let DG = GC = y cm, then from right angled triangle DGC,

⇒ DC² = DG² + GC²

⇒ 10² = y² + y²

⇒ 2y² = 100

⇒ y² = 50

⇒ y = $\sqrt{50} = 5\sqrt{2}$ cm

Area of right angled △DCG = $\dfrac{1}{2}$ × base × height

$= \dfrac{1}{2} \times y \times y = \dfrac{1}{2}y^2 \\[1em] = \dfrac{1}{2} \times 50 \\[1em] = 25 \text{ cm}^2$

Since, HAD and BFC are equilateral triangle with side = 8 cm.

Area of HAD = Area of BFC = $\dfrac{\sqrt{3}}{4}$ × (side)²

= $\dfrac{\sqrt{3}}{4} \times (8)^2$

= $16\sqrt{3}$ cm²

Area of rectangle ABCD = l × b = AB × CD

= 10 × 8 = 80 cm²

From figure,

Area of shaded region = Area of (△DGC + △BFC + △AEB + △HAD + rectangle ABCD)

= $25 + 16\sqrt{3} + 25 + 16\sqrt{3} + 80 = 130 + 32\sqrt{3}$ cm².

Perimeter of figure = (AE + EB + BF + FC + CG + GD + DH + HA)

= $(5\sqrt{2} + 5\sqrt{2} + 8 + 8 + 5\sqrt{2} + 5\sqrt{2} + 8 + 8)$

= $20\sqrt{2} + 32$ cm.

Hence, area of shaded region = $130 + 32\sqrt{3}$ and perimeter = $20\sqrt{2} + 32$ cm.

Find the area enclosed by the figure (i) given below, where ABC is an equilateral triangle and DEFG is an isosceles trapezium. All measurements are in centimeters.

Answer

In right angle triangle ECF,

Using pythagoras theorem,

⇒ EF² = EC² + CF²

⇒ 5² = EC² + 3²

⇒ EC² = 5² - 3²

⇒ EC² = 25 - 9 = 16

⇒ EC = $\sqrt{16}$ = 4 cm.

Since, DEFG is an isosceles trapezium.

∴ GD = EF= 5 cm.

Since, BDEC is a rectangle,

∴ BD = EC = 4 cm and BC = DE = 6 cm.

In right angle triangle DBG,

Using pythagoras theorem,

⇒ GD² = BD² + GB²

⇒ 5² = 4² + GB²

⇒ GB² = 5² - 4²

⇒ GB² = 25 - 16 = 9

⇒ GB = $\sqrt{9}$ = 3 cm.

In trapezium,

GF = GB + BC + CF = 3 + 6 + 3 = 12 cm.

Area of trapezium DEFG = $\dfrac{1}{2} \times$ (sum of parallel sides) × distance between them

= $\dfrac{1}{2} \times (12 + 6) \times 4$

= 18 × 2

= 36 cm².

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

= $\dfrac{\sqrt{3}}{4} \times (6)^2$

= $\dfrac{\sqrt{3}}{4} \times 36$

= 1.732 × 9

= 15.59 cm²

Area of figure = Area of trapezium DEFG + Area of equilateral triangle ABC

= 36 + 15.59 = 51.59 cm².

Hence, area of figure = 51.59 cm².

Find the area enclosed by the figure (ii) given below. All measurements are in centimeters.

Answer

From figure,

BJ = 2 + 2 + 2 + 2 = 8 cm.

Area of rectangle ABJK = l × b

= AB × BJ = 2 × 8

= 16 cm².

From figure,

JH = KH - KI = 6 - 2 = 4 cm.

Area of trapezium FGHI = $\dfrac{1}{2} \times$ (sum of parallel sides) × distance between them

= $\dfrac{1}{2}$ × (FI + GH) × JH

= $\dfrac{1}{2}$ × (2 + 2) × 4

= $\dfrac{1}{2}$ × 4 × 4 = 8 cm².

Area of trapezium CDEF = $\dfrac{1}{2} \times$ (sum of parallel sides) × distance between them

= $\dfrac{1}{2}$ × (CF + DE) × BD

= $\dfrac{1}{2}$ × (2 + 2) × 4

= $\dfrac{1}{2}$ × 4 × 4 = 8 cm².

Total area enclosed = Area of rectangle ABJK + Area of trapezium FGHI + Area of trapezium CDEF

= 16 + 8 + 8

= 32 cm².

Hence, area enclosed by figure = 32 cm².

In the figure (iii) given below, from a 24 cm × 24 cm piece of cardboard, a block in the shape of letter M is cut off. Find the area of the cardboard left over, all measurements are in centimetres.

Answer

From figure,

Area of rectangle (I) = length × breadth

= 24 × 6

= 144 cm².

Area of rectangle (II) = length × breadth

= 24 × 6

= 144 cm².

Area of parallelogram (III) = base × height

= 8 × 6

= 48 cm².

Area of parallelogram (IV) = base × height

= 8 × 6

= 48 cm².

Area of figure M = Area of rectangle (I) + Area of rectangle (II) + Area of parallelogram (III) + Area of parallelogram (IV)

= 144 + 144 + 48 + 48

= 384 cm².

