Circumference & Area of a Circle

Number of disc along the length of sheet :

= $\frac{\text{Length of sheet}}{\text{Diameter of disc}} = \frac{11}{0.5}$ = 22 discs.

Number of disc along the width of sheet :

= $\frac{\text{Width of sheet}}{\text{Diameter of disc}} = \frac{2}{0.5}$ = 4 discs.

Total number of discs = 22 × 4 = 88.

Hence, number of discs formed = 88.

Circumference of circle = 2πr = 2 × π × 17.5

= 2 × $\frac{22}{7}$ × 17.5

= 2 × 22 × 2.5

= 22 × 5 = 110 cm.

Area of circle = πr²

= $\frac{22}{7}$ × (17.5)²

= $\frac{22}{7}$ × 17.5 × 17.5

= 22 × 2.5 × 17.5

= 962.5 cm².

Hence, circumference = 110 cm and area = 962.5 cm².

Circumference of circle = 2πr = 2 × π × 15

= 2 × 3.14 × 15

= 3.14 × 30

= 94.2 cm.

Area of circle = πr²

= 3.14 × (15)²

= 3.14 × 225

= 706.5 cm².

Hence, circumference = 94.2 cm and area = 706.5 cm².

Let the radius of circle be r cm.

(i) Circumference of circle = 2πr

⇒ 123.2 = 2 × $\frac{22}{7}$ × r

⇒ r = $\frac{123.2 × 7}{2 × 22}$

⇒ r = $\frac{862.4}{44}$ = 19.6 cm.

Hence, radius of circle = 19.6 cm.

(ii) Area of circle = πr²

= $\frac{22}{7}$ × (19.6)²

= $\frac{22}{7}$ × 19.6 × 19.6

= 22 × 2.8 × 19.6

= 1207.36

≈ 1207 cm².

Hence, area of circle = 1207 cm².

(iii) Area of circle = π(radius)²

If radius is double then it will become 2r.

New area = π(2r)²

= π × 4r²

= 4 × πr².

Hence, new area becomes 4 times the old area.

Given,

Grazed area = 9856 m²

Radius of circular field = Length of the rope

A cow tethered to a fixed point by a rope will graze in a circular pattern.

Area of circle = πr²

⇒ 9856 = $\frac{22}{7}$ × r²

⇒ r² = $\frac{9856 × 7}{22}$

⇒ r² = $\frac{68992}{22}$

⇒ r² = 3136

⇒ r = $\sqrt{3136}$ = 56 m.

Hence, length of the rope = 56 m.

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

Given,

Area = 394.24 cm²

Area of circle = πr²

⇒ 394.24 = $\frac{22}{7}$ × r²

⇒ r² = $\frac{394.24 × 7}{22}$

⇒ r² = 125.44

⇒ r = $\sqrt{125.44}$

⇒ r = 11.2 cm.

Hence, radius = 11.2 cm.

(ii) Circumference of circle = 2πr

= 2 × $\frac{22}{7}$ × 11.2

= 2 × 22 × 1.6

= 70.4 cm.

Hence, circumference = 70.4 cm.

Given,

Radius = 25 cm

Area of full circle = πr²

= 3.14 × (25)²

= 3.14 × 625

= 1962.5 cm².

Area of semicircle = $\frac{1}{2}$ × 1962.5

= 981.25 cm².

Arc length of semi-circle = $\frac{1}{2}$ × 2πr = πr

Perimeter = Circumference of semi-circle + diameter

= πr + 2r

= 3.14 × 25 + 2 × 25

= 78.5 + 50 = 128.5 cm.

Hence, area = 981.25 cm² and perimeter = 128.5 cm respectively.

Perimeter of semicircular plate = Circumference of semicircle + diameter

= πr + 2r

= r(π + 2).

Thus,

$$\begin{aligned}\Rightarrow 86.4 = r(π + 2) \\[1em] \Rightarrow r = \frac{86.4}{π + 2} \\[1em] \Rightarrow r = \frac{86.4}{3.14 + 2} \\[1em] \Rightarrow r = \frac{86.4}{5.14} \approx 16.8 \text{ cm}.\end{aligned}$$

Area of semi-circular plate = $\frac{1}{2}$ πr²

= $\frac{1}{2}$ × 3.14 × (16.8)²

= 443.12 cm².

Hence, radius of plate = 16.8 cm and area = 443.12 cm².

(i) Given,

Circumference of a circle exceeds diameter by 180 cm.

So, Circumference = Diameter + 180

Circumference - Diameter = 180

$$\begin{aligned}\Rightarrow 2\pi r - 2r = 180 \\[1em] \Rightarrow 2r(\pi - 1) = 180 \\[1em] \Rightarrow 2r\Big(\frac{22}{7} - 1\Big) = 180 \\[1em] \Rightarrow 2r\Big(\frac{22 - 7}{7}\Big) = 180 \\[1em] \Rightarrow 2r\Big(\frac{15}{7}\Big) = 180 \\[1em] \Rightarrow 30r = 180 × 7 \\[1em] \Rightarrow 30r = 1260 \\[1em] \Rightarrow r = \frac{1260}{30} \\[1em] \Rightarrow r = 42 \text{ cm}.\end{aligned}$$

Hence, radius = 42 cm.

(ii) As calculated,

Radius = 42 cm.

Circumference of circle = 2πr

= 2 × $\frac{22}{7}$ × 42

= 2 × 22 × 6

= 44 × 6 = 264 cm.

Hence, circumference of circle = 264 cm.

(iii) By formula,

Area of circle = πr²

= $\frac{22}{7}$ × (42)²

= $\frac{22}{7}$ × 1764

= 22 × 252

= 5544 cm².

Hence, area of circle = 5544 cm².

By formula,

Area of square = (side)²

Given,

Area of square = 272.25 cm².

∴ (side)² = 272.25

⇒ side = $\sqrt{272.25}$

⇒ side = 16.5 cm.

Perimeter of square = 4 × side = 4 × 16.5 = 66 cm.

Circumference of circle of same wire = Perimeter of square of same wire.

Let radius of circle be r cm.

$$\begin{aligned}\therefore 2πr = 66 \\[1em] \Rightarrow 2 \times \frac{22}{7} \times r = 66 \\[1em] \Rightarrow 44 \times r = 66 \times 7 \\[1em] \Rightarrow r = \frac{66 \times 7}{44} \\[1em] \Rightarrow r = \frac{21}{2} = 10.5 \text{ cm}.\end{aligned}$$

By formula,

Area of circle = πr²

= $\frac{22}{7}$ × (10.5)²

= $\frac{22}{7}$ × 10.5 × 10.5

= 22 × 1.5 × 10.5

= 346.5 cm².

Hence, area enclosed by the wire in the form of a circle = 346.5 cm².

By formula,

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

Given,

Area of equilateral triangle = $121\sqrt{3}$ cm²

$\therefore 121\sqrt{3} = \frac{\sqrt{3}}{4} \text{ (side)}^2$

⇒(side)² = $\frac{121 \sqrt{3} × 4}{\sqrt{3}}$

⇒ (side)² = 121 × 4

⇒ (side)² = 484

⇒ side = $\sqrt{484}$

⇒ side = 22 cm.

Perimeter of equilateral triangle = 3 x side = 3 x 22 = 66 cm.

Circumference of circle of same wire = Perimeter of triangle of same wire

Let radius of circle be r cm.

