Question

4. A chord of a circle of radius 10 cm subtends a right angle at the centre. Find the
the corresponding: (i) minor segment (ii) major sector. (Use pie = 3.14)
​

Answer 1

area of minor segment= area of minor sector -area of triangle

Answer 2

Answer:

Area of minor sector

$\alpha \div360 \times \pi \: {r}^{2}$

90÷360×3.14×10×10

1÷4×314

78.5cm^2

Area of triangle

$1 \div 2b \times h$

1÷2×10×10

50cm^2

Area of segment= area of minor segment- area of triangle

=78.5-50.0

=28.5m^2

Area of major sector

$360 - \alpha \div 360 \times \pi \: {r}^{2}$

360-90/360×3.14×10×10

270×314

84780cm^2