Area and Perimeter of Plane Figures

Length of a side and corresponding altitude of a triangle is doubled. The area of triangle will become :

1. double

2. half

3. four times

4. one-fourth

Answer

The area of a triangle is given by:

Area = $\dfrac{1}{2}$ x length x altitude

When the length of the side and the corresponding altitude of the triangle are doubled:

Area₁ = $\dfrac{1}{2}$ x 2 x length x 2 x altitude

= $4 \times \Big( \dfrac{1}{2} \times \text{length} \times \text{altitude} \Big)$

= 4 x Area

Area₁ = 4 x Area

Thus, the new area of triangle becomes four times the original area.

Hence, option 3 is the correct option.

If each side of an equilateral triangle is halved, its area will be :

1. halved

2. four times

3. unaltered

4. one-fourth

Answer

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

If each side of the equilateral triangle is halved, then the new area will be:

New area = $\dfrac{\sqrt{3}}{4} \times \Big(\dfrac{side}{2}\Big)^2$

= $\dfrac{1}{4}\Big(\dfrac{\sqrt{3}}{4} \times side^2\Big)$

= $\dfrac{1}{4}$ x Original area

Therefore, the new area of the equilateral triangle will be one-fourth of the original area.

Hence, option 4 is the correct option.

The area of a triangle is 37.5 cm². If its base is 12.5 cm; the corresponding altitude is :

1. 3 cm

2. 25 cm

3. 6 cm

4. 12 cm

Answer

Given:

Area = 37.5 cm²

Base = 12.5 cm

Let h be the altitude of triangle.

Area of triangle = $\dfrac{1}{2}$ x base x altitude

⇒ 37.5 = $\dfrac{1}{2}$ x 12.5 x h

⇒ h = $\dfrac{37.5 \times 2}{12.5}$

⇒ h = $\dfrac{75}{12.5}$

⇒ h = 6 cm

Hence, option 3 is the correct option.

ABC is a triangle with AB = AC = 12 cm and ∠A = 90°, the area of the triangle ABC is :

1. 144 cm²

2. 36 cm²

3. 72 cm²

4. 108 cm²

Answer

Given:

AB = AC = 12 cm

∠A = 90°

Area of triangle = $\dfrac{1}{2}$ x base x altitude

= $\dfrac{1}{2}$ x 12 x 12 cm²

= $\dfrac{1}{2}$ x 144 cm²

= 72 cm²

Hence, option 3 is the correct option.

The sides of a triangle are 9 cm, 12 cm and 15 cm; the area of the triangle is :

1. 54 cm²

2. 96 cm²

3. 108 cm²

4. 135 cm²

Answer

Let the sides of the triangle be:

a = 9 cm, b = 12 cm and c = 15 cm.

The semi-perimeter s:

$∵ s = \dfrac{a + b + c}{2}\\[1em] = \dfrac{9 + 12 + 15}{2}\\[1em] = \dfrac{36}{2}\\[1em] = 18$

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

= $\sqrt{18(18 - 9)(18 - 12)(18 - 15)}$ cm²

= $\sqrt{18 \times 9 \times 6 \times 3}$ cm²

= $\sqrt{2,916}$ cm²

= 54 cm²

Hence, option 1 is the correct option.

Find the area of a triangle whose sides are 18 cm, 24 cm and 30 cm.

Also, find the length of altitude corresponding to the largest side of the triangle.

Answer

Let the sides of the triangle be:

a = 18 cm, b = 24 cm and c = 30 cm.

The semi-perimeter s:

$∵ s = \dfrac{a + b + c}{2}\\[1em] = \dfrac{18 + 24 + 30}{2}\\[1em] = \dfrac{72}{2}\\[1em] = 36$

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

= $\sqrt{36(36 - 18)(36 - 24)(36 - 30)}$ cm²

= $\sqrt{36 \times 18 \times 12 \times 6}$ cm²

= $\sqrt{46,656}$ cm²

= 216 cm²

Using the area formula to find the altitude corresponding to the largest side (base = 30 cm):

Area = $\dfrac{1}{2}$ x base x altitude

Let h be the altitude:

⇒ 216 = $\dfrac{1}{2}$ x 30 x h

⇒ 216 = 15 x h

⇒ h = $\dfrac{216}{15}$

⇒ h = 14.4 cm

Hence, the area of the triangle is 216 cm² and the length of altitude corresponding to the largest side is 14.4 cm.

The lengths of the sides of a triangle are in the ratio 3 : 4 : 5. Find the area of the triangle if its perimeter is 144 cm.

Answer

It is given that the lengths of the sides of a triangle are in the ratio 3 : 4 : 5.

Let the lengths of the sides be 3a, 4a and 5a.

The perimeter of the triangle is 144 cm.

Perimeter = sum of all sides of triangle

⇒ 144 = 3a + 4a + 5a

⇒ 144 = 12a

⇒ a = $\dfrac{144}{12}$

⇒ a = 12

So, the sides of triangle = 3a, 4a and 5a

= 3 x 12, 4 x 12 and 5 x 12

= 36, 48 and 60

Let a = 36 cm, b = 48 cm and c = 60 cm.

The semi-perimeter s:

$∵ s = \dfrac{a + b + c}{2}\\[1em] = \dfrac{36 + 48 + 60}{2}\\[1em] = \dfrac{144}{2}\\[1em] = 72$

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

= $\sqrt{72(72 - 36)(72 - 48)(72 - 60)}$ cm²

= $\sqrt{72 \times 36 \times 24 \times 12}$ cm²

= $\sqrt{746,496}$ cm²

= 864 cm²

Hence, the area of the triangle is 864 cm².

ABC is a triangle in which AB = AC = 4 cm and ∠A = 90°. Calculate :

(i) the area of Δ ABC,

(ii) the length of perpendicular from A to BC.

Answer

(i) Given:

AB = AC = 4 cm

∠A = 90°

Area = $\dfrac{1}{2}$ x base x height

= $\dfrac{1}{2}$ x 4 x 4 cm²

= $\dfrac{1}{2}$ x 16 cm²

= 8 cm²

(ii) By using the Pythagoras theorem,

AB² + AC² = BC²

⇒ (4)² + (4)² = BC²

⇒ 16 + 16 = BC²

⇒ BC² = 32

⇒ BC = $\sqrt{32}$

⇒ BC = 4 $\sqrt{2}$ cm

Now considering BC as the base of the triangle, altitude AD will be its height.

Area of the triangle will remain the same as before i.e., 8 cm².

Let h be the altitude of triangle

Area = $\dfrac{1}{2}$ x base x height

∴ 8 = $\dfrac{1}{2}$ x BC x AD

⇒ 8 = $\dfrac{1}{2}$ x $4 \sqrt{2}$ x h

⇒ 8 = $2 \sqrt{2}$ x h

⇒ h = $\dfrac{8}{2 \sqrt{2}}$

⇒ h = $\dfrac{4}{\sqrt{2}}$

⇒ h = 2.83 cm

Hence, the length of perpendicular from A to BC is 2.83 cm.

The area of an equilateral triangle is $36\sqrt{3}\text{ sq. cm}$. Find its perimeter.

Answer

Given:

Area = $36\sqrt{3}\text{ sq. cm}$

Let s be the side of equilateral triangle.

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

$⇒ 36\sqrt{3} = \dfrac{\sqrt{3}}{4} \times s^2\\[1em] ⇒ 36\cancel{\sqrt{3}} = \dfrac{\cancel{\sqrt{3}}}{4} \times s^2\\[1em] ⇒ 36 = \dfrac{1}{4} \times s^2\\[1em] ⇒ s^2 = 36 \times 4\\[1em] ⇒ s^2 = 144\\[1em] ⇒ s = \sqrt{144}\\[1em] ⇒ s = 12$

Perimeter = 3 x side

= 3 x 12 cm

= 36 cm

Hence, the perimeter is 36 cm.

Find the area of an isosceles triangle with perimeter 36 cm and base 16 cm.

Answer

Given:

Perimeter = 36 cm

Base = 16 cm

Let a be the length of the equal sides of the triangle.

Perimeter = 2 x Equal side + Base

⇒ 36 = 2a + 16

⇒ 2a = 36 - 16

⇒ 2a = 20

⇒ a = $\dfrac{20}{2}$

⇒ a = 10 cm

Thus, the triangle has two equal sides, each 10 cm long, and a base of 16 cm.

a = 10 cm, b = 10 cm and c = 16 cm.

The semi-perimeter is:

$∵ s = \dfrac{a + b + c}{2}\\[1em] = \dfrac{10 + 10 + 16}{2}\\[1em] = \dfrac{36}{2}\\[1em] = 18$

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

= $\sqrt{18(18 - 10)(18 - 10)(18 - 16)}$ cm²

= $\sqrt{18 \times 8 \times 8 \times 2}$ cm²

= $\sqrt{2,304}$ cm²

= 48 cm²

Hence, the area is 48 cm².