∴ 2πr = 66

⇒ 2 × $\frac{22}{7}$ × r = 66

⇒ 44 × r = 66 × 7

⇒ r = $\frac{66 \times 7}{44}$

⇒ r = $\frac{21}{2}$ = 10.5 cm

Area of circle = πr²

= $\frac{22}{7}$ × (10.5)²

= $\frac{22}{7}$ × 10.5 × 10.5

= 22 × 1.5 × 10.5 = 346.5 cm².

Hence, area = 346.5 cm².

Let r₁ and r₂ be the radii of the two circles.

Given,

Sum of the radii of two circles is 140 cm.

r₁ + r₂ = 140 .......(1)

Circumference of circle₁ - Circumference of circle₂ = 88

⇒ 2πr₁ - 2πr₂ = 88

⇒ 2π(r₁ - r₂) = 88

⇒ r₁ - r₂ = $\frac{88}{2π}$

⇒ r₁ - r₂ = $\frac{88}{2 \times \frac{22}{7}}$

⇒ r₁ - r₂ = $\frac{88 \times 7}{2 \times 22}$

⇒ r₁ - r₂ = 14 ..........(2)

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

⇒ r₁ + r₂ + r₁ - r₂ = 140 + 14

⇒ 2r₁ = 140 + 14

⇒ 2r₁ = 154

⇒ r₁ = $\frac{154}{2}$

= 77 cm.

Substituting value of r₁ in equation (1), we get :

⇒ 77 + r₂ = 140

⇒ r₂ = 140 - 77

⇒ r₂ = 63 cm.

Hence, radii of the circles = 77 cm and 63 cm.

Let r₁ and r₂ be the radii of the two circles.

Given,

Sum of the radii of two circles is 84 cm.

r₁ + r₂ = 84 .......(1)

Given,

Difference of their areas is 5544 cm².

Area of circle₁ - Area of circle₂ = 5544

⇒ πr₁² - πr₂² = 5544

⇒ $\frac{22}{7}$ (r₁² - r₂²) = 5544

⇒ r₁² - r₂² = $\frac{5544 × 7}{22}$

⇒ r₁² - r₂² = $\frac{38808}{22}$

⇒ r₁² - r₂² = 1764

⇒ (r₁ + r₂) (r₁ - r₂) = 1764

Substituting the value from equation (1) in above equation, we get :

⇒ 84(r₁ - r₂) = 1764

⇒ r₁ - r₂ = $\frac{1764}{84}$

⇒ r₁ - r₂ = 21 ......(2)

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

⇒ r₁ + r₂ + r₁ - r₂ = 84 + 21

⇒ 2r₁ = 84 + 21

⇒ 2r₁ = 105

⇒ r₁ = $\frac{105}{2}$ = 52.5 cm

Substituting value of r₁ in equation (1), we get :

⇒ r₁ + r₂ = 84

⇒ 52.5 + r₂ = 84

⇒ r₂ = 84 - 52.5

⇒ r₂ = 31.5 cm.

Hence radii of the two circles = 52.5 cm and 31.5 cm.

Let x and y be the radii of the of two circles with centers A and B respectively.

By formula,

Area of a circle = π (radius)²

Given,

Sum of areas = 117π cm²

πx² + πy² = 117π

x² + y² = 117 .......(1)

Given,

Distance between centers = 15 cm

x + y = 15

x = 15 - y .......(2)

Substituting value of x from equation (2) in (1), we get :

⇒ (15 - y)² + y² = 117

⇒ 225 - 30y + y² + y² = 117

⇒ 2y² - 30y + 225 - 117 = 0

⇒ 2y² - 30y + 108 = 0

⇒ 2(y² - 15y + 54) = 0

⇒ y² - 15y + 54 = 0

⇒ y² - 9y - 6y + 54 = 0

⇒ y(y - 9) - 6(y - 9) = 0

⇒ (y - 6)(y - 9) = 0

⇒ y - 6 = 0 or y - 9 = 0

⇒ y = 6 cm or y = 9 cm.

If y = 6 cm, then x = 15 - y = 9 cm.

If y = 9 cm, then x = 15 - y = 6 cm.

Since, x is the radius of circle with center A and from figure it is clear that A is the circle with greater radius thus, x = 9 cm and y = 6 cm.

Hence, the radii of two circles with centers A and B = 9 cm and 6 cm respectively.

Let R be the radius of bigger circle and r be the radius of smaller circle.

Given,

Distance between centres (R - r) = 4 cm

Sum of areas = 170π cm²

If two circles touch internally then,

Distance between centres = R - r

∴ R - r = 4.....(1)

Given,

Sum of their areas = 170π

⇒ πR² + πr² = 170π

⇒ π(R² + r²) = 170π

⇒ R² + r² = 170

Substituting the value R = r + 4

⇒ (r + 4)² + r² = 170

⇒ r² + 16 + 2 × r × 4 + r² = 170

⇒ 2r² + 8r + 16 - 170 = 0

⇒ 2r² + 8r - 154 = 0

⇒ r² + 4r - 77 = 0

⇒ r² + 11r - 7r - 77 = 0

⇒ r(r + 11) - 7(r + 11) = 0

⇒ (r - 7)(r + 11) = 0

⇒ r = 7 cm.

Substituting r = 7 in equation (1), we get :

R - 7 = 4

R = 4 + 7 = 11 cm.

Hence, radii of the two circles = 7 cm and 11 cm.

Given,

Outer radius (R) = 19 cm

Inner radius (r) = 16 cm

Area of ring = Area of big circle − Area of small circle

Area of ring = πR² - πr²

= π(R² - r²)

= π(19² - 16²)

= π(361 - 256)

= π(105)

= $\frac{22}{7} \times 105$

= 22 × 15

= 330 cm².

Hence, area of the ring = 330 cm².

Let radius of smaller circle be r cm and radius of bigger circle be R cm.

Given

Area of smaller circle = 962.5 cm²

Area of bigger circle = 1386 cm²

Width of ring = R − r

Area of bigger circle = πR²

$$\begin{aligned}\Rightarrow 1386 = \frac{22}{7} \times R^2 \\[1em] \Rightarrow R^2 = \frac{1386 × 7}{22} \\[1em] \Rightarrow R^2 = \frac{9702}{22} \\[1em] \Rightarrow R^2 = 441 \\[1em] \Rightarrow R = \sqrt{441} \\[1em] = 21 \text{ cm}.\end{aligned}$$

Calculating the area of smaller circle,

$$\begin{aligned}\text{Area of smaller circle} = πr^2 \\[1em] \Rightarrow 962.5 = \frac{22}{7} r^2 \\[1em] \Rightarrow r^2 = \frac{962.5 × 7}{22} \\[1em] \Rightarrow r^2 = \frac{6737.5}{22} \\[1em] \Rightarrow r^2 = 306.25 \\[1em] \Rightarrow r = \sqrt{306.25} \\[1em] = 17.5 \text{ cm}.\end{aligned}$$

Width of the ring = R - r

= 21 - 17.5 = 3.5 cm.

Hence, width of the ring = 3.5 cm.

Let radius of inner circle be r cm.

Given,

Area enclosed between two concentric circles is 770 cm²

Area enclosed between two concentric circles = Area of bigger circle - Area of smaller circle

⇒ 770 = π(21)² - πr²

⇒ 770 = 441π - πr²

⇒ 770 = $\frac{22}{7}$ (441 - r²)

⇒ 441 - r² = $\frac{770 × 7}{22}$

⇒ 441 - r² = 35 × 7

⇒ 441 - r² = 245

⇒ r² = 441 - 245

⇒ r² = 196

⇒ r = $\sqrt{196}$ = 14 cm.

Hence, radius of inner circle = 14 cm.

Let R be the radius of outer circle and r be the radius of inner circle.

