Question 4: A chord of a circle of radius 10 cm subtends a right angle at the centre - Maths - Areas Related to Circles

Question

Question 4:

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
[Use π = 3 14]
please write in a simple method

Chetna Khanna · Accepted Answer

Let AB be the chord of circle of radius 10 cm Let O be the centre of the circle

Radius OA = OB = 10 cm âˆ AOB = 90Â° $Area of minor sector AOB = \frac{90 °}{360 °} \times π r^{2} [Area = \frac{θ}{360 °} \times π r^{2}] = \frac{90 °}{360 °} \times \frac{22}{7} \times (10)^{2} {cm}^{2} = \frac{1}{4} \times \frac{22}{7} \times 100 {cm}^{2} = \frac{2200}{28} {cm}^{2} = 78 57 {cm}^{2} Area of Δ AOB = \frac{1}{2} \times AO \times OB [Area = \frac{1}{2} \times Base \times Height] = \frac{1}{2} \times 10 \times 10 {cm}^{2} = 50 {cm}^{2}$ (i) Area of minor segment = Area of minor sector AOB â€“ Area of âˆ†AOB = 78 57 cm2 â€“ 50cm2 = 28 57 cm2 (ii) Area of major sector = Area of circle â€“ Area of minor sector Ï€r2 â€“ 78 57 cm2 $= \frac{22}{7} \times (10)^{2} {cm}^{2} - 78 57 {cm}^{2} = \frac{2200}{7} {cm}^{2} - 78 57 {cm}^{2}$ = 314 29 cm2 â€“ 78 57 cm2 = 235 72 cm2

Question 4: 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 [Use π = 3.14] please write in a simple method

Question 4:

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

[Use π = 3.14]

please write in a simple method