The base of an isosceles triangle is 24 cm and its area is 192 sq. cm. Find its perimeter.

Answer

Given:

Base (b) = 24 cm

Area = 192 cm²

Let s be the length of the equal sides of the triangle.

Area of an isosceles triangle = $\dfrac{1}{4} \times b \times \sqrt{4s^2 - b^2}$

$⇒ 192 = \dfrac{1}{4} \times 24 \times \sqrt{4s^2 - 24^2}\\[1em] ⇒ 192 = 6 \times \sqrt{4s^2 - 576}\\[1em] ⇒ \sqrt{4s^2 - 576} = \dfrac{192}{6}\\[1em] ⇒ \sqrt{4s^2 - 576} = 32\\[1em] ⇒ 4s^2 - 576 = 1024\\[1em] ⇒ 4s^2 = 1024 + 576\\[1em] ⇒ 4s^2 = 1600\\[1em] ⇒ s^2 = \dfrac{1600}{4}\\[1em] ⇒ s^2 = 400\\[1em] ⇒ s = \sqrt{400}\\[1em] ⇒ s = 20$

Perimeter = sum of all sides of triangle

= 20 + 20 + 24 cm

= 64 cm

Hence, the perimeter is 64 cm.

The given figure shows a right-angled triangle ABC and an equilateral triangle BCD. Find the area of the shaded portion.

Answer

For triangle ABC (right-angled triangle),

Using the Pythagoras theorem,

Base² + Height² = Hypotenuse²

⇒ 8² + height² = 16²

⇒ 64 + height² = 256

⇒ height² = 256 - 64

⇒ height² = 192

⇒ height = $\sqrt{192}$

⇒ height = 8 $\sqrt{3}$

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

= $\dfrac{1}{2} \times 8 \times 8 \sqrt{3}$

= $4 \times 8 \sqrt{3}$

= $32 \sqrt{3}$ cm²

For triangle BCD,

Area of an equilateral triangle = $\dfrac{\sqrt{3}}{4}$ x side²

= $\dfrac{\sqrt{3}}{4}$ x 8²

= $\dfrac{\sqrt{3}}{4}$ x 64

= 16 $\sqrt{3}$ cm²

Area of Δ ABD (shaded portion) = Area of Δ ABC - Area of Δ BDC

= 32 $\sqrt{3}$ - 16 $\sqrt{3}$ cm²

= 16 $\sqrt{3}$ cm²

Hence, the area of the shaded portion is 16$\sqrt{3}$ cm² = 27.712 cm².

Find the area and the perimeter of quadrilateral ABCD, given below; if, AB = 8 cm, AD = 10 cm, BD = 12 cm, DC = 13 cm and ∠DBC = 90°.

Answer

Given:

AB = 8 cm, AD = 10 cm, BD = 12 cm, DC = 13 cm and ∠DBC = 90°

In Δ BCD,

By using the Pythagoras theorem,

Base² + Height² = Hypotenuse²

⇒ BC² + 12² = 13²

⇒ BC² + 144 = 169

⇒ BC² = 169 - 144

⇒ BC² = 25

⇒ BC = $\sqrt{25}$

⇒ BC = 5 cm

Area of Δ BCD = $\dfrac{1}{2}$ x base x height

= $\dfrac{1}{2}$ x 12 x 5 cm²

= $\dfrac{1}{2}$ x 60 cm²

= 30 cm²

For Δ ABD,

Let AD = a = 10 cm, BD = b = 12 cm and AB = c = 8 cm.

$∵ s = \dfrac{a + b + c}{2}\\[1em] = \dfrac{10 + 12 + 8}{2}\\[1em] = \dfrac{30}{2}\\[1em] = 15$

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

= $\sqrt{15(15 - 10)(15 - 12)(15 - 8)}$ cm²

= $\sqrt{15 \times 5 \times 3 \times 7}$ cm²

= $\sqrt{1,575}$ cm²

= 39.7 cm²

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

= 39.5 + 30 cm²

= 69.5 cm²

Perimeter of quadrilateral ABCD = Sum of all sides of quadrilateral

= AB + BC + CD + DA

= 8 + 10 + 13 + 5

= 36 cm

Hence, the area of quadrilateral is 69.7 cm² and the perimeter is 36 cm.

The base of a triangular field is three times its height. If the cost of cultivating the field at ₹ 36.72 per 100 m² is ₹ 49,572; find its base and height.

Answer

Given:

Cost of cultivating the field = ₹ 36.72 per 100 m²

Total cost = ₹ 49,572

Total cost = Area x Cost of cultivating per 100 m²

Area = $\dfrac{\text{Total cost}}{\text{Cost of cultivating}}$

$= \dfrac{49,572 \times 100}{36.72}\\[1em] = \dfrac{49,572 \times 10,000}{3672}\\[1em] = 1,35,000 \text{ m}^2$

Let the height of the triangle be h.

Since the base of the field is three times its height, we have:

Base = 3h

Area = $\dfrac{1}{2}$ x base x height

$⇒ 1,35,000 = \dfrac{1}{2} \times 3h \times h\\[1em] ⇒ 1,35,000 = \dfrac{3}{2} \times h^2\\[1em] ⇒ h^2 = \dfrac{1,35,000 \times 2}{3}\\[1em] ⇒ h^2 = \dfrac{2,70,000}{3}\\[1em] ⇒ h^2 = 90,000\\[1em] ⇒ h = \sqrt{90,000}\\[1em] ⇒ h = 300$

Thus, the height of the triangle is 300 m.

Base = 3 x height = 3 x 300 m = 900 m

Hence, the height is 300 m and the base is 900m.

The sides of a triangular field are in the ratio 5 : 3 : 4 and its perimeter is 180 m. Find :

(i) its area.

(ii) altitude of the triangle corresponding to its largest side.

(iii) the cost of levelling the field at the rate of ₹ 10 per square metre.

Answer

(i) Given:

The sides of a triangular field are in the ratio 5 : 3 : 4.

Perimeter = 180 m

Let the sides of field be 5a, 3a and 4a.

Perimeter = Sum of all sides of triangular field

⇒ 180 = 5a + 3a + 4a

⇒ 180 = 12a

⇒ a = $\dfrac{180}{12}$

⇒ a = 15

Thus, sides of field = 5a , 3a and 4a

= 5 x 15, 3 x 15 and 4 x 15

= 75 m, 45 m and 60 m

Let a = 75 m, b = 45 m and c = 60 m.

$∵ s = \dfrac{a + b + c}{2}\\[1em] = \dfrac{75 + 45 + 60}{2}\\[1em] = \dfrac{180}{2}\\[1em] = 90$

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

= $\sqrt{90(90 - 75)(90 - 45)(90 - 60)}$ m²

= $\sqrt{90 \times 15 \times 45 \times 30}$ m²

= $\sqrt{18,22,500}$ m²

= 1,350 m²

Hence, the area is 1,350 m².

(ii) Area = $\dfrac{1}{2}$ x base x altitude

Base = 75 m

Let h be the altitude of triangle corresponding to the largest side 75 m,

⇒ 1350 = $\dfrac{1}{2}$ x 75 x h

⇒ h = $\dfrac{1350 \times 2}{75}$

⇒ h = $\dfrac{2,700}{75}$

⇒ h = 36 m

Hence, the altitude of the triangle corresponding to its largest side is 36 m.

(iii) Cost of levelling = ₹ 10 per square metre

Total cost = Area x Cost of levelling

= ₹ 1,350 x 10

= ₹ 13,500

Hence, the total cost of levelling is ₹ 13,500.

Each of equal sides of an isosceles triangle is 4 cm greater than its height. If the base of the triangle is 24 cm; calculate the perimeter and the area of the triangle.

Answer

Each of equal sides of an isosceles triangle is 4 cm greater than its height.

Let h be the height of the triangle.

Equal sides: AB = AC = h + 4

Base: BC = 24 cm

In Δ ABD and Δ ACD,

AD = AD [∵ Common Side]

∠ ADB = ∠ ADC [∵ Both are 90°]

AB = AC [∵ Δ ABC is isosceles]

∴ Δ ABD ≅ Δ ACD [RHS axiom]

∴ BD = CD [C.P.C.T]

∴ BD = CD = $\dfrac{\text{BC}}{2}$ = ${\dfrac{24}{2}}$ = 12 cm

By using the Pythagoras theorem in Δ ABD,

BD² + AD² = AB²

⇒ 12² + h² = (h + 4)²

⇒ 144 + h² = h² + 4² + 2 x h x 4

⇒ 144 + $\cancel{h^2}$= $\cancel{h^2}$+ 16 + 8h

⇒ 144 - 16 = 8h

⇒ 128 = 8h

⇒ h = $\dfrac{128}{8}$

⇒ h = 16 cm

⇒ a = h + 4 = 16 + 4 = 20 cm

Perimeter of triangle = Sum of all sides

= AB + AC + BC

= (20 + 20 + 24) cm

= 64 cm

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

= $\dfrac{1}{2}$ x 24 x 16

= 12 x 16

= 192 cm²

Hence, the perimeter is 64 cm and the area is 192 cm².