Area of cardboard = 24 × 24

= 576 cm².

Area of cardboard left = Area of cardboard - Area of figure M

= 576 - 384

= 192 cm².

Hence, area of cardboard left = 192 cm².

The figure (i) given below shows the cross-section of the concrete structure with the measurements as given. Calculate the area of cross-section.

Answer

From figure,

The figure consist of a trapezium and a rectangle.

Area of trapezium = $\dfrac{1}{2} \times$ (sum of parallel sides) × distance between them

= $\dfrac{1}{2}$ × (0.6 + 1.5) × (1.2 + 2.4)

= $\dfrac{1}{2}$ × 2.1 × 3.6

= 2.1 × 1.8

= 3.78 m².

Area of rectangle = l × b

= 2.4 × 0.3 = 0.72 m².

Area of cross section = Area of trapezium + Area of rectangle

= 3.78 + 0.72 = 4.5 m².

Hence, area of cross-section = 4.5 m².

The figure (ii) given below shows a field with the measurements given in metres. Find the area of the field.

Answer

From figure,

Area of right angled △AXB = $\dfrac{1}{2}$ × base × height

= $\dfrac{1}{2}$ × BX × AX

= $\dfrac{1}{2}$ × 30 × 12

= 180 m².

Area of trapezium XZCB = $\dfrac{1}{2} \times$ (sum of parallel sides) × distance between them

= $\dfrac{1}{2}$ × (BX + CZ) × 15

= $\dfrac{1}{2}$ × (30 + 25) × 15

= $\dfrac{1}{2}$ × 55 × 15

= 412.5 cm².

Area of right angled △CZD = $\dfrac{1}{2}$ × base × height

= $\dfrac{1}{2}$ × CZ × ZD

= $\dfrac{1}{2}$ × 25 × 10

= 125 m².

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

= $\dfrac{1}{2}$ × AD × EY

= $\dfrac{1}{2}$ × 37 × 20

= 370 m².

Area of field = Area of right angled △AXB + Area of trapezium XZCB + Area of right angled △CZD + Area of △AED

= 180 + 412.5 + 125 + 370 = 1087.5 m².

Hence, area of field = 1087.5 m².

Calculate the area of the pentagon ABCDE shown in fig (iii) below, given that AX = BX = 6 cm, EY = CY = 4 cm, DE = DC = 5 cm, DX = 9 cm and DX is perpendicular to EC and AB.

Answer

From figure,

In right angled △DEY,

⇒ DE² = DY² + EY²

⇒ 5² = DY² + 4²

⇒ DY² = 5² - 4²

⇒ DY² = 25 - 16 = 9

⇒ DY = $\sqrt{9}$ = 3 cm.

Area of right angled △DEY = $\dfrac{1}{2}$ × base × height

= $\dfrac{1}{2}$ × EY × DY

= $\dfrac{1}{2}$ × 4 × 3

= 6 cm².

Area of right angle △DYC = $\dfrac{1}{2}$ × base × height

= $\dfrac{1}{2}$ × CY × DY

= $\dfrac{1}{2}$ × 4 × 3

= 6 cm².

From figure,

XY = DX - DY = 9 - 3 = 6 cm.

Area of trapezium ECBA = $\dfrac{1}{2} \times$ (sum of parallel sides) × distance between them

= $\dfrac{1}{2}$ × (EC + AB) × XY

= $\dfrac{1}{2}$ × [(EY + CY) + (AX + BX)] × XY

= $\dfrac{1}{2}$ × [(4 + 4) + (6 + 6)] × 6

= $\dfrac{1}{2}$ × 20 × 6

= 60 cm².

Area of pentagon = Area of right angled △DEY + Area of right angled △DYC + Area of trapezium ECBA

= 6 + 6 + 60

= 72 cm².

Hence, area of trapezium = 72 cm².

If the length and the breadth of a room are increased by 1 metre, the area is increased by 21 square metres. If the length is increased by 1 metre and breadth is decreased by 1 metre the area is decreased by 5 square metres. Find the perimeter of the room.

Answer

Let length = l metres and breadth = b metres.

Area = lb m²

Given,

If the length and the breadth of a room are increased by 1 metre, the area is increased by 21 square metres,

∴ (l + 1)(b + 1) - lb = 21

⇒ lb + l + b + 1 - lb = 21

⇒ l + b = 21 - 1

⇒ l + b = 20 ..........(1)

Given,

If the length is increased by 1 metre and breadth is decreased by 1 metre the area is decreased by 5 square metres.

∴ lb - (l + 1)(b - 1) = 5

⇒ lb - (lb - l + b - 1) = 5

⇒ lb - lb + l - b + 1 = 5

⇒ l - b = 4 ..........(2)

Adding equation (1) and (2) we get,

⇒ l + b + l - b = 20 + 4

⇒ 2l = 24

⇒ l = 12 m.

Substituting value of l in (2) we get,

⇒ 12 - b = 4

⇒ b = 12 - 4 = 8 m.

Perimeter of room = 2(l + b) = 2 × 20 = 40 m.

Hence, perimeter of room = 40 m.