(i) Given,

Circumference of outer circle = 242 cm.

$$\begin{aligned}\therefore 2πR = 242 \\[1em] \Rightarrow 2 \times \frac{22}{7} \times R = 242 \\[1em] \Rightarrow \frac{44}{7} \times R = 242 \\[1em] \Rightarrow R = \frac{242 × 7}{44} \\[1em] = 38.5 \text{ cm}.\end{aligned}$$

Area enclosed between two concentric circles = Area of bigger circle - Area of smaller circle

= πR² - πr²

= π(R² - r²)

Given,

Area enclosed between two concentric circles = 808.5 cm²

⇒ 808.5 = $\frac{22}{7}$ (R² - r²)

⇒ (38.5)² - r² = $\frac{808.5 × 7}{22}$

⇒ 1482.25 - r² = $\frac{5659.5}{22}$

⇒ 1482.25 - r² = 257.25

⇒ r² = 1482.25 - 257.25

⇒ r² = 1225

⇒ r = $\sqrt{1225}$ = 35 cm.

Hence, radius of inner circle = 35 cm.

(ii) Width of ring = R - r

= 38.5 - 35

= 3.5 cm.

Hence, width of the ring = 3.5 cm.

Given,

Shaded region = 308 cm²

Area of circle = πr²

Two perpendicular diameters divide the circle into 4 quadrants.

Area of each quadrant = $\frac{1}{4}$ Area of circle

Since, 2 quadrants are shaded. Thus,

$$\begin{aligned}\therefore \text{Shaded area} = 2 \times \frac{1}{4} \times \text{Area of circle} \\[1em] \Rightarrow 308 = \frac{1}{2} \times \text{Area of circle} \\[1em] \Rightarrow \text{Area of circle} = 616 \\[1em] \Rightarrow πr^2 = 616 \\[1em] \Rightarrow \frac{22}{7} r^2 = 616 \\[1em] \Rightarrow r^2 = \frac{616 × 7}{22} \\[1em] \Rightarrow r^2 = \frac{4312}{22} \\[1em] \Rightarrow r^2 = 196 \\[1em] \Rightarrow r = \sqrt{196} = 14 \text{ cm}.\end{aligned}$$

Diameter of circle = 2r = 28 cm.

(i) Length of AC = 28 cm.

Hence, length of AC = 28 cm.

(ii) Circumference of circle = 2πr

= 2 × $\frac{22}{7}$ × 14

= 2 × 22 × 2 = 88 cm.

Hence, circumference of circle = 88 cm.

Given,

AC = 16 cm (diameter)

Radius (r) = $\frac{16}{2}$ = 8 cm.

Two perpendicular diameters divide the circle into 4 quadrants.

Area of each quadrant = $\frac{1}{4}$ Area of circle

Since, 2 quadrants are shaded. Thus,

$$\begin{aligned}\therefore \text{Area of shaded part} = 2 \times \frac{4}{2} πr^2 \\[1em] = \frac{1}{2} \times 3.14 \times 8^2 \\[1em] = \frac{1}{2} \times 3.14 \times 64 \\[1em] = 3.14 \times 32 \\[1em] = 100.48 \text{ cm}^2.\end{aligned}$$

Full circumference = 2πr = 2 × 3.14 × 8 = 50.24 cm.

Only quarter arc is shaded.

AD = $\frac{1}{4}$ × 50.24 = 12.56.

Perimeter of the shaded part = 2(OA + OD + arc AD)

Perimeter = 2(8 + 8 + 12.56)

= 2(16 + 12.56)

= 2(28.56) = 57.12 cm.

Hence, area = 100.48 cm² and perimeter = 57.12 cm.

Given,

Side of an equilateral triangle = 15 cm.

For an equilateral,

Radius of circumcircle = $\frac{\text{Length of each side of triangle}}{\sqrt{3}}$

R = $\frac{15}{\sqrt{3}} = 5\sqrt{3}$ cm.

Area of circle = πR²

= $\pi(5\sqrt{3})^2$

= π × 25 × ($\sqrt{3}$)²

= π × 25 × 3

= 3.14 × 75 = 235.5 cm².

Hence, area of circle = 235.5 cm².

For an equilateral triangle,

Radius of incircle = $\frac{\text{Length of each side of triangle}}{2\sqrt{3}}$

$$\begin{aligned}= \frac{18}{2\sqrt{3}} \\[1em] = \frac{9}{\sqrt{3}} \\[1em] = 3\sqrt{3} \text{ cm}.\end{aligned}$$

Area of circle = πr²

= $3.14 \times (3\sqrt{3})^2$

= 3.14 × 27

= 84.78 cm².

Hence, area of circle = 84.78 cm².

Since,

OB = OD = 60 cm

Thus, radius of circle (r) = 60 cm

∠BOD = 90°.

(i) Area of circle = πr²

= 3.14 × (60)²

= 3.14 × 3600 = 11304 cm².

Area of sector of 90° = $\frac{90}{360}$ × πr²

= $\frac{1}{4}$ × 11304 = 2826 cm².

From figure,

Area of top of the table = Area of circle - Area of sector 90°

Area of top of the table = 11304 - 2826

= 8478 cm^2.

Hence, area of top of the table = 8478 cm².

(ii) Circumference of circle = 2πr

= 2 × 3.14 × 60 = 376.8 cm.

Length of 270° arc = $\frac{270}{360}$ × 376.8

= $\frac{3}{4}$ × 376.8

= 3 × 94.2 = 282.6 cm.

From figure,

Perimeter of table top = Circumference of major arc (of 270°) + OB + OD

= 282.6 + 60 + 60

= 282.6 + 120

= 402.6 cm.

Hence, perimeter of the table = 402.6 cm.

(i) Given,

Side of a square = 5 cm.

From figure,

Diagonal of square = Diameter of circle

Diagonal of square = $\sqrt{2}$ × Side of square

= $5\sqrt{2}$ cm

∴ Radius = $\frac{\text{Diameter}}{2} = \frac{5\sqrt{2}}{2}$ cm.

Hence, radius of circle = $\frac{5\sqrt{2}}{2}$ cm.

(ii) Area of shaded region = Area of circle - Area of square

$$\begin{aligned}= 3.14 \times \Big(\frac{5\sqrt{2}}{2}\Big)^2 - 5^2 \\[1em] = 3.14 \times \frac{50}{4} - 25 \\[1em] = 3.14 \times \frac{25}{2} - 25 \\[1em] = 39.25 - 25 \\[1em] = 14.25 \text{ cm}^2.\end{aligned}$$

Hence area of shaded region = 14.25 cm².

(i) Given,

Rectangle sides = 8 cm and 6 cm.

Let AB = 8 cm and BC = 6 cm

By pythagoras theorem,

AC² = AB² + BC²

AC² = 8² + 6²

AC² = 64 + 36

AC² = 100

AC = $\sqrt{100}$ = 10 cm.

Diameter of circle = 10 cm

Radius = $\frac{10}{2}$ = 5 cm.

Hence, radius of circle = 5 cm.

(ii) Shaded area = Area of circle - Area of rectangle

Area of circle = πr²

= 3.14 × 5²

= 3.14 × 25 = 78.5 cm².

Area of rectangle = 8 × 6 = 48 cm².

Shaded area = 78.5 - 48 = 30.5 cm².

Hence, shaded area = 30.5 cm².