Calculate the area and the height of an equilateral triangle whose perimeter is 60 cm.

Answer

Given:

Perimeter = 60 cm

Let a be the sides of equilateral triangle.

Perimeter = Sum of all sides

⇒ 60 = a + a + a

⇒ 60 = 3a

⇒ a = $\dfrac{60}{3}$

⇒ a = 20 cm

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

$= \dfrac{\sqrt{3}}{4} \times 20^2\\[1em] = \dfrac{\sqrt{3}}{4} \times 400\\[1em] = 100 \sqrt{3} \\[1em] = 173.2 cm^2$

Height of equilateral triangle = $\dfrac{\sqrt{3}}{2} \times side$

$= \dfrac{\sqrt{3}}{2} \times 20\\[1em] = 10 \sqrt{3}\\[1em] = 17.32 cm$

Hence, the area is 173.2 cm² and the perimeter is 17.32 cm.

In triangle ABC; angle A = 90°, side AB = x cm, AC = (x + 5) cm and area = 150 cm². Find the sides of the triangle.

Answer

Δ ABC is shown in the figure below:

Given:

Area = 150 cm²

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

$⇒ 150 = \dfrac{1}{2} \times x \times (x + 5)\\[1em] ⇒ 150 \times 2 = x^2 + 5x\\[1em] ⇒ 300 = x^2 + 5x\\[1em] ⇒ x^2 + 5x - 300 = 0\\[1em] ⇒ x^2 + 20x - 15x - 300 = 0\\[1em] ⇒ x(x + 20) - 15(x + 20) = 0\\[1em] ⇒ (x + 20)(x - 15) = 0\\[1em] ⇒ x = -20 \text{ or } 15$

Since length cannot be negative, AB = 15 cm.

Thus, AC = x + 5 = 15 + 5 cm = 20 cm

By using the Pythagoras theorem,

AB² + AC² = BC²

⇒ 15² + 20² = BC²

⇒ 225 + 400 = BC²

⇒ 625 = BC²

⇒ BC = $\sqrt{625}$

⇒ BC = 25

Hence, the sides of the triangle are 15 cm, 20 cm and 25 cm.

If the difference between the sides of a right angled triangle is 3 cm and its area is 54 cm²; find its perimeter.

Answer

Given:

Area = 54 cm²

Let the two sides of the right-angled triangle be x cm and (x - 3) cm.

Area = $\dfrac{1}{2}$ x base x height

⇒ 54 = $\dfrac{1}{2}$ x BC x AB

⇒ 54 = $\dfrac{1}{2} \times$ (x - 3) $\times$ x

⇒ 54 $\times$ 2 = (x - 3) $\times$ x

⇒ 108 = x² - 3x

⇒ x² - 3x - 108 = 0

⇒ x² - 12x + 9x - 108 = 0

⇒ x(x - 12) + 9(x - 12) = 0

⇒ (x - 12)(x + 9) = 0

⇒ x = 12 or -9

Since length cannot be negative, AB is 12 cm.

BC = (x - 3) cm = 12 - 3 cm = 9 cm

By using the Pythagoras theorem,

AB² + BC² = AC²

⇒ 12² + 9² = AC²

⇒ 144 + 81 = AC²

⇒ 225 = AC²

⇒ AC = $\sqrt{225}$

⇒ AC = 15

Perimeter of right angled triangle = AB + BC + AC

= 12 + 9 + 15 cm

= 36 cm

Hence, the perimeter of the triangle is 36 cm.

The area of the given figure is :

1. AC x BD

2. $\dfrac{1}{2}\times AC \times BD$

3. $\dfrac{1}{2}\times AB \times BD$

4. $\dfrac{1}{2}\times (AD + BC) \times BD$

Answer

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

For Δ ABD,

Area = $\dfrac{1}{2}$ x BD x AO

For Δ CBD,

Area = $\dfrac{1}{2}$ x BD x CO

Area of quadrilateral ABCD = Area of Δ ABD + Area of Δ CBD

= $\dfrac{1}{2}$ x BD x AO + $\dfrac{1}{2}$ x BD x CO

= $\dfrac{1}{2}$ x BD x (AO + CO)

= $\dfrac{1}{2}$ x BD x AC

Hence, option 2 is the correct option.

If two adjacent sides and a diagonal of a rectangle are x, y and d respectively. The area of the rectangle is :

1. $x \times y$

2. $\dfrac{1}{2} \times\ d^2$

3. $\dfrac{1}{2} \times\ x \times d$

4. $\dfrac{1}{2} \times\ y \times d$

Answer

Given:

Adjacent sides of rectangle = x and y

Area of rectangle = base x height

= $x \times y$

Hence, option 1 is the correct option.

A square DEFG of side 8 cm is inscribed in a rectangle of adjacent sides 20 cm and 16 cm as shown in the figure.

The area of the shaded portion is :

1. 64 cm²

2. 128 cm²

3. 256 cm²

4. 320 cm²

Answer

Area of shaded portion = Area of rectangle ABCD - Area of square DEFG

Area of square = Side²

Area of rectangle = Base x Height

Area of shaded portion = 20 x 16 - 8² cm²

= 320 - 64 cm²

= 256 cm²

Hence, option 3 is the correct option.

The perimeter of a square is 72 cm, its area is :

1. $900\sqrt{2} \text{ cm}^2$

2. $30\sqrt{3} \text{ cm}^2$

3. 324 cm²

4. 356 cm²

Answer

Given:

Perimeter = 72 cm

Let s be the side of square.

Perimeter = 4 x side

⇒ 4 x s = 72

⇒ s = $\dfrac{72}{4}$

⇒ s = 18 cm

Area = Side²

= 18² cm²

= 324 cm²

Hence, option 3 is the correct option.

Area of a rhombus is 360 cm². If one diagonal of it is 20 cm; the other diagonal is :

1. 24 cm

2. 18 cm

3. 40 cm

4. 36 cm

Answer

Given:

Area = 360 cm².

One diagonal = 20 cm

Let d be the other diagonal.

Area of rhombus = $\dfrac{1}{2}$ x product of diagonals

⇒ $\dfrac{1}{2}$ x 20 x d = 360

⇒ 10 x d = 360

⇒ d = $\dfrac{360}{10}$

⇒ d = 36 cm

Hence, option 4 is the correct option.

Find the area of a quadrilateral one of whose diagonals is 30 cm long and the perpendiculars from the other two vertices are 19 cm and 11 cm respectively.

Answer

The quadrilateral is shown in the figure below:

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

For Δ ABD,

Area = $\dfrac{1}{2}$ x BD x AM

= $\dfrac{1}{2}$ x 30 x 19 cm²

= 15 x 19 cm²

= 285 cm²

For Δ CBD,

Area = $\dfrac{1}{2}$ x BD x CN

= $\dfrac{1}{2}$ x 30 x 11 cm²

= 15 x 11 cm²

= 165 cm²

Area of quadrilateral ABCD = Area of Δ ABD + Area of Δ CBD

= 285 + 165 cm²

= 450 cm²

Hence, the area of quadrilateral is 450 cm².

The diagonals of a quadrilateral are 16 cm and 13 cm. If they intersect each other at right angles; find the area of the quadrilateral.

Answer

Since the diagonals of the quadrilateral intersect at right angles, the area of the quadrilateral is given by:

Area = $\dfrac{1}{2}$ x Product of diagonals

= $\dfrac{1}{2}$ x 16 x 13 cm²

= 8 x 13 cm²

= 104 cm²

Hence, the area of the quadrilateral is 104 cm².

Calculate the area of quadrilateral ABCD, in which ∠ABD = 90°, triangle BCD is an equilateral triangle of side 24 cm and AD = 26 cm.

Answer

For Δ ABD,

By using the Pythagoras theorem,

Base² + Height² = Hypotenuse²

⇒ AB² + BD² = AD²

⇒ AB² + (24)² = (26)²

⇒ AB² + 576 = 676

⇒ AB² = 676 - 576

⇒ AB² = 100

⇒ AB = $\sqrt{100}$

⇒ AB = 10 cm

Area of Δ ABD = $\dfrac{1}{2}$ x AB x BD

= $\dfrac{1}{2}$ x 10 x 24 cm²

= 5 x 24 cm²

= 120 cm²

Area of equilateral triangle BCD = $\dfrac{\sqrt{3}}{4}$ x side²

= $\dfrac{\sqrt{3}}{4}$ x 24²

= $\dfrac{\sqrt{3}}{4}$ x 576

= 144 ${\sqrt{3}}$ cm²

= 249.41 cm²

Total area of quadrilateral ABCD = Δ ABD + Δ BCD

= 120 + 249.41 cm²

= 369.41 cm²

Hence, the area of quadrilateral ABCD is 369.41 cm².

