Question

A chord of a circle of radius 10 cm subtends a right angle at the centre. Find the area of the corresponding (i) minor segment (ii) major sector.

Answer 1

Given Data :

1. θ = 90°
2. radius (r) = 10 cm

1. Minor Segment :

θ/360° × 2πr = 90/360 × 2 × 3.14 × 10 = ¼×2×10×3.14 = 5×3.14 = 15.7 cm

2. Major Sector :

θ/360 × πr² = ¼ × 3.14 × 100 = 3.14 × 50 = 157 cm²

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Answer 2

Radius of circle = 10 cm

Angle subtended by chord at the centre = 90°

We know,

$\rm{Area \: of \: segment = }\tt{πr² \frac{θ}{360} - \frac{ {r}^{2} }{2} \sin \theta }$

$\rm{ Area \: of \: \: sector }= \tt{πr² \frac{θ}{360} }$

Here,

$\rm{Area \: of \: minor \: segment = } \pi {r}^{2} \times \frac{90}{360} - \frac{ {(10)}^{2} }{2} \sin90 \degree \\ \\ \to \tt\pi {(10)}^{2} \times \frac{1}{4} - \frac{100}{2} \times 1 \\ \\ \to \tt\pi \times 25 - 50 \\ \to \tt \green{25\pi - 50} \: \: {cm}^{2}$

And,

$\rm{ Area \: of \: minor\: sector }= \tt{πr² \times \frac{90}{360} } \\ \\ \to \tt \pi \times {(10)}^{2} \times \frac{1}{4} \\ \\ \to \tt \pi \times 100 \times \frac{1}{4} \\ \\ \to \tt \green{25\pi} \: \: {cm}^{2}$

So,

Area of major sector = πr² – Area of minor sector

→ πr² – 25π

→ π(r² - 25)

→ π(10² - 25)

→ π(100 - 25)

→ 75π cm²

Hence,

• Area of minor segment = 25(π – 2) cm²
• Area of major sector = 75π cm²