Given,

Diameter of semicircle = 14 cm

radius = $\frac{14}{2}$ = 7 cm

Calculating the area of semi-circle EDF,

$$\begin{aligned}\text{Area of semi-circle EDF} = \frac{1}{2} \times πr^2 \\[1em] = \frac{1}{2} \times \frac{22}{7} \times 7^2 \\[1em] = \frac{11}{7} \times 49 \\[1em] = 11 \times 7 \\[1em] = 77 \text{ cm}^2.\end{aligned}$$

From figure,

AC = ED = 14 cm

Since, AB = BC

Thus, AB = BC = $\frac{AC}{2} = \frac{14}{2}$ = 7 cm.

From figure,

AE = AB = 7 cm.

Area of rectangle ACDE = AC × AE

= 14 × 7 = 98 cm².

ABE and BCD are two quadrants each with radius 7 cm.

Calculating the area of two quadrants,

$$\begin{aligned}\text{Area of two quadrants} = 2 \times \frac{1}{4}πr^2 \\[1em] = \frac{1}{2} \times \frac{22}{7} \times 7^2 \\[1em] = \frac{11}{7} \times 49 \\[1em] = 11 \times 7 \\[1em] = 77 \text{ cm}^2.\end{aligned}$$

Area of shaded region = Area of semi-circle EDF + Area of rectangle ACDE - Area of two quadrants

= 77 + 98 - 77 = 98 cm².

Hence, area of shaded region = 98 cm².

Given,

Radius of circle = 3 cm.

From figure,

Length of rectangle = Diameter of first circle + Diameter of second circle + Radius of third circle

= 6 + 6 + 3

= 15 cm.

Breadth of rectangle = Diameter of a circle = 6 cm.

Area of rectangle = length × bredath

= 15 × 6 = 90 cm².

Area of 2 full circles = 2 × πr²

= 2 × 3.14 × 3²

= 6.28 × 9

= 56.52 cm².

Area of 1 semicircle = $\frac{1}{2}$ × πr²

= $\frac{1}{2}$ × 3.14 × 9 = 14.13 cm².

From figure,

Area of unshaded portion = Area of rectangle - Area of 2 full circle - Area of semicircle

= 90 - 56.52 - 14.

= 90 - 70.65

= 19.35 cm².

Hence area of unshaded region = 19.35 cm².

Given,

Equilateral triangle side = 14 cm.

BC = 14 cm is diameter of semi-circle.

∴ Radius = $\frac{14}{2}$ = 7 cm.

Area of shaded region = Area of equilateral △ABC + Area of semi-circle BDC

Calculating the area of equilateral triangle ABC,

$$\begin{aligned}\text{Area of equilateral triangle ABC} = \frac{\sqrt{3}}{4} \times \text{(side)}^2 \\[1em] = \frac{\sqrt{3}}{4} \times 14^2 \\[1em] = \frac{\sqrt{3}}{4} \times 196 \\[1em] = 1.732 \times 49 \\[1em] = 84.868 \text{ cm}^2.\end{aligned}$$

Calculating the area of semi-circle BDC,

$$\begin{aligned}\text{Area of semi-circle BDC} = \frac{1}{2} πr^2 \\[1em] = \frac{1}{2} \times \frac{22}{7} \times 7^2 \\[1em] = \frac{1}{2} \times \frac{22}{7} \times 49 \\[1em] = 11 \times 7 \\[1em] = 77 \text{ cm}^2.\end{aligned}$$

Area of shaded region = Area of equilateral triangle ABC + Area of semi-circle BDC

= 84.868 + 77 = 161.868 cm².

Hence, area of shaded region = 161.868 cm².

In the figure,

AB and CD both are diameter of circle and they intersect at 90°.

OA = OB = OC = OD = radius of big circle (R) = 7 cm.

∠BOC = 90°

In triangle BOC,

By pythagoras theorem,

BC² = OB² + OC²

BC² = 7² + 7²

BC² = 49 + 49

BC² = 98

BC = $\sqrt{98} = 7\sqrt{2}$ cm.

In figure,

The smaller circle has diameter = OA = 7 cm, so radius (r) = $\frac{7}{2}$ cm.

Calculating,

$$\begin{aligned}\text{Area of smaller circle} = πr^2 \\[1em] = \frac{22}{7} \times \Big(\frac{7}{2}\Big)^2 \\[1em] = \frac{22}{7} \times \frac{49}{4} \\[1em] = \frac{11 × 7}{2} \\[1em] = \frac{77}{2} \\[1em] = 38.5 \text{ cm}^2.\end{aligned}$$

Calculating area of semi-circle CBD,

$$\begin{aligned}\text{Area of semi-circle CBD} = \frac{1}{2}πR^2 \\[1em] = \frac{1}{2} \times \frac{22}{7} \times 7^2 \\[1em] = \frac{11}{7} \times 49 \\[1em] = 11 \times 7 \\[1em] = 77 \text{ cm}^2.\end{aligned}$$

Calculating area of triangle DBC,

$$\begin{aligned}\text{Area of △DBC} = \frac{1}{2} \times CD \times OB \\[1em] = \frac{1}{2} \times 14 \times 7 \\[1em] = 49 \text{ cm}^2.\end{aligned}$$

From figure,

Area of shaded area = Area of smaller circle + Area of semicircle - Area of triangle

= 38.5 + 77 - 49

= 115.5 - 49 = 66.5 cm².

Hence, area of shaded region = 66.5 cm².

Given,

PS = 12 cm

PQ = QR = RS = $\frac{12}{3}$ = 4 cm, QS = 8 cm.

Perimeter of shaded region = Arc PTS + Arc PBQ + Arc QES

Radius of semi-circle PTS = $\frac{12}{2}$ = 6 cm.

Arc PTS = $\frac{1}{2}$ × 2π.radius

= πr

= 3.14 × 6 = 18.84 cm.

Radius of semi-circle PBQ = $\frac{4}{2}$ = 2 cm.

Arc PBQ = π.radius

= 3.14 × 2 = 6.28 cm.

Radius of semi-circle QES = $\frac{8}{2}$ = 4 cm.

Arc QES = π.radius

= 3.14 × 4 = 12.56 cm.

Perimeter of shaded region = Arc PTS + Arc PBQ + Arc QES

= 18.84 + 6.28 + 12.56

= 37.68 cm.

Area of shaded region = Area of big semicircle + Area of small semicircle on PQ - Area of semicircle on QS

Calculating area of big semicircle,

$$\begin{aligned}\text{Area of big semicircle} = \frac{1}{2}π.\text{(radius)}^2 \\[1em] = \frac{1}{2} \times 3.14 \times 6^2 \\[1em] = \frac{1}{2} \times 3.14 \times 36 \\[1em] = 3.14 × 18 \\[1em] = 56.52 \text{ cm}^2.\end{aligned}$$

Area of small semicircle on PQ = $\frac{1}{2}$ π.(radius)²

= $\frac{1}{2}$ × 3.14 × 2^2

= 3.14 × 2 = 6.28 cm².

Area of semicircle on QS = $\frac{1}{2}$ π.(radius)²

= $\frac{1}{2}$ × 3.14 × 4²

= 3.14 × 8 = 25.12 cm²

Area of shaded region = 56.52 + 6.28 - 25.12

= 37.68 cm².

Hence, area of shaded region = 37.68 cm² and perimeter = 37.68 cm.

Given,

AB ∥ DC

∠ABC = 90°

DC = 4.2 cm

BC = 4.2 cm

AE = 2 cm

Quarter circle BEFC is removed.

Since it is a quarter circle with centre at B:

So, radius = BC = 4.2 cm

From figure,

AB = AE + EB

AB = 2 + 4.2 = 6.2 cm.