Calculate the area of quadrilateral ABCD in which AB = 32 cm, AD = 24 cm, ∠A = 90° and BC = CD = 52 cm.

Answer

Quadrilateral ABCD is shown in the figure below:

Area of Δ DAB = $\dfrac{1}{2}$ x base x height

= $\dfrac{1}{2}$ x DA x AB

= $\dfrac{1}{2}$ x 24 x 32 cm²

= 12 x 32 cm²

= 384 cm²

By using the Pythagoras theorem,

AD² + AB² = BD²

⇒ 24² + 32² = BD²

⇒ 576 + 1,024 = BD²

⇒ 1,600 = BD²

⇒ BD = $\sqrt{1,600}$

⇒ BD = 40 cm

In triangle BCD,

Let the sides of the triangle be:

a = 40 cm, b = 52 cm and c = 52 cm.

The semi-perimeter s:

$∵ s = \dfrac{a + b + c}{2}\\[1em] = \dfrac{40 + 52 + 52}{2}\\[1em] = \dfrac{144}{2}\\[1em] = 72$

∵ Area of Δ BCD = $\sqrt{s(s - a)(s - b)(s - c)}$

= $\sqrt{72(72 - 40)(72 - 52)(72 - 52)}$ cm²

= $\sqrt{72 \times 32 \times 20 \times 20}$ cm²

= $\sqrt{921,600}$ cm²

= 960 cm²

Therefore, area of quadrilateral ABCD = Area of Δ DAB + Area of triangle BCD

= 384 + 960 cm²

= 1344 cm²

Hence, the area of quadrilateral ABCD is 1344 cm².

The perimeter of a rectangular field is $\dfrac{3}{5}$ km. If the length of the field is twice its width; find the area of the rectangle in sq. metres.

Answer

Given:

Perimeter = $\dfrac{3}{5}$ km.

Length of the field = Twice its width.

Let a be the width of the field.

So, the length = 2a

Perimeter of a rectangle = 2(length + width)

$⇒ \dfrac{3}{5} = 2(2a + a)\\[1em] ⇒ \dfrac{3}{5} = 2 \times 3a\\[1em] ⇒ \dfrac{3}{5} = 6a\\[1em] ⇒ a = \dfrac{3}{5 \times 6}\\[1em] ⇒ a = \dfrac{3}{30}\\[1em] ⇒ a = \dfrac{1}{10}\\[1em]$

Thus, width = $\dfrac{1}{10}$ km

= $\dfrac{1}{10} \times 1,000$ m

= 100 m

Length = 2a = 2 x 100 m

= 200 m

Area = length x width

= 200 x 100 m²

= 20,000 m²

Hence, the area of the rectangle is 20,000 m².

A rectangular plot 85 m long and 60 m broad is to be covered with grass leaving 5 m all around. Find the area to be laid with grass.

Answer

Given:

The length of the rectangular field is 85 m.

The breadth of the rectangular field is 60 m.

The width of the path to be covered with grass is 5 m.

The length of the inner rectangular field = 85 m - 5 m - 5 m = 75 m

The breadth of the inner rectangular field = 60 m - 5 m - 5 m = 50 m

The area of a rectangle = length x breadth

⇒ Area of the inner field = 75 x 50 m² = 3,750 m²

Hence, the area to be covered with grass is 3,750 m².

The length and the breadth of a rectangle are 6 cm and 4 cm respectively. Find the height of a triangle whose base is 6 cm and area is 3 times that of the rectangle.

Answer

Given:

Length of the rectangle = 6 cm

Breadth of the rectangle = 4 cm

Area of the rectangle = length x breadth

= 6 x 4 cm²

= 24 cm²

Let h be the height of the triangle.

Base of the triangle = 6 cm

It is given that area of triangle is 3 times the area of the rectangle.

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

⇒ 3 x 24 = $\dfrac{1}{2}$ x 6 x h

⇒ 3 x 24 = 3 x h

⇒ $\cancel{3}$ x 24 = $\cancel{3}$ x h

⇒ h = 24 cm

Hence, the height of the triangle is 24 cm.

How many tiles, each of area 400 cm², will be needed to pave a footpath which is 2 m wide and surrounds a grass plot 25 m long and 13 m wide ?

Answer

Given:

Dimensions of grass plot (ABCD) = 25 x 13 m

Width of footpath = 2 m

Area of footpath = Area of ABQP + Area of QNOR + Area of RCDS + Area of SHMP

= 2Area of ABQP + 2Area of QNOR

= 2(Area of ABQP + Area of QNOR)

Area of ABQP = l x b = AB x AP

= 25 x 2 m²

= 50 m²

Area of QNOR = l x b = DN x QN

= 17 x 2 m²

= 34 m²

Area of footpath = 2(50 + 34) m²

= 2 x 84 m²

= 168 m²

= 1680000 sq. cm

Total area = Number of tiles x Area covered by 1 tiles

⇒ 1680000 = Number of tiles x 400

⇒ Number of tiles = $\dfrac{1680000}{400}$

= 4200

Hence, the number of tiles = 4200.

The cost of enclosing a rectangular garden with a fence all round, at the rate of 75 paise per metre, is ₹ 300. If the length of the garden is 120 metres, find the area of the field in square metres.

Answer

Given:

Length of the garden = 120 m

Cost of fencing = 75 paise per metre = ₹ 0.75 per m

Total cost = ₹ 300

Total cost of fencing = Perimeter x Cost per metre

⇒ ₹ 300 = Perimeter x ₹ 0.75

⇒ Perimeter = $\dfrac{300}{0.75}$ m

⇒ Perimeter = $\dfrac{30000}{75}$ m

⇒ Perimeter = 400 m

Let b be the breadth of the garden.

Perimeter = 2(l + b)

⇒ 2(120 + b) = 400

⇒ 120 + b = $\dfrac{400}{2}$

⇒ 120 + b = 200

⇒ b = 200 - 120

⇒ b = 80 m

Area of the rectangular garden = Length x Breadth

= 120 x 80 m²

= 9600 m²

Hence, the area of the garden is 9600 sq. m.

The width of a rectangular room is $\dfrac{4}{7}$ of its length, x, and its perimeter is y. Write an equation connecting x and y. Find the length of the room when the perimeter is 4400 cm.

Answer

Given:

Length = x

Width = $\dfrac{4}{7}$ of the length = $\dfrac{4}{7} \times x$

Perimeter = $y$

Perimeter of a rectangle = 2(l + b)

$⇒ 2\Big(x + \dfrac{4}{7}x\Big) = y\\[1em] ⇒ 2\Big(\dfrac{7 + 4}{7}x\Big) = y\\[1em] ⇒ 2\Big(\dfrac{11}{7}x\Big) = y\\[1em] ⇒ \dfrac{22}{7}x = y\\[1em]$

The length when y = 4400 cm,

$⇒ \dfrac{22}{7}x = 4400\\[1em] ⇒ x = \dfrac{4400 \times 7}{22}\\[1em] ⇒ x = \dfrac{30800}{22}\\[1em] ⇒ x = 1400 cm = 14 m$

Hence, the equation connecting x and y is $y = \dfrac{22}{7}x$ and the length of the room is 14 m.

The length of a rectangular verandah is 3 m more than its breadth. The numerical value of its area is equal to the numerical value of its perimeter.

(i) Taking x as the breadth of the verandah, write an equation in x that represents the above statement.

(ii) Solve the equation obtained in (i) above and hence find the dimensions of the verandah.

Answer

(i) Given:

Breadth of the verandah = x

Length of the verandah = x + 3

It is also given that the numerical value of the area is equal to the numerical value of the perimeter.

⇒ l x b = 2(l + b)

⇒ x(x + 3) = 2(x + x + 3)

⇒ x² + 3x = 2(2x + 3)

⇒ x² + 3x = 4x + 6

⇒ x² + 3x - 4x - 6 = 0

⇒ x² - x - 6 = 0

Hence, the equation is x² - x - 6.

(ii) From (i),

⇒ x² - x - 6 = 0

⇒ x² - 3x + 2x - 6 = 0

⇒ x(x - 3) + 2(x - 3) = 0

⇒ (x - 3)(x + 2) = 0

⇒ x = 3 or - 2

Since breadth cannot be negative, x = 3 m.

Length = x + 3 = 3 + 3 m = 6 m

Hence, length = 6 m and breadth = 3 m.

The diagram, given below, shows two paths drawn inside a rectangular field 80 m long and 45 m wide. The widths of the two paths are 8 m and 15 m as shown. Find the area of the shaded portion.

