Area & Perimeter of Plane Figures — Exercise 19(C)

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 = $\dfrac{\text{Distance}}{\text{Time}}$

= $\dfrac{1.584}{\dfrac{1}{60}}$

= $\dfrac{1.584 \times 60}{1}$

= 95.04 km/hr

Hence, the speed of the train is 95.04 km/hr.

The minute hand of a clock is 8 cm long. Find the area swept by the minute hand between 8.30 a.m. and 9.05 a.m.

Answer

Length of the minute hand (radius of the circle) = 8 cm

Time interval between 9.05 a.m. and 8.30 a.m. = 35 minutes

Area swept by the minute hand in 1 hr = πr²

$= \dfrac{22}{7} \times 8^2\\[1em] = \dfrac{22}{7} \times 64\\[1em] = \dfrac{1,408}{7} \text{ cm}^2$

Area of the circle in 60 minutes = $\dfrac{1,408}{7}$ cm²

Area of the circle in 1 minute = $\dfrac{1,408}{7 \times 60}$ cm²

= $\dfrac{352}{105}$ cm²

Area of the circle in 35 minutes = $\dfrac{352 \times 35}{105}$ cm²

= $\dfrac{352}{3}$ cm²

= $117\dfrac{1}{3}$ cm²

Hence, the area swept by the minute hand between 8:30 a.m. and 9:05 a.m. is $117\dfrac{1}{3}$ cm².

The shaded portion of the figure, given alongside, shows two concentric circles.

If the circumference of the two circles be 396 cm and 374 cm, find the area of the shaded portion.

Answer

Circumference of the circle = 2πr

For outer circle,

$⇒ 2 \times \dfrac{22}{7} \times R = 396\\[1em] ⇒ \dfrac{44}{7} \times R = 396\\[1em] ⇒ R = \dfrac{396 \times 7}{44}\\[1em] ⇒ R = \dfrac{9 \times 7}{1}\\[1em] ⇒ R = 63 \text{ cm}$

For inner circle,

$⇒ 2 \times \dfrac{22}{7} \times r = 374\\[1em] ⇒ \dfrac{44}{7} \times r = 374\\[1em] ⇒ r = \dfrac{374 \times 7}{44}\\[1em] ⇒ r = \dfrac{17 \times 7}{2}\\[1em] ⇒ r = 59.5 \text{ cm}$

Area of shaded portion = π(R² - r²)

= π(63² - 59.5²) cm²

= π(3,969 - 3,540.25) cm²

= π x 428.75 cm²

= $\dfrac{22}{7}$ x 428.75 cm²

= 22 x 61.25 cm²

= 1,347.5 cm²

Hence, the area of the shaded portion is 1,347.5 cm².