Chapter 11: Areas Related to Circles

Radius of first circle = r₁ = 15 cm

Radius of second circle = r₂ = 18 cm

∴ Circumference of first circle = 2πr₁ = 30π cm

Circumference of second circle = 2πr₂ = 36π cm

Let the radius of the circle = R

According to the question,

Circumference of circle = Circumference of first circle + Circumference of second circle

2πR= 2πr₁+ 2πr₂

⇒ 2πR = 30π + 36π

⇒ 66π ⇒ R = 33

⇒ Radius = 33 cm

Hence, required radius of a circle is 33 cm.

Let a be the side of square.

∴ Diameter of a circle = Diagonal of the square = 8 cm

In right angled triangle ABC,

Using Pythagoras theorem,

(AC)² = (AB)² + (BC)²

∴ (8)²= a² +a²

⇒ 64= 2a²

⇒ a²= 32

Hence,

area of square = a²= 32 cm²

∴ Radius of the circle = Diameter/2 = 4 cm

∴ Area of the circle = πr² = π(4)² = 16 cm²

Therefore, the area of the shaded region = Area of circle – Area of square

The area of the shaded region = 16π – 32

= 16 × (22/7) – 32

= 128/7

= 18.286 cm²

Area of a sector of a circle = (1/2)r²θ,

(Here r = radius and θ = angle in radians subtended by the arc at the center of the circle)

Here, Radius of circle = 28 cm

Angle subtended at the center = 45°

Angle subtended at the center (in radians) = θ 45π/180 = π/4

∴ Area of a sector of a circle = ½ r²θ

= ½ × (28)² × (π/4)

= 28 × 28 × (22/8×7)

= 308 cm²

Hence, the required area of a sector of a circle is 308 cm^2.

Radius of wheel = r = 35 cm

1 revolution of the wheel = Circumference of the wheel

= 2πr

= 2 × (22/7) × 35

= 220 cm

But, given that,

Speed of the wheel = 66 km/hr

= (66×1000×100)/60 cm/min

= 110000 cm/min

∴ Number of revolutions in 1 min = 110000/220 = 500

Hence, the required number of revolutions per minute is 500.

Let ABCD be a rectangular field.

Length of field = 20 m

Breadth of the field = 16 m

According to the question,

A cow is tied at a point A.

Let the length of the rope be AE = 14 m = l.

The angle subtended at the centre of the sector = 90°

The angle subtended at the centre (in radians) θ = 90π/180 = π/2

∴ Area of a sector of a circle = ½ r²θ

= ½ × (14)² × (π/2)

= 154 m²

Hence, the required area of a sector of a circle is 154 m^2.

According to the given figure,

The length and breadth of the rectangular portion AFDC of the flower bed are 38 cm and 10 cm, respectively.

We know that,

Area of the flower bed = Area of the rectangular portion + Area of the two semi-circles.

∴ Area of rectangle AFDC = Length × Breadth

= 38 × 10 = 380 cm²

Both ends of the flower bed are semi-circle in shape.

∴ Diameter of the semi-circle = Breadth of the rectangle AFDC = 10 cm

∴ Radius of the semi-circle = 10/2 = 5 cm

Area of the semi-circle = πr²/2 = 25π/2 cm²

Since there are two semi-circles in the flower bed,

∴ Area of two semi-circles = 2 × (πr²/2) = 25π cm²

Total area of flower bed = (380 + 25π) cm²

According to the question,

AC = 6cm and BC = 8 cm

A triangle in a semi-circle with a hypotenuse as the diameter is a right-angled triangle.

Using the Pythagoras theorem in right-angled triangle ACB,

(AB)² = (AC)² + (CB)²

(AB)² = (6)² + (8)²

⇒(AB)² = 36 + 64

⇒(AB)² = 100 ⇒(AB)= 10

∴ Diameter of the circle = 10 cm

Thus, the radius of the circle = 5 cm

Area of circle = πr²

= π(5)²

= 25π cm²

= 25 × 3.14 cm²

= 78.5 cm²

We know that,

Area of the right-angled triangle = ( ½ ) × Base × Height

= (½) × AC × CB

= (½) × 6 × 8

= 24 cm²

Now, the area of the shaded region = Area of the circle – Area of the triangle

= (78.5-24)cm²

= 54.5cm²