Answer

Given:

Length of the rectangular field = 80 m

Width of the rectangular field = 45 m

Area of shaded path = Area of path ABCD + Area of cross path EFGH - Area of path IJKL

Area of path ABCD = l x b = AB x BC

= 15 x 45 m²

= 675 m²

Area of path EFGH = l x b = EF x FG

= 8 x 80 m²

= 640 m²

Area of path IJKL = l x b = IJ x KL

= 15 x 8 m²

= 120 m²

Area of shaded path = 675 + 640 - 120 m²

= 1,195 m²

Hence, the area of shaded portion is 1,195 m².

The rate for a 1.20 m wide carpet is ₹ 40 per metre; find the cost of covering a hall 45 m long and 32 m wide with this carpet. Also, find the cost of carpeting the same hall if the carpet, 80 cm wide, is at ₹ 25 per metre.

Answer

Given:

Dimensions of the hall = 45 m x 32 m

Area = l x b

= 45 x 32 m²

= 1,440 m²

Cost per metre of carpet = ₹40

Width of the carpet = 1.20 m

Cost per square metre = $\dfrac{40}{1.20}$ = $\dfrac{400}{12}$ = ₹ 33.33

Total cost = Cost per square metre x area of hall

= 33.33 x 1440

= ₹ 48,000

Width of the carpet in metres: 80 cm = 0.8 m

Cost per metre of carpet = ₹25

Cost per square metre = $\dfrac{25}{0.8}$ = $\dfrac{250}{8}$ = ₹ 31.25

Total cost = Cost per square metre x area of hall

= 31.25 x 1440

= ₹ 45,000

Hence, the cost of covering the hall with a 1.20 m wide carpet at ₹ 40 per metre is ₹ 48,000 and the cost of carpeting the hall with an 80 cm wide carpet at ₹ 25 per metre is ₹ 45,000.

Find the area and perimeter of a square plot of land, the length of whose diagonal is 15 metres. Give your answer correct to 2 places of decimals.

Answer

Given:

Diagonal of the square = 15 m

Let a be the length of side of the square.

Using the Pythagoras theorem for a square,

⇒ Side² + Side² = Diagonal²

⇒ 2 x Side² = Diagonal²

⇒ 2 x a² = 15²

⇒ 2a² = 225

⇒ a² = $\dfrac{225}{2}$

⇒ a² = 112.5

⇒ a = $\sqrt{112.5}$

⇒ a = 10.61 m

Area of square = side²

= 10.61² m²

= 112.5 m²

Perimeter of square = 4 x side

= 4 x 10.60 m

= 42.43 m

Hence, the area of the square plot is 112.5 m² and the perimeter is 42.43 m.

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

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

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

Answer

Given:

Length of the lawn = 30 m

Breadth of the lawn = 12 m

Width of flowerbed = 2 m

New length of lawn = 30 - 2 - 2 m = 30 - 4 m = 26 m

New breadth of lawn = 12 - 2 m = 10 m

Hence, the length of lawn = 26 m and the breadth of lawn = 10 m.

(ii) Area of flowerbed = Area of AMPX + Area of XDCY + Area of NBYO

Area of AMPX = l x b = AM x MP

= 10 x 2 m²

= 20 m²

Area of XDCY = l x b = XD x DC

= 30 x 2 m²

= 60 m²

Area of NBYO = l x b = NB x BY

= 10 x 2 m²

= 20 m²

Area of flowerbed = 20 + 60 + 20 m²

= 100 m²

Hence, the area of flowerbed is 100 m².

A floor which measures 15 m x 8 m is to be laid with tiles measuring 50 cm x 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 left uncovered.

Answer

Given:

Dimensions of the floor = 15 m x 8 m

Dimensions of the tile = 50 cm x 25 cm = 0.5 m x 0.25 m

Area of floor = l x b

= 15 x 8 m²

= 120 m²

Area of tile = l_tile x b_tile

= 0.5 x 0.25 m²

= 0.125 m²

Let n be the number of tiles.

Area of floor = Area of tile x Number of tiles

⇒ 120 = 0.125 x n

⇒ n = $\dfrac{120}{0.125}$

⇒ n = $\dfrac{120000}{125}$

⇒ n = 960

Now, a carpet is laid on the floor, leaving a space of 1 m between its edges and the edges of the floor.

Length of the carpet = 15 - 1 - 1 m = 13 m

Breadth of the carpet = 8 - 1 - 1 m = 6 m

Area of uncovered floor = Area of floor - Area of carpet

= 15 x 8 - 13 x 6 m²

= 120 - 78 m²

= 42 m²

Fraction of floor left uncovered = $\dfrac{42}{120}$ = $\dfrac{7}{20}$

Hence, the number of tiles needed is 960 and the fraction of the floor left uncovered is $\dfrac{7}{20}$.

Two adjacent sides of parallelogram are 24 cm and 18 cm. If the distance between the longer sides is 12 cm; find the distance between the shorter sides.

Answer

Step 1 Finding the area of the parallelogram.

Length of one side of the parallelogram (longer side) = 24 cm

Length of the adjacent side (shorter side) = 18 cm

The distance between the longer sides = 12 cm

Area of parallelogram = base x height

= AB x DE

= 24 x 12 cm²

= 288 cm²

Step 2 Finding the sides of the parallelogram.

Let a be the distance between the shorter sides.

Area = base x height

= BC x AF

= 18 x a

Since area of the parallelogram will be same, if we consider AB as the base or BC as the base.

⇒ AB x DE = BC x AF

⇒ 288 = 18 x a

⇒ a = $\dfrac{288}{18}$

⇒ a = 16 cm

Hence, the distance between the shorter sides of the parallelogram is 16 cm.

Two adjacent sides of a parallelogram are 28 cm and 26 cm. If one diagonal of it is 30 cm long; find the area of the parallelogram. Also, find the distance between its shorter sides.

Answer

By joining diagonal BD, the parallelogram is divided into two triangles. Let the sides of Δ ABD be:

a = 28 cm, b = 26 cm and c = 30 cm.

The semi-perimeter s:

$∵ s = \dfrac{a + b + c}{2}\\[1em] = \dfrac{28 + 26 + 30}{2}\\[1em] = \dfrac{84}{2}\\[1em] = 42$

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

= $\sqrt{42(42 - 28)(42 - 26)(42 - 30)}$ cm²

= $\sqrt{42 \times 14 \times 16 \times 12}$ cm²

= $\sqrt{112,896}$ cm²

= 336 cm²

Area of parallelogram ABCD = 2 x area of Δ ABD

= 2 x 336 cm²

= 672 cm²

Let h be the distance between the shorter sides.

Area of the parallelogram = base x height

⇒ 26 x h = 672

⇒ h = $\dfrac{672}{26}$

⇒ h = 25.84 cm

Hence, the area of the parallelogram is 672 cm² and the distance between the shorter sides is 25.84 cm.

The area of a rhombus is 216 sq. cm. If its one diagonal is 24 cm; find :

(i) length of its other diagonal,

(ii) length of its side,

(iii) perimeter of the rhombus.

Answer

(i) Given:

Area of rhombus = 216 sq. cm

One diagonal = 24 cm

Let d be the other diagonal of rhombus.

Area = $\dfrac{1}{2}$ x product of diagonals

⇒ 216 = $\dfrac{1}{2}$ x 24 x d

⇒ 216 = 12 x d

⇒ d = $\dfrac{216}{12}$

⇒ d = 18

Hence, the length of other diagonal is 18 cm.

(ii) The rhombus is shown in the figure below:

Diagonal AC = 24 cm

Then, OA = OC = $\dfrac{24}{2}$ = 12 cm

Diagonal BD = 18 cm

Then, OB = OD = $\dfrac{18}{2}$ = 9 cm

Since the diagonals of a rhombus bisect at 90°.

Applying pythagoras theorem in Δ AOB, we get:

AB² = OA² + OB²

⇒ AB² = (12)² + (9)²

⇒ AB² = 144 + 81

⇒ AB² = 225

⇒ AB = $\sqrt{225}$

⇒ AB = 15 cm

Hence, the length of each side of the rhombus is 15 cm.

(iii) Perimeter of the rhombus = 4 x side

= 4 x 15 cm

= 60 cm

Hence, the perimeter of the rhombus 60 cm.

The perimeter of a rhombus is 52 cm. If one diagonal is 24 cm; find :

(i) the length of its other diagonal,

(ii) its area.

Answer

(i) Given:

Perimeter of the rhombus = 52 cm

One diagonal BD = 24 cm

Let a be the length of a side of the rhombus.

Perimeter of a rhombus = 4 x Side

⇒ 4 x a = 52

⇒ a = $\dfrac{52}{4}$

⇒ a = 13 cm

BD = 24 cm

Then, OB = OD = $\dfrac{24}{2}$ = 12 cm

Since the diagonals of a rhombus bisect at 90°.

