Chapter 12: Areas Related to Circles

Key Formulas:
• Area of circle = πr²
• Perimeter (circumference) = 2πr
• Area of sector = θ/360 × πr² (or ½r²θ in radians)
• Length of arc = θ/360 × 2πr
• Area of segment = Area of sector - Area of corresponding triangle

Exercise 12.1

Q1. The radii of two circles are 19cm and 9cm. Find radius of circle having circumference equal to sum of circumferences of two circles.

Solution

2π(19)+2π(9)=2πr
19+9=r
r=28 cm

Q2. Radii are 8cm and 6cm. Find radius of circle with area equal to sum of areas.

Solution

π(8²)+π(6²)=πr²
64+36=r²
r²=100
r=10 cm

Q3. Diameter of goldsmith's circular sheet is 3cm. How many sheets of diameter 3mm are made from it?

Solution

π(30²) = n × π(0.3²)
900 = n × 0.09
n = 900/0.09 = 10000 sheets

Q4. Find π from: √(5/7)(r+21)=r-5

Solution

√(5/7)(r+21)=r-5
5(r+21)/7=(r-5)²
5r+105=7(r²-10r+25)
5r+105=7r²-70r+175
7r²-75r+70=0
r=(75±√(5625-1960))/14
r=(75±√3665)/14
Since r must be positive and valid:
r≈ 4.5 cm (taking positive root that satisfies equation)

Exercise 12.2

Q1. Find area of sector with radius 14cm and angle 90°.

Solution

Area = θ/360 × πr²
= 90/360 × π × 14²
= 1/4 × 196π
= 49π cm²

Q2. Find area of sector with radius 21cm and angle 120°.

Solution

Area = 120/360 × π × 21²
= 1/3 × 441π
= 147π cm²

Q3. Find area of quadrant of circle with circumference 22cm.

Solution

2πr=22 → r=22/2π=7/2 cm
Area of quadrant = 1/4 × π × (7/2)²
= π/4 × 49/4
= 49π/16 cm²

Q4. Find area of sector with radius 15cm and angle 60°.

Solution

Area = 60/360 × π × 15²
= 1/6 × 225π
= 75π/2 cm²

Exercise 12.3

Q1. Find area of shaded region where PQ=24cm, PR=7cm, O is centre of circle.

Solution

In ΔPQR: ∠R=90° (angle in semicircle)
RQ = √(24²-7²) = √(576-49) = √527...
Actually PQ is diameter=24, so radius=12cm
PR=7cm, RQ=√(24²-7²)=√527
Area of ΔPQR = 1/2 × 7 × √527...

Let me recalculate: PQ=24 (diameter), so radius=12
Area of semicircle = π × 12²/2 = 72π
Area of ΔPQR = 1/2 × PR × RQ
RQ=√(24²-7²)=√527
Area Δ = 1/2 × 7 × √527
Shaded area = 72π - 1/2 × 7 × √527

Q2. Find area of shaded region where AQ=7cm, ∠PAQ=60°, P and Q are on circle with centre O, radius 7cm.

Solution

ΔPAQ is equilateral (PA=AQ=7cm, ∠A=60°)
Area of sector PAQ = 60/360 × π × 7²
= 1/6 × 49π = 49π/6
Area of ΔPAQ = √3/4 × 7² = 49√3/4
Shaded area = 49π/6 - 49√3/4

Q3. Find area of shaded region where AB=28cm, ∠AOB=90° (quadrant).

Solution

r = 28/2 = 14cm
Area of quadrant = 1/4 × π × 14²
= 1/4 × 196π = 49π cm²

Q4. Find area of shaded region in square of side 14cm (quadrants at corners with radius 7cm, circle in middle).

Solution

Area of square = 14² = 196 cm²
Area of 4 quadrants = 4 × 1/4 × π × 7² = 49π
Area of inner circle = π × 7² = 49π (if circle inscribed)
Shaded = 196 - 49π (area between square and 4 quadrants)

Q5. Find area of shaded region: quadrant OAQB with ∠PAQ=60°, radius 7cm.

Solution

Area of quadrant = 1/4 × π × 7² = 49π/4
Area of ΔPAQ = 1/2 × 7 × 7 × sin 60°
= 1/2 × 49 × √3/2 = 49√3/4
Shaded area = 49π/4 - 49√3/4
= 49/4(π-√3) cm²

Q7. Find area of shaded region: circle radius 6cm, ∠AQB=60°.

Solution

Area of minor segment = Area of sector - Area of Δ
Area of sector = 60/360 × π × 6²
= 1/6 × 36π = 6π
Area of Δ = √3/4 × 6² = 9√3
Area of segment = 6π - 9√3
Shaded area = π × 6² - (6π-9√3) = 36π-6π+9√3
= 30π+9√3 cm²

Q8. Find area of shaded region: ∠AOD=60° (O is centre, radius 10cm).

Solution

Area of sector = 60/360 × π × 10²
= 1/6 × 100π = 50π/3
Area of ΔAOD = 1/2 × 10 × 10 × sin 60²
= 50 × √3/2 = 25√3
Area of segment = 50π/3 - 25√3

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