Calculating the area of trapezium ABCD,

$$\begin{aligned}\text{Area of trapezium ABCD} = \frac{1}{2} \times \text{(sum of parallel sides)} \times \text{height} \\[1em] = \frac{1}{2} \times (6.2 + 4.2) \times 4.2 \\[1em] = \frac{1}{2} \times (10.4) \times 4.2 \\[1em] = 5.2 \times 4.2 \\[1em] = 21.84 \text{ cm}^2.\end{aligned}$$

Calculating the area of quarter circle,

$$\begin{aligned}\text{Area of quarter circle} = \frac{1}{4}πr^2 \\[1em] = \frac{1}{4} \times \frac{22}{7} \times (4.2)^2 \\[1em] = \frac{11}{2 × 7} \times 17.64 \\[1em] = 11 \times 1.26 \\[1em] = 13.86 \text{ cm}^2.\end{aligned}$$

From figure,

Area of remaining piece of cardboard = Area of trapezium - Area of quarter circle

= 21.84 - 13.86

= 7.98 cm².

Hence, area of remaining piece of cardboard = 7.98 cm².

4 quadrants = 1 full circle

Circumference of circle = 2πr

= 2 × 3.14 × 1

= 6.28 cm.

Length of each shaded straight sides = 4 - 1 - 1 = 2 cm.

Total length = 4 × 2 = 8 cm.

Central circle boundary :

radius = $\frac{2}{2}$ = 1 cm.

Circumference of circle = 2πr

= 2 × 3.14 × 1

= 6.28 cm.

Total perimeter = 6.28 + 8 + 6.28 = 20.56 cm.

Area of shaded region = Area of the square - (Area removed at the corners - Area of central circle)

Area of square = (side)² = 4² = 16 cm².

Area removed at the corners :

Area of 1 quadrant = $\frac{1}{4}$ πr²

= $\frac{1}{4}$ × 3.14 × 1² = 0.785

∴ For 4 quadrants = 4 × 0.785 = 3.14 cm².

Area of central circle = πr²

= 3.14 × 1² = 3.14 cm².

Area of shaded region = Area of square - Area of central circle - Area of 4 quadrants

= 4 × 4 - (3.14 + 3.14)

= 16 - 6.28 = 9.72 cm².

Hence, perimeter = 20.56 cm and area of shaded region = 9.72 cm².

Given,

AB = 12 cm

BC = 16 cm

AC is the diameter of the semicircle.

Angle in a semicircle is a right angle, thus, ABC = 90°

Using Pythagoras theorem in △ABC:

⇒ AC² = AB² + BC²

⇒ AC² = 12² + 16²

⇒ AC² = 144 + 256

⇒ AC² = 400

⇒ AC = $\sqrt{400}$ = 20 cm.

So, radius = $\frac{20}{2}$ = 10 cm.

Semicircle arc = πr = 3.14 × 10 = 31.4 cm

Perimeter = Semicircle arc + side AB + side BC

Perimeter = 31.4 + 12 + 16

= 59.4 cm

Area of shaded region = Area of semicircle - Area of triangle

Calculating the area of semicircle,

$$\begin{aligned}\text{Area of semicircle} = \frac{1}{2} πr^2 \\[1em] = \frac{1}{2} \times 3.14 \times 10^2 \\[1em] = \frac{1}{2} \times 3.14 \times 100 \\[1em] = 3.14 \times 50 \\[1em] = 157 \text{ cm}^2.\end{aligned}$$

Area of triangle = $\frac{1}{2}$ × AB × BC

= $\frac{1}{2}$ × 12 × 16

= 6 × 16 = 96 cm².

Area of shaded region = 157 - 96 = 61 cm².

Hence, perimeter of shaded region = 59.4 cm and area of shaded region = 61 cm².

Given,

ABCP is a quadrant of radius = 14 cm

AB = BC = 14 cm

AC is the diameter of semicircle.

Using Pythagoras theorem in △ABC,

⇒ AC² = AB² + BC²

⇒ AC² = 14² + 14²

⇒ AC² = 196 + 196

⇒ AC² = 392

⇒ AC = $\sqrt{392} = 14\sqrt{2}$ cm.

So, radius of semicircle = $\frac{14\sqrt{2}}{2} = 7\sqrt{2}$ cm.

Calculating area of semicircle :

$$\begin{aligned}\text{Area of semicircle} = \frac{1}{2}πr^2 \\[1em] = \frac{1}{2} \times \frac{22}{7} \times (7\sqrt{2})^2 \\[1em] = \frac{11}{7} \times 49 \times 2 \\[1em] = 11 \times 7 \times 2 \\[1em] = 154 \text{ cm}^2.\end{aligned}$$

Calculating area of quadrant :

$$\begin{aligned}\text{Area of quadrant} = \frac{1}{4}πr^2 \\[1em] = \frac{1}{4} \times \frac{22}{7} \times 14^2 \\[1em] = \frac{11}{14} \times 196 \\[1em] = 154 \text{ cm}^2.\end{aligned}$$

Calculating area of triangle :

Area of triangle ABC = $\frac{1}{2}$ × AB × BC

= $\frac{1}{2}$ × 14 × 14

= 7 × 14 = 98 cm².

Shaded area = (Area of semicircle AQC) - [(Area of quadrant APCB) - (Area of triangle ABC)]

Area of shaded region = 154 - [(154 - 98)]

= 154 - 56 = 98 cm².

Hence, area of shaded region = 98 cm².

Given,

ABCD is a square.

Side = 14 cm

Arcs are drawn from A, B, C, D as centres.

Radius of each arc = 7 cm

Thus, each arc is a quarter circle of radius 7 cm.

Area of square = (side)²

= 14² = 196 cm².

Area of four quarter circles:

$$\begin{aligned}\text{Area of one quarter circle} = \frac{1}{4}πr^2 \\[1em] = \frac{1}{4} \times π \times 7^2 \\[1em] = \frac{49π}{4} \text{ cm}^2.\end{aligned}$$

For 4 quarter circles, area = 4 × $\frac{49π}{4}$

= 49 × $\frac{22}{7}$

= 22 × 7 = 154

So, Area of four quarter circles = 154 cm².

Area of shaded region = Area of square - Area of four quarter circles

Area of shaded region = 196 - 154 = 42 cm².

Hence, area of shaded region = 42 cm².

Given,

ABCD is a square.

Side = 7 cm

Distance between centres of each circle = 7 cm

2r = 7

r = $\frac{7}{2}$ = 3.5 cm.

Area of 4 circles:

Area of one circle = πr²

= $\frac{22}{7}$ × (3.5)²

= $\frac{22}{7}$ × 12.25

= 22 × 1.75 = 38.5

∴ For 4 circles, area = 4 × 38.5 = 154 cm².

Area of square :

(side)² = 7² = 49 cm².

Area of 4 quadrants = Area of one circle = 38.5 cm².

Area of shaded region = Area of square + Area of 4 circle - Area of 4 quadrants

= 49 + 154 - 38.5

= 164.5 cm².

Hence, area of shaded region = 164.5 cm².

Given,

BC = 14 cm

AB = $\frac{1}{2}BC = \frac{1}{2}$ × 14 = 7 cm.

BC is a diameter of the semicircle

So, radius = $\frac{1}{2}$ × 14 = 7 cm.

(i) Area of quadrilateral AEFD:

Height (AE) = AB + BE = 7 + 7 = 14 cm.

Width : AD = BC = 14

Area of quad. AEFD = 14 × 14 = 196 cm².

Hence, area of quad. AEFD = 196 cm².

(ii) Area of △ABC:

Area = $\frac{1}{2}$ × base × height

= $\frac{1}{2}$ × BC × AB

= $\frac{1}{2}$ × 14 × 7

= 7 × 7 = 49 cm².

Hence area of △ABC = 49 cm².