Applying pythagoras theorem in triangle AOB, we get:

AB² = OA² + OB²

⇒ (13)² = OA² + (12)²

⇒ 169 = OA² + 144

⇒ OA² = 169 - 144

⇒ OA² = 25

⇒ OA = $\sqrt{25}$

⇒ OA = 5 cm

AC = 2 x OA = 2 x 5 cm = 10 cm

Hence, the length of the other diagonal is 10 cm.

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

= $\dfrac{1}{2}$ x 24 x 10 cm²

= 12 x 10 cm²

= 120 cm²

Hence, the area of the rhombus is 120 cm².

The perimeter of a rhombus is 46 cm. If the height of the rhombus is 8 cm; find its area.

Answer

Given:

Perimeter of the rhombus = 46 cm

Height of the rhombus = 8 cm

Let a be the length of a side of the rhombus.

Perimeter = 4 x Side

⇒ 46 = 4 x a

⇒ a = $\dfrac{46}{4}$

⇒ a = 11.5 cm

Area = base x height

= 11.5 x 8 cm²

= 92 cm²

Hence, the area of the rhombus is 92 cm².

The given figure shows the cross-section of a concrete structure. Calculate the area of cross-section if AB = 1.8 m, CD = 0.6 m, DE = 0.8 m, EF = 0.3 m and AF = 1.2 m.

Answer

Given:

AB = 1.8 m, CD = 0.6 m, DE = 0.8 m, EF = 0.3 m and AF = 1.2 m

Area of ABCDEF = Area of rectangle AGEF + Area of rectangle GHCD + Area of triangle HBC.

Area of rectangle AGEF = l x b = AF x EF

= 1.2 x 0.3 m²

= 0.36 m²

Area of rectangle GHCD = l x b = GH x HC

= 0.6 x 2 m² (HC = AF + ED = 1.2 + 0.8 = 2 cm)

= 1.2 m²

Area of triangle HBC = $\dfrac{1}{2}$ x b x h = $\dfrac{1}{2}$ x HB x HC

= $\dfrac{1}{2}$ x 0.9 x 2 m² (HB = AB - AH = 1.8 - 0.9 = 0.9)

= 0.9 x 1 m²

= 0.9 m²

Now, area of ABCDEF = 0.36 m² + 1.2 m² + 0.9 m²

= 2.46 m²

Hence, the area of cross-section is 2.46 sq. m.

Calculate the area of the figure given below which is not drawn to scale.

Answer

Area of figure ABCEF = Area of trapezium ABCF + Area of triangle CEF

Area of triangle CEF = $\dfrac{1}{2}$ x b x h = $\dfrac{1}{2}$ x CF x DE

= $\dfrac{1}{2}$ x 25 x 12 m²

= 25 x 6 m²

= 150 m²

In triangle BGC,

Base² + Height² = Hypotenuse²

⇒ CG² + BG² = CB²

Let h be the length of BG.

⇒ 10² + h² = 26² (∵CG = CF - GF = CF - BA = 25 - 15 = 10)

⇒ 100 + h² = 676

⇒ h² = 676 - 100

⇒ h² = 576

⇒ h = $\sqrt{576}$

⇒ h = 24 cm

Area of trapezium = $\dfrac{1}{2}$ (sum of parallel sides) x distance between the parallel sides

= $\dfrac{1}{2}$ (AB + CF) x BG

= $\dfrac{1}{2}$ (15 + 25) x 24 m²

= $\dfrac{1}{2}$ x 40 x 24 m²

= 20 x 24 m²

= 480 m²

So, total area of figure ABCEF = 150 + 480 m²

= 630 m²

Hence, the area is 630 m².

The following diagram shows a pentagonal field ABCDE in which the lengths of AF, FG, GH and HD are 50 m, 40 m, 15 m and 25 m respectively; and the lengths of perpendiculars BF, CH and EG are 50 m, 25 m and 60 m respectively. Determine the area of the field.

Answer

Area of figure ABCDE = Area of Δ ABF + Area of Δ AED + Area of Δ DHC + Area of trapezium BFHC

Area of Δ ABF = $\dfrac{1}{2}$ x b x h = $\dfrac{1}{2}$ x AF x BF

= $\dfrac{1}{2}$ x 50 x 50 m²

= 25 x 50 m²

= 1250 m²

Area of Δ AED = $\dfrac{1}{2}$ x b x h = $\dfrac{1}{2}$ x AD x GE

= $\dfrac{1}{2}$ x (AF + FG + GH + HD) x GE

= $\dfrac{1}{2}$ x (50 + 40 + 15 + 25) x 60 m²

= 130 x 30 m²

= 3900 m²

Area of Δ DHC = $\dfrac{1}{2}$ x b x h = $\dfrac{1}{2}$ x HD x CH

= $\dfrac{1}{2}$ x 25 x 25 m²

= 12.5 x 25 m²

= 312.5 m²

Area of trapezium BFHC = $\dfrac{1}{2}$ (sum of parallel sides) x distance between the parallel sides

= $\dfrac{1}{2}$ (BF + CH) x FH

= $\dfrac{1}{2}$ (BF + CH) x (FH + GH)

= $\dfrac{1}{2}$ (50 + 25) x (40 + 15) m²

= $\dfrac{1}{2}$ x 75 x 55 m²

= 37.5 x 55 m²

= 2062.5 m²

Thus, area of figure ABCDE = 1250 + 3900 + 312.5 + 2062.5 m²

= 7525 m²

Hence, the area is 7525 sq. m.

In the given figure, AB = BC, ∠ABC = 90°, AC = 14 ${\sqrt2}$ cm and BPC (shaded portion) is semi-circle. If $π = \dfrac{22}{7}$, the area of shaded portion is :

1. 77 cm²

2. 308 cm²

3. 231 cm²

4. 154 cm²

Answer

Let AB = BC = a cm and triangle ABC is a right triangle,

Using the Pythagoras theorem,

Base² + Height² = Hypotenuse²

⇒ BC² + AB² = AC²

⇒ a² + a² = ($14\sqrt{2}$)²

⇒ 2a² = 392

⇒ a² = $\dfrac{392}{2}$

⇒ a² = 196

⇒ a = $\sqrt{196}$

⇒ a = 14 cm

Thus, AB = BC = 14 cm.

The diameter of the semicircle is BC = 14 cm, so the radius r is:

r = $\dfrac{d}{2}$ = $\dfrac{14}{2}$ = 7 cm

Area of semi-circle = $\dfrac{1}{2}$ πr²

$= \dfrac{1}{2} \times \dfrac{22}{7} \times 7^2\\[1em] = \dfrac{1 \times 22}{2 \times 7}\times 49\\[1em] = \dfrac{22}{14}\times 49\\[1em] = \dfrac{1,078}{14}\\[1em] = 77 \text{cm}^2$

Thus, the area of the shaded portion (the semicircle) is 77 cm².

Hence, option 1 is the correct option.

The diameter of a circle is 14 cm. If it is doubled, the perimeter of the resulting circle will become:

1. four times

2. doubled

3. halved

4. three times

Answer

Given:

Diameter of the original circle = 14 cm

Radius of the original circle = r = $\dfrac{d}{2}$ = $\dfrac{14}{2}$ = 7 cm

Perimeter of the original circle = 2πr

$= 2 \times \dfrac{22}{7} \times 7\\[1em] = \dfrac{44}{7} \times 7\\[1em] = \dfrac{44}{\cancel{7}} \times \cancel{7}\\[1em] = 44 \text{ cm}$

When diameter is doubled, new diameter = 2 x 14 cm = 28 cm

New radius, r = $\dfrac{d}{2}$ = $\dfrac{28}{2}$ = 14 cm

New perimeter of circle:

$= 2 \times \dfrac{22}{7} \times 14\\[1em] = \dfrac{44}{7} \times 14\\[1em] = \dfrac{616}{7}\\[1em] = 88 \text{ cm}\\[1em]$

Thus, the new perimeter is doubled compared to the original perimeter.

Hence, option 2 is the correct option.

In the given figure, OABC is a square of side 14 cm. Taking $π = \dfrac{22}{7}$, the area of the shaded portion is :

1. 154 cm²

2. 196 cm²

3. 42 cm²

4. 52 cm²

Answer

Side of the square = Radius of the circle = 14 cm

Area of shaded portion = Area of square - $\dfrac{1}{4}$ Area of quarter circle

$= Side^2 - \dfrac{1}{4} πr^2\\[1em] = 14^2 - \dfrac{1}{4} \times \dfrac{22}{7} \times 14^2\\[1em] = 196 - \dfrac{22}{28}\times 14 \times 14\\[1em] = 196 - \dfrac{22}{2} \times 14\\[1em] = 196 - \dfrac{308}{2} \\[1em] = 196 - 154\\[1em] = 42\text{cm}^2$

The area of the shaded portion is 42 cm².

Hence, option 3 is the correct option.

