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A chord of a circle of radius $ 14 \mathrm{~cm} $ makes a right angle at the centre. Find the areas of the minor and major segments of the circle.
Given:
A chord of a circle of radius \( 14 \mathrm{~cm} \) makes a right angle at the centre.
To do:
We have to find the areas of the minor and major segments of the circle.
Solution:
Radius of the circle $r = 14\ cm$
Angle at the centre $\theta = 90^o$
Area of the circle $=\pi r^{2}$
$=\frac{22}{7} \times(14)^{2}$
$=616 \mathrm{~cm}^{2}$
$\mathrm{AB}$ is the chord.
Area of the minor segment $\mathrm{ACB}=(\frac{\pi \theta}{360^{\circ}}-\sin \frac{\theta}{2} \cos \frac{\theta}{2}) r^{2}$
$=(\frac{\pi \times 90^{\circ}}{360^{\circ}}-\sin \frac{90^{\circ}}{2} \cos \frac{90^{\circ}}{2})(14)^{2}$
$=(\frac{\pi}{4}-\sin 45^{\circ} \cos 45^{\circ}) \times 196$
$=196(\frac{22}{7 \times 4}-\frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{2}})$
$=(\frac{22 \times 196}{28}-196 \times \frac{1}{2})$
$=154-98$
$=56 \mathrm{~cm}^{2}$
Therefore,
Area of the major segment $\mathrm{ADB}=$ Area of the circle $-$ Area of the minor segment
$=616-56$
$=560 \mathrm{~cm}^{2}$
The areas of the minor and major segments of the circle are $56\ cm^2$ and $560 \mathrm{~cm}^{2}$ respectively.