(iii) Calculating,

$$\begin{aligned}\text{Area of semicircle} = \frac{1}{2}πr^2 \\[1em] = \frac{1}{2} \times \frac{22}{7} \times 7^2. \\[1em] = \frac{11}{7} \times 49 \\[1em] = 11 \times 7 \\[1em] = 77 \text{ cm}^2.\end{aligned}$$

Hence, area of semicircle = 77 cm².

Area of shaded region = Area of quad. AEFD - (Area of triangle ABC + Area of semicircle)

= 196 - (49 + 77)

= 196 - 126

= 70 cm².

Hence, area of shaded region = 70 cm².

Given,

Inside perimeter of the track = 312 m

Straight portions = 90 m each

Width of track = 2 m

Let inner radius be 'r' and outer radius be 'R'.

Inside perimeter consists of :

Two straights lengths = 90 + 90 = 180

Two semicircles = one full circle

So,

⇒ 180 + 2πr = 312

⇒ 2πr = 312 - 180

⇒ 2πr = 132

⇒ 2 × $\frac{22}{7}$ × r = 132

⇒ r = $\frac{132 \times 7}{2 \times 22}$

⇒ r = 21 m.

Outer radius (R) = r + 2

= 21 + 2 = 23 m.

Area of straight rectangular parts:

Area = 2 × (90 × 2)

= 360 m².

Area of curved part :

Area = π(R² - r²)

= π(23² - 21²)

= π(529 - 441)

= 88π

= 88 × $\frac{22}{7}$ = 276.57 m².

Total area of track = 360 + 276.57 = 636.57 m².

Hence, area of track = 636.57 m².

Given,

Diameter of wheel (d) = 1.26 m

Number of revolutions (n) = 500

Distance traveled in one revolution = Circumference of wheel

C = 2πr

= πd

= $\frac{22}{7}$ × 1.26

= 3.96 m.

∴ For 500 revolutions distance is :

500 × 3.96 = 1980 m

Hence, distance covered = 1980 m.

Given,

Circumference of wheel = $4\frac{2}{7} \text{ m} = \frac{30}{7}$ m.

Number of revolutions = 7

Time = 3 seconds

We know that,

Distance = circumference × revolutions

= $\frac{30}{7}$ × 7

= 30 m.

So the train moves 30 m in 3 seconds.

Speed = $\frac{Distance}{Time}$

= $\frac{30}{3}$ = 10 m/s.

1 m/s = 3.6 km/h

10 m/s = 10 × 3.6 km/h = 36 km/h

Hence, speed of the train = 36 km per hour.

Given,

Diameter of toothed larger wheel = 50 cm

Diameter of smaller wheel = 30 cm

Revolutions of larger wheel = 30

Distance covered by toothed larger wheel :

Distance in one revolution = circumference

C = 2πr = πd

C₁ = π × 50

Distance covered in 30 revolutions = 30 × 50π

= 1500π cm

Distance covered by smaller wheel:

C₂ = π × 30

Distance covered by smaller wheel in 1 revolution = 1 × 30π = 30π

Number of revolutions of the smaller vehicle

= $\frac{\text{Distance moved by toothed wheel in 30 revolutions}}{\text{Distance moved by smaller wheel in 1 revolution}}$

= $\frac{1500π}{30π}$

= 50.

Hence, number of revolutions = 50.

Given,

Distance covered = 88 km = 88000 m

Number of revolutions = 1000

We know that,

Distance in one revolution = Circumference of wheel

C = 2πr

Circumference = $\frac{\text{Total distance}}{\text{Number of revolutions}}$

= $\frac{88000}{1000}$ = 88

So, 2πr = 88

⇒ 2 × $\frac{22}{7}$ × r = 88

⇒ r = $\frac{88 × 7}{44}$

⇒ r = 2 × 7 = 14 m.

Hence, radius of the wheel = 14 m.

Area of circle = πr².

Given,

Ratio of areas of two circles = 25 : 4

πR₁² : πR₂² = 25 : 4

R₁² : R₂² = 25 : 4

Taking square root on both terms

$\sqrt{R_1^2} : \sqrt{R_2^2} = \sqrt{25} : \sqrt{4}$

R₁ : R₂ = 5 : 2

Let R₁ = 5a and R₂ = 2a.

Diameter = 2 × Radius

D₁ = 2 × R₁ = 10a

D₂ = 2 × R₂ = 4a

D₁ : D₂ = 10a : 4a = 5 : 2.

Hence, option 1 is the correct option.

Given,

Diameter of circle = side of square = s

Area of square = s²

Radius = $\frac{\text{Diameter}}{2} = \frac{s}{2}$

Area of circle = πr²

= $\pi \Big(\frac{s}{2}\Big)^2$

= $\frac{πs^2}{4}$

Area of square : Area of circle

= s² : $\frac{πs^2}{4}$

= $\frac{s^2}{\frac{πs^2}{4}}$

= $\frac{4}{π}$

= 4 : π.

Hence, option 3 is the correct option.

Circumference of semicircle = πr = 5π.

$$\begin{aligned}\text{Area of semicircle} = \frac{1}{2}πr^2 \\[1em] = \frac{1}{2} \times π \times 5^2 \\[1em] = \frac{25π}{2}\end{aligned}$$

Required ratio = $\frac{5π}{\frac{25π}{2}}$

= $\frac{10π}{25π}$

= $\frac{2}{5}$

= 2 : 5.

Hence, option 2 is the correct option.

Given,

Circumference of circle = 2πr

Perimeter of square = 4a

Given,

Perimeter of square = Circumference of circle

⇒ 4a = 2πr

⇒ 2a = πr

⇒ a = $\frac{πr}{2}$

Area of circle = πr²

Area of square = a²

= $\Big(\frac{πr}{2}\Big)^2$

= $\frac{π^2r^2}{4}$

⇒ Area of circle : Area of square

= πr² : $\frac{π^2r^2}{4}$

= $\frac{πr^2}{\frac{π^2r^2}{4}}$

= $\frac{4πr^2}{π^2r^2}$

= $\frac{4}{π}$

= 4 : π.

Hence, option 2 is the correct option.

ABCD is a square with diagonal 'd' units and side 'p' units.

From figure,

Diameter of the circle = diagonal of the square.

Side of square = p units.

$$\begin{aligned}\text{Diagonal of square (d)} = \sqrt{p^2 + p^2} \\[1em] = \sqrt{2p^2} \\[1em] = p\sqrt{2}\end{aligned}$$

Radius of circle = $\frac{d}{2} = \frac{p\sqrt{2}}{2}$ units.

Calculating,

$$\begin{aligned}\text{Area of circle} = πr^2 \\[1em] = π\Big(\frac{p\sqrt{2}}{2}\Big)^2 \\[1em] = π \times \frac{p^2 × 2}{4} \\[1em] = \frac{πp^2}{2} \text{ sq units}.\end{aligned}$$

Hence, option 3 is the correct option.

Let A₁ and A₂ be the areas of two circles.

Area = πr²

Areas of the two circles:

⇒ A₁ = π.(7)² = 49π.

⇒ A₂ = π.(24)² = 576π

∴ Sum of the areas of the two circles = 49π + 576π = 625π.

Let the radius of new circle be 'R' cm.

⇒ πR² = 625π

⇒ R² = 625

⇒ R = $\sqrt{625}$ = 25 cm.

Hence, option 4 is the correct option.

Let A₁ and A₂ be the areas of two circles.

Area = πr²

Areas of the two circles:

⇒ A₁ = π.(9)² = 81π.

