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)

(i) Let AB be the chord subtending right angle at the center.

Given,

Radius (r) = 10 cm

From figure,

APB is the minor segment.

We know that,

Area of sector of angle θ and radius r = $\dfrac{θ}{360°} \times πr^2$

Area of right angle triangle = $\dfrac{1}{2}$ × base × height

Area of minor segment APB = Area of sector AOBP - Area of right angle triangle AOB

Substituting values we get :

$= \dfrac{90°}{360°} \times 3.14 \times 10^2 - \dfrac{1}{2} \times AO \times BO \\[1em] = \dfrac{1}{4} \times 3.14 \times 100 - \dfrac{1}{2} \times r \times r \\[1em] = \dfrac{314}{4} - \dfrac{1}{2} \times 10 \times 10 \\[1em] = 78.5 - 50 \\[1em] = 28.5 \text{ cm}^2.$

Hence, area of corresponding minor segment = 28.5 cm².

(ii) Angle subtended by major sector = 360° - 90° = 270°.

We know that,

Area of sector of angle θ and radius r = $\dfrac{θ}{360°} \times πr^2$

Substituting values we get :

$\Rightarrow \text{Area of major sector } = \dfrac{270°}{360°} \times 3.14 \times 10^2 \\[1em] = \dfrac{3}{4} \times 314 \\[1em] = 3 \times 78.5 \\[1em] = 235.5 \text{ cm}^2.$

Hence, area of major sector = 235.5 cm².