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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 segment.
Given:
A chord of a circle of radius $10\ cm$ subtends a right angle at the center.
To do:
We have to find the area of the corresponding minor sector.
Solution:
Let $AB$ be the chord which subtends an angle $90^o$ at the center $O$.
(i) Area of the minor sector$=$Area of the segment $AOB-$Area of $\vartriangle AOB$
$=\frac{\pi\theta}{360^{\circ}}\times OA\times OB-\frac{1}{2}\times OA\times OB$
$=\frac{3.14\times 90^{\circ}}{360^{\circ}}\times 10\times 10-0.5\times 10\times10$
$=78.5-50$
$=28.5\ cm^2$
Therefore, the area of the minor sector is $28.5\ cm^2$.
(ii) As the radius of the circle$=OA=OB=10\ cm$
Angle subtended by the chord $=\angle AOB=\theta=90^{\circ}$
Area of the major sector$=$ Area of the circle$-$Area of the minor sector
$=\pi r^2-\frac{\pi\theta}{360^{\circ}}\times OA\times OB$
$=3.14\times10\times10-3.14\times\frac{90^{\circ}}{360^{\circ}}\times10\times10$
$=314-78.5$
$=235.5\ cm^2$
Therefore, the area of the major sector is $235.5\ cm^2$.
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