⇒ A₂ = π.(12)² = 144π

∴ Sum of the areas of the two circles = 81π + 144π = 225π.

Let the radius of new circle be R cm.

⇒ πR² = 225π

⇒ R² = 225

⇒ R = $\sqrt{225}$ = 15 cm.

Hence, option 4 is the correct option.

Given,

Area of circle = 55.44 m².

By formula,

$$\begin{aligned}\text{Area of circle} = πr^2 \\[1em] \Rightarrow 55.44 = πr^2 \\[1em] \Rightarrow \frac{22}{7} \times r^2 = 55.44 \\[1em] \Rightarrow r^2 = \frac{55.44 × 7}{22} \\[1em] \Rightarrow r^2 = 17.64 \\[1em] \Rightarrow r = \sqrt{17.64} \\[1em] \Rightarrow r = 4.2 \text{ m}.\end{aligned}$$

Wire length = Circumference of the circle

= 2πr

= 2 × $\frac{22}{7}$ × 4.2

= 26.4 m.

Hence, option 4 is the correct option.

Given,

The sum of the lengths of a semicircular bow and its string is 360 cm.

∴ Arc length + Diameter = 360

$$\begin{aligned}\Rightarrow πr + 2r = 360 \\[1em] \Rightarrow \frac{22}{7} \times r + 2r = 360 \\[1em] \Rightarrow \frac{22r + 14r}{7} = 360 \\[1em] \Rightarrow \frac{36r}{7} = 360 \\[1em] \Rightarrow r = 70 \text{ cm}.\end{aligned}$$

Length of the bow = πr

= $\frac{22}{7}$ × 70

= 220 cm.

Hence, option 4 is the correct option.

Given,

Each side of a square formed by a wire is 14 cm.

Total length of wire = Perimeter of square = 4 × 14 = 56 cm.

The circle formed with the wire will be having the circumference = 56 cm.

Let radius of circle be r cm.

2πr = 56

r = $\frac{56}{2π} = \frac{28}{π}$

Area of circle = πr²

$$\begin{aligned}= \pi \times \left(\frac{28}{π}\right)^2 \\[1em] = \pi \times \frac{28}{π} \times \frac{28}{π} \\[1em] = \frac{28 \times 28}{π} \\[1em] = \frac{28 \times 28}{\frac{22}{7}} \\[1em] = \frac{28 \times 28 \times 7}{22} \\[1em] = 249.45 \text{ cm}^2\end{aligned}$$

Hence, option 4 is the correct option.

Given,

Circumference of circle - Diameter of circle = 30

$$\begin{aligned}\Rightarrow 2πr - 2r = 30 \\[1em] \Rightarrow 2r(π - 1) = 30 \\[1em] \Rightarrow 2r\Big(\frac{22}{7} - 1\Big) = 30 \\[1em] \Rightarrow 2r(\frac{22 - 7}{7}) = 30 \\[1em] \Rightarrow 2r \times \frac{15}{7} = 30 \\[1em] \Rightarrow r = \frac{30 \times 7}{2 \times 15} \text{ cm}.\end{aligned}$$

Circumference of circle = 2πr

= 2 × $\frac{22}{7}$ × 7

= 2 × 22

= 44 cm.

Hence, option 1 is the correct option.

ABCD is a square with inscribed circle having radius 'r' and center O.

Given,

Area of circle = 154 cm².

We know that,

$$\begin{aligned}\text{Area of circle} = πr^2 \\[1em] \Rightarrow πr^2 = 154 \\[1em] \Rightarrow \frac{22}{7} \times r^2 = 154 \\[1em] \Rightarrow r^2 = \frac{154 × 7}{22} \\[1em] \Rightarrow r^2 = 49 \\[1em] \Rightarrow r = \sqrt{49} = 7 \text{ cm}.\end{aligned}$$

For a circle inscribed in a square,

Diameter of circle = Side of square

Side = 2r = 2 × 7 = 14 cm.

Area of square = (side)²

= (14)²

= 196 cm².

Hence, option 3 is the correct option.

Given,

Rate of mowing = ₹ 1.20 per square metre.

Total cost = ₹4,620

Area = $\frac{\text{Total cost}}{\text{Rate}}$

= $\frac{4620}{1.20}$ = 3850 m².

Let radius of circular field be r meters.

$$\begin{aligned}\Rightarrow πr^2 = 3850 \\[1em] \Rightarrow \frac{22}{7} \times r^2 = 3850 \\[1em] \Rightarrow r^2 = \frac{3850 \times 7}{22} \\[1em] \Rightarrow r^2 = 1225 \\[1em] \Rightarrow r = \sqrt{1225} \\[1em] \Rightarrow r = 35 \text{ m}.\end{aligned}$$

Circumference of circle = 2πr

= 2 × $\frac{22}{7}$ × 35

= 220 m.

Cost of fencing = Circumference of field × Rate of fencing

= 220 × 4 = ₹ 880.

Hence, option 2 is the correct option.

Outer circumference (C₁) = 328 m

Inner circumference (C₂) = 200 m

Circumference = 2π.radius

Let outre radius be R meters and inner radius be r meters.

Calculating outer circumference,

$$\begin{aligned}\Rightarrow \text{Outer circumference} = 2 \times \frac{22}{7} \times R \\[1em] \Rightarrow 328 = 2 \times \frac{22}{7} \times R \\[1em] \Rightarrow R = \frac{328 \times 7}{44} = \frac{574}{11} \text{ m}\end{aligned}$$

Calculating inner circumference,

$$\begin{aligned}\Rightarrow \text{Inner circumference} = 2 \times \frac{22}{7} \times r \\[1em] \Rightarrow 200 = 2 \times \frac{22}{7} \times r \\[1em] \Rightarrow r = \frac{200 \times 7}{44} \\[1em] \Rightarrow r = \frac{350}{11} \text{ m}.\end{aligned}$$

Area = π(R² - r²)

= π(R + r)(R - r)

$$\begin{aligned}= \frac{22}{7} \times \Big(\frac{574}{11} + \frac{350}{11}\Big) \times \Big(\frac{574}{11} - \frac{350}{11}\Big) \\[1em] = \frac{22}{7} \times \frac{924}{11} \times \frac{224}{11} \\[1em] = 2 \times 84 \times 32 \\[1em] = 5376 \text{ m}^2.\end{aligned}$$

Hence, option 1 is the correct option.

Circumference of circle = 2πr = πd

Front wheel circumference :

C₁ = π × 70

Rear wheel circumference :

C₂ = π × 168

Distance covered by front wheel in 600 revolutions = 600 × π × 70

Let rear wheel revolutions = x

Since both cover the same distance

∴ x × π × 168 = 600 × π × 70

x = $\frac{600 × 70}{168}$

x = $\frac{42000}{168}$ = 250.

Hence, option 2 is the correct option.

1. Area of plot left for parking is : ADEF

Area of rectangle ADEF = length × breadth

= 3 × 21 = 63 m².

Hence, option (d) is the correct option.

2. The vertical line FE = 21 m.

Two semicircles are placed one above the other and are congruent.

Diameter of each = $\frac{21}{2}$ = 10.5 m.

Hence, option (b) is the correct option.

3. Two semi-circles = one full circle

Diameter = 10.5

Radius = $\frac{10.5}{2}$ = 5.25 m

Area of circle = πr²

= $\frac{22}{7}$ × (5.25)²

= $\frac{22}{7}$ × 27.5625

= 86.625 m².

Hence, option (c) is the correct option.

4. Since,

BF = FE = EC = BC = 21 m.

Thus, BCEF is a square.

Area of square BCEF = (side)²

= (21)²

= 441 m².