If the radius of a circle is doubled, then its area will become :

1. doubled

2. halved

3. four times

4. three times

Answer

Let r be the radius of circle.

Area of circle = πr²

When the radius is doubled, new radius = 2r

Area of new circle = π x (2r)²

= π x 4r²

= 4πr²

The new area is four times the area of the original circle.

Hence, option 3 is the correct option.

The radii of two circles are 14 cm and 28 cm. If $π = \dfrac{22}{7}$ the area of the shaded portion is :

1. 1848 cm²

2. 1644 cm²

3. 486 cm²

4. 702 cm²

Answer

Area of shaded portion = Area of bigger circle - Area of smaller circle

$= π \times R^2 - π \times r^2\\[1em] = \dfrac{22}{7} \times 28^2 - \dfrac{22}{7} \times 14^2\\[1em] = \dfrac{22}{7} \times (28^2 - 14^2)\\[1em] = \dfrac{22}{7} \times (784 - 196)\\[1em] = \dfrac{22}{7} \times 588\\[1em] = \dfrac{12,936}{7}\\[1em] = 1,848 \text{ cm}^2$

The area of shaded portion is 1848 cm².

Hence, option 1 is the correct option.

The diameter of a circle is 28 cm. Find its :

(i) circumference

(ii) area

Answer

Given:

Diameter = d = 28 cm

Radius = r = $\dfrac{d}{2}$ = $\dfrac{28}{2}$ = 14 cm

(i) Circumference of a circle = 2πr

$= 2 \times \dfrac{22}{7} \times 14\\[1em] = \dfrac{44}{7} \times 14 \\[1em] = \dfrac{616}{7}\\[1em] = 88 \text{ cm}$

Hence, the circumference of a circle is 88 cm.

(ii) Area of a circle = πr²

$= \dfrac{22}{7} \times 14^2\\[1em] = \dfrac{22}{7} \times 196\\[1em] = \dfrac{4,312}{7}\\[1em] = 616 \text{ cm}^2$

Hence, the area of a circle is 616 cm².

The circumference of a circular field is 308 m. Find its :

(i) radius

(ii) area.

Answer

(i) Let r be the radius of the circle.

The circumference of a circle = 308 m

As we know, the circumference of a circle = 2πr

$⇒ 2 \times \dfrac{22}{7} \times r = 308\\[1em] ⇒ \dfrac{44}{7} \times r = 308\\[1em] ⇒ r = \dfrac{308 \times 7}{44}\\[1em] ⇒ r = \dfrac{2,156}{44}\\[1em] ⇒ r = 49 \text{ m}^2$

Hence, the radius of a circle is 49 m.

(ii) Area of a circle = πr²

$= \dfrac{22}{7} \times 49^2\\[1em] = \dfrac{22}{7} \times 2,401\\[1em] = \dfrac{52,822}{7}\\[1em] = 7,546 \text{ m}^2$

Hence, the area of a circle is 7,546 m².

The sum of the circumference and diameter of a circle is 116 cm. Find its radius.

Answer

Given:

Circumference of a circle + Diameter of a circle = 116 cm

$⇒ 2πr + 2r = 116\\[1em] ⇒ 2r(π + 1) = 116\\[1em] ⇒ 2r\Big(\dfrac{22}{7} + 1\Big) = 116\\[1em] ⇒ 2r\Big(\dfrac{22 + 7}{7}\Big) = 116\\[1em] ⇒ 2r\Big(\dfrac{29}{7}\Big) = 116\\[1em] ⇒ r\Big(\dfrac{58}{7}\Big) = 116\\[1em] ⇒ r = \dfrac{116 \times 7}{58}\\[1em] ⇒ r = \dfrac{812}{58}\\[1em] ⇒ r = 14 \text{ cm}$

Hence, the radius of a circle is 14 cm.

The radii of two circles are 25 cm and 18 cm. Find the radius of the circle which has circumference equal to the sum of circumferences of these two circles.

Answer

Circumference of a circle = 2πr

For the first circle,

r₁ = 25 cm

Circumference₁ = 2 x π x 25 = 50π cm

For the second circle,

r₂ = 18 cm

Circumference₂ = 2 x π x 18 = 36π cm

Total circumference = Circumference₁ + Circumference₂

= 50π + 36π

= 86π cm

Let the radius of the new circle be R.

Circumference of the new circle = 2πR

86π = 2πR

86 = 2R

R = $\dfrac{86}{2}$

R = 43 cm

Hence, the radius of the new circle is 43 cm.

The radii of two circles are 48 cm and 13 cm. Find the area of the circle which has its circumference equal to the difference of the circumferences of the given two circles.

Answer

Circumference of a circle = 2πr

For the first circle,

r₁ = 48 cm

Circumference₁ = 2 x π x 48 = 96π cm

For the second circle,

r₂ = 13 cm

Circumference₂ = 2 x π x 13 = 26π cm

Total circumference = Circumference₁ - Circumference₂

= 96π - 26π

= 70π cm

Let the radius of the new circle be R.

Circumference of the new circle = 2πR

70π = 2πR

70 = 2R

R = $\dfrac{70}{2}$

R = 35 cm

And, area of the new circle = πr²

$= \dfrac{22}{7} \times 35^2\\[1em] = \dfrac{22}{7} \times 1,225\\[1em] = \dfrac{26,950}{7}\\[1em] = 3,850 \text{ cm}^2$

Hence, the area of new circle is 3,850 cm².

The diameters of two circles are 32 cm and 24 cm. Find the radius of the circle having its area equal to sum of the areas of the two given circles.

Answer

For the first circle,

d₁ = 32 cm

r₁ = $\dfrac{d_1}{2}$ = $\dfrac{32}{2}$ = 16 cm

Area₁ = π x 16² = 256π cm²

For the second circle,

d₂ = 24 cm

r₂ = $\dfrac{d_2}{2}$ = $\dfrac{24}{2}$ = 12 cm

Area₂ = π x 12² = 144π cm²

Total area = Area₁ + Area₂

= 256π + 144π

= 400π cm

Let the radius of the new circle be R.

Area of the new circle = πR²

400π = πR²

400 = R²

R = $\sqrt{400}$

R = 20 cm

Hence, the radius of the new circle is 20 cm.

The radius of a circle is 5 m. Find the circumference of the circle whose area is 49 times the area of the given circle.

Answer

Given:

Radius of original circle = r = 5 m

Area of new circle = 49 times area of original circle.

Let R be the radius of new circle.

$⇒ πR^2 = 49 \times πr^2\\[1em] ⇒ \dfrac{22}{7} \times R^2 = 49 \times \dfrac{22}{7} \times 5^2\\[1em] ⇒ \cancel{\dfrac{22}{7}} \times R^2 = 49 \times \cancel{\dfrac{22}{7}} \times 5^2\\[1em] ⇒ R^2 = 49 \times 25\\[1em] ⇒ R^2 = 1,225\\[1em] ⇒ R = \sqrt{1,225}\\[1em] ⇒ R = 35 \text{ m}^2$

Circumference of new circle = 2πR

$= 2 \times \dfrac{22}{7} \times 35\\[1em] = \dfrac{44}{7} \times 35\\[1em] = \dfrac{1,540}{7}\\[1em] = 220 \text{ m}^2$

Hence, the circumference of new circle is 220 m.

A circle of largest area is cut from a rectangular piece of card-board with dimensions 55 cm and 42 cm. Find the ratio between the area of the circle cut and the area of the remaining card-board.

Answer

Given:

The dimensions of rectangular piece of card-board are:

Length = 55 cm

Width = 42 cm

The largest circle that can be cut from the rectangle will have a diameter equal to the shorter side of the rectangle.

Diameter = Width = 42 cm

∵ Radius = r = $\dfrac{d}{2}$ = $\dfrac{42}{2}$ = 21 cm

Area of the circle = πr²

$= \dfrac{22}{7} \times 21^2\\[1em] = \dfrac{22}{7} \times 441\\[1em] = \dfrac{9,704}{7}\\[1em] = 1,386 \text{ cm}^2$

Area of the rectangular piece of cardboard = 55 x 42 cm²

= 2,310 cm²

Therefore, area of remaining cardboard = Area of rectangle - Area of circle

= (2,310 - 1,386) cm²

= 924 cm²

So, the ratio between the area of the circle cut and the area of the remaining card-board = 1,386 : 924

= 231 : 154

= 3 : 2

Hence, the ratio between the area of the circle cut and the area of the remaining cardboard is 3 : 2.

The following figure shows a square card-board ABCD of side 28 cm. Four identical circles of largest possible size are cut from this card as shown below.

Find the area of the remaining card-board.