Hence, option (a) is the correct option.

5. Two semicircles = one circle

∴ Total length of two semicircular arcs = Circumference of one circle = 2πr = πd.

= $\frac{22}{7}$ × 10.5

= 22 × 1.5

= 33 m.

Hence, option (a) is the correct option.

1. OA = OD + AD = 7 + 3 = 10 cm

OB = OC + CB = 7 + 3 = 10 cm

Area of right triangle △AOB = $\frac{1}{2} × OA × OB$

= $\frac{1}{2}$ × 10 × 10

= 50 cm².

Hence, option (a) is the correct option.

2. Arc CD subtends 90° at the centre.

∴ Arc length CD = $\frac{90°}{360°}$ × 2πr

= $\frac{1}{4} × 2 × \frac{22}{7}$ × 7

= 11 cm.

Hence, option (d) is the correct option.

3. Calculating,

$$\begin{aligned}\Rightarrow \text{Area of quadrant OCD} = \frac{90°}{360°} \times πr^2 \\[1em] \Rightarrow \frac{1}{4} \times \frac{22}{7} \times 7^2 \\[1em] \Rightarrow \frac{11 × 7}{2} \\[1em] \Rightarrow 38.5 \text{ cm}^2.\end{aligned}$$

Hence, option (c) is the correct option.

4. Major sector angle:

360° - 90° = 270°

$$\begin{aligned}\Rightarrow \text{Area of major sector} = \frac{270}{360} \times πr^2 \\[1em] = \frac{3}{4} \times \frac{22}{7} \times 7^2 \\[1em] = \frac{3}{4} \times 22 \times 7 \\[1em] = \frac{3 × 154}{4} \\[1em] = 115.5 \text{ cm}^2.\end{aligned}$$

Hence, option (d) is the correct option.

5. Area of plated region = Area of triangle AOB - Area of quadrant OCD

= 50 - 38.5

= 11.5 cm²

Total cost of silver plating = ₹50 × 11.5 = ₹ 575.

Hence, option (a) is the correct option.

Total surface area of semicircle formula = $\frac{1}{2}πr^2$

Considering radius = 14 cm

Total surface area of semicircle = $\frac{1}{2} × \frac{22}{7}$ × 14²

= 308 cm².

But the given assertion says 18.48 cm².

∴ Both Assertion and Reason are false.

Hence, option 4 is the correct option.

Diameter of wheel = 21 cm

Distance covered in 1 revolution = Circumference of the wheel = 2πr

∴ Reason (R) is true.

We know that,

Circumference of wheel = 2πr = πd

= $\frac{22}{7}$ × 21

= 66 cm.

Distance covered in 2000 revolutions

= 2000 × 66 = 132000 cm = $\frac{132000}{100000}$ = 1.32 km

∴ Assertion (A) is true.

Both Assertion and Reason are true.

Hence, option 3 is the correct option.

Given,

Diameter of wheel (d) = $\frac{7}{11}$ m

Distance = 4 km = 4000 m.

Distance covered in 1 round = Circumference

= 2πr

= πd

= $\frac{22}{7} × \frac{7}{11}$

= 2 m.

Number of rounds = $\frac{\text{Total distance}}{\text{Distance covered in one round}}$

= $\frac{4000}{2}$

= 2000.

Hence, option 4 is the correct option.

Given,

Radius = 7 cm

From figure,

Distance between centres of two touching circles = 2r = 2 × 7 = 14 cm.

So side of the square : AB = BC = CD = DA = 14 cm.

Area of square = (side)²

= (14)² = 196 cm².

At each corner of the square there is a quarter circle of radius 7 cm.

Four quarter circles together make one full circle.

Area of one full circle = πr²

= $\frac{22}{7}$ × 7²

= 22 × 7 = 154 cm².

Area of enclosed space = Area of square - Area of four quarter circles

= 196 - 154

= 42 cm².

Hence, option 1 is the correct option.

There are 9 equal circles each with radius = 5 cm.The shaded region is the space between the four middle touching circles.

The centres of a the four middle circles form a square.

Diameter = 2r = 2 × 5 = 10 cm

Side of a square = 10 cm.

Area of square = (side)²

= 10² = 100 cm².

Inside the square there are 4 quarter circles, which together to form one full circle of radius 5 cm.

∴ Area of circle = πr²

= π × 5² = 25π.

Area of one shaded region = 100 - 25π.

∴ For 4 identical shaded region

Total area = 4(100 - 25π)

= 400 - 100π cm².

Hence, option 3 is the correct option.

Given,

Sum of the areas of two circles with radii r₁ and r₂ is equal to the area of circle of radius r.

π(r₁)² + π(r₂)² = πr²

r₁² + r₂² = r²

Hence, option 4 is the correct option.

Given,

Area of sector = 770 cm²

Central angle = 200°

We know that,

$$\begin{aligned}\Rightarrow \text{Area of sector} = \frac{\theta}{360°} \times πr^2 \\[1em] \Rightarrow 770 = \frac{200°}{360°} \times \frac{22}{7} \times r^2 \\[1em] \Rightarrow 35 = \frac{5}{9} \times \frac{1}{7} \times r^2 \\[1em] \Rightarrow r^2 = 7 \times 9 \times 7 \\[1em] \Rightarrow r^2 = 441 \\[1em] \Rightarrow r = \sqrt{441} \\[1em] \Rightarrow r = 21 \text{ cm}.\end{aligned}$$

Calculating the arc length :

$$\begin{aligned}\Rightarrow \text{Arc length} = \frac{\theta}{360°} \times 2πr \\[1em] = \frac{200°}{360°} \times 2 \times \frac{22}{7} \times 21 \\[1em] = \frac{5}{9} \times 132 \\[1em] = \frac{660}{9} \\[1em] = 73\frac{1}{3} \text{ cm}.\end{aligned}$$

Hence, option 2 is the correct option.

We know that,

Arc length = $\frac{\theta}{360}$ × 2πr

Let L₁ and L₂ be the arc length of first circle and second circle and θ₁ and θ₂ be the central angle of the sector for the first circle and second circle respectively.

Arc length for first circle (L₁) = $\frac{\theta_1}{360°}$ × 2πr

Arc length for second circle (L₂) = $\frac{\theta_2}{360}°$ × 2π(2r)

L₂ = $\frac{\theta_2}{360°}$ × 4πr

According to question :

L₁ = L₂

$\frac{\theta_1}{360°} × 2πr = \frac{\theta_2}{360°}$ × 4πr

2θ₁ = 4θ₂

θ₁ = 2θ₂

∴ Angle of the corresponding sector of the first circle is double the angle of the corresponding sector of the other circle.

Hence, the statement is True.

Given,

Side of square = 5 cm.

Let A₁ and A₂ be the areas of the inner and outer circle respectively.

Inner circle (inscribed in the square):

Diameter of the circle = side of the square

Diameter = 5 cm

So, radius = $\frac{5}{2}$ = 2.5 cm.

Area of inner circle (A₁) = πr²

= π(2.5)²

= 6.25π cm²

Outer circle (circumscribing the square)

Diameter of the outer circle = Diagonal of the square

Diameter of the outer circle = $a\sqrt{2} = 5\sqrt{2}$ cm.

Radius of the outer circle = $\frac{5\sqrt{2}}{2}$

Area of outer circle (A₂) = πr²

= $π\Big(\frac{5\sqrt{2}}{2}\Big)^2$

= π × $\frac{25 × 2}{4}$

= 12.5π cm².

Ratio of areas :

Outer area : Inner area

A₂ : A₁

12.5π : 6.25π

2 : 1.

Hence, ratio = 2 : 1.