Answer

Given:

Side of square ABCD = 28 cm

Side of square = 2 x diameter of circle

Diameter of circle = $\dfrac{28}{2}$ = 14 cm

Radius of circle = $\dfrac{d}{2}$ = $\dfrac{14}{2}$ = 7 cm

Area of the remaining card-board = Area of square - 4 x Area of 1 circle

= side² - 4 x πr²

$= 28^2 - 4 \times \dfrac{22}{7} \times 7^2\\[1em] = 28^2 - 4 \times \dfrac{22}{7} \times 7 \times 7\\[1em] = 28^2 - 4 \times \dfrac{22}{\cancel{7}} \times \cancel{7} \times 7\\[1em] = 28^2 - 4 \times 22 \times 7\\[1em] = 784 - 616\\[1em] = 168 \text{ cm}^2$

Hence, the area of remaining cardboard is 168 cm².

The radii of two circles are in the ratio 3 : 8. If the difference between their areas is 2695 π cm², find the area of the smaller circle.

Answer

Let the two radii of two circles be 3a and 8a.

The difference between their areas = 2695 π cm²

⇒ π x (8a)² - π x (3a)² = 2695 π

⇒ π x 64a² - π x 9a² = 2695 π

⇒ π x (64a² - 9a²) = 2695 π

⇒ $\cancel{π}$ x (64a² - 9a²) = 2695 $\cancel{π}$

⇒ 64a² - 9a² = 2695

⇒ 55a² = 2695

⇒ a² = $\dfrac{2695}{55}$

⇒ a² = 49

⇒ a = $\sqrt{49}$

⇒ a = 7 cm

The radii are 3a and 8a = 3 x 7 cm and 8 x 7 cm = 21 cm and 56 cm

Area of smaller circle = π x (21)²

$= \dfrac{22}{7} \times 441\\[1em] = \dfrac{9,702}{7} \\[1em] = 1,386 \text{ cm}^2$

Hence, the area of smaller circle is 1,386 cm².

The diameters of three circles are in the ratio 3 : 5 : 6. If the sum of the circumferences of these circles be 308 cm; find the difference between the areas of the largest and the smallest of these circles.

Answer

Let the diameters of the three circles be 3a, 5a and 6a.

Radius of three circles = $\dfrac{3a}{2}$, $\dfrac{5a}{2}$ and $\dfrac{6a}{2}$

Circumference of a circle = 2πr

For the first circle,

r₁ = $\dfrac{3a}{2}$ cm

Circumference₁ = 2 x π x $\dfrac{3a}{2}$ = 3aπ cm

For the second circle,

r₂ = $\dfrac{5a}{2}$ cm

Circumference₂ = 2 x π x $\dfrac{5a}{2}$ = 5aπ cm

For the third circle,

r₂ = $\dfrac{6a}{2}$ cm

Circumference₂ = 2 x π x $\dfrac{6a}{2}$ = 6aπ cm

Total circumference = Circumference₁ + Circumference₂ + Circumference₃

⇒ 3aπ + 5aπ + 6aπ = 308

⇒ 14aπ = 308

⇒ 14 x a x $\dfrac{22}{7}$ = 308

⇒ 2 x a x 22 = 308

⇒ 44a = 308

⇒ a = $\dfrac{308}{44}$

⇒ a = 7 cm

Radius of three circles = $\dfrac{3}{2}$ x 7 cm, $\dfrac{5}{2}$ x 7 cm and $\dfrac{6}{2}$ x 7 cm

= 10.5 cm, 17.5 cm and 21 cm

Difference between the area of the largest and the smallest circles = π(21)² - π(10.5)²

= 441π - 110.25π cm²

= 330.75π cm²

= 330.75 x $\dfrac{22}{7}$ cm²

= 47.25 x 22 cm²

= 1039.5 cm²

Hence, the difference in the area = 1039.5 cm².

Find the area of a ring shaped region enclosed between two concentric circles of radii 20 cm and 15 cm.

Answer

Given:

r₁ = 20 cm

r₂ = 15 cm

Area of ring shaped region = π(r₁² - r₂²)

= π(20² - 15²) cm²

= π(400 - 225) cm²

= π x 175 cm²

= $\dfrac{22}{7}$ x 175 cm²

= 22 x 25 cm²

= 550 cm²

Hence, the area of ring shaped region is 550 cm².

The circumference of a given circular park is 55 m. It is surrounded by a path of uniform width 3.5 m. Find the area of the path.

Answer

Let r be the radius of the circular park.

Circumference = 2πr

⇒ 2πr = 55

⇒ 2 x $\dfrac{22}{7}$ x r = 55

⇒ $\dfrac{44}{7}$ x r = 55

⇒ r = $\dfrac{7 \times 55}{44}$

⇒ r = $\dfrac{7 \times 5}{4}$

⇒ r = $\dfrac{35}{4}$

⇒ r = 8.75 m

Radius of outer park = radius of park + width of the path

= 8.75 + 3.5 m

= 12.25 m

Area of the path = Area of outer park - Area of circular park

= π (12.25)² - π (8.75)² m²

= 150.0625π - 76.5625π m²

= 73.5 π m²

= 73.5 x $\dfrac{22}{7}$ m²

= 10.5 x 22 m²

= 231 m²

Hence, the area of the path is 231 m².

There are two circular gardens A and B. The circumference of garden A is 1.760 km and the area of garden B is 25 times the area of garden A. Find the circumference of garden B.

Answer

Let r be the radius of the circular garden A.

Circumference of garden A = 2πr

⇒ 2πr = 1.760 km

⇒ 2πr = 1760 m

⇒ 2 x $\dfrac{22}{7}$ x r = 1760 m

⇒ $\dfrac{44}{7}$ x r = 1760 m

⇒ r = $\dfrac{7 \times 1760}{44}$ m

⇒ r = 7 x 40 m

⇒ r = 280 m

Area of garden A = πr²

= π x (280)²

= 78400π

Let R be the radius of garden B.

It is given that the area of garden B is 25 times the area of garden A.

⇒ πR² = 25 x πr²

⇒ πR² = 25 x 78400π

⇒ $\cancel{π}$ R² = 25 x 78400 $\cancel{π}$

⇒ R² = 25 x 78400

⇒ R² = 1960000

⇒ R = $\sqrt{1960000}$

⇒ R = 1400 m

Circumference of garden B = 2πR

= 2 x $\dfrac{22}{7}$ x 1400 m

= 2 x 22 x 200 m

= 8800 m = 8.8 km

Hence, the circumference of garden B is 8.8 km.

A wheel has diameter 84 cm. Find how many complete revolutions must it make to cover 3.168 km.

Answer

Diameter of the wheel = 84 cm

Radius of the wheel = $\dfrac{d}{2}$ = $\dfrac{84}{2}$ = 42 cm

Circumference of the wheel = 2πr

$= 2 \times \dfrac{22}{7} \times 42\\[1em] = 2 \times 22 \times 6\\[1em] = 264 \text{ cm}$

Let the wheel make n revolutions.

Total distance traveled by the wheel = n x Circumference of the wheel

⇒ 3.168 km = n x 264 cm

⇒ 316,800 cm = n x 264 cm

⇒ n = $\dfrac{316,800}{264}$

⇒ n = 1200

Hence, the wheel makes 1,200 revolutions.

Each wheel of a car is of diameter 80 cm. How many complete revolutions does each wheel make in 10 minutes when the car is travelling at a speed of 66 km per hour ?

Answer

Diameter of wheel = 80 cm

Radius of the wheel = $\dfrac{d}{2}$ = $\dfrac{80}{2}$ = 40 cm

Circumference of the wheel = 2πr

$= 2 \times \dfrac{22}{7} \times 40\\[1em] = \dfrac{1,760}{7} \text{ cm}$

Let the wheel make n number of revolutions.

Speed of the car = 66 km per hour

Time = 10 min

Total distance = Speed x Time

= $\dfrac{66 \times 100,000}{60}$ x 10 cm

= 11 x 100,000 cm

= 1,100,000

Total distance = Number of revolutions x Circumference of the wheel

⇒ 1,100,000 cm = n x $\dfrac{1,760}{7}$ cm

⇒ n = $\dfrac{1,100,000 x 7}{1,760}$

⇒ n = $\dfrac{7,700,000 }{1,760}$

⇒ n = 4,375

Hence, the wheel makes 4,375 complete revolutions.

An express train is running between two stations with a uniform speed. If the diameter of each wheel of the train is 42 cm and each wheel makes 1200 revolutions per minute, find the speed of the train.

Answer

Diameter of wheel = 42 cm

Radius of wheel = $\dfrac{d}{2}$ = $\dfrac{42}{2}$ = 21 cm

Circumference of wheel = 2πr

$= 2 \times \dfrac{22}{7} \times 21\\[1em] = 2 \times 22 \times 3\\[1em] = 132 \text{ cm}$

Total distance covered by one wheel in 1 minute = Number of revolutions x Circumference of wheel

= 1,200 x 132 cm

= 158,400 cm

= 1.584 km

Speed of the train = $\dfrac{\text{Distance}}{\text{Time}}$