A chord of a circle of radius $10 c m$ subtends a right angle at the centre. Find the area of the corresponding:
minor segment

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Solution

$R a d i u s o f t h e c i r c l e,$ $r = 10 c m$ .
$A n g l e a t t h e c e n t r e o f c i r c l e,$ $θ = 90^{o}$
$1) A r e a o f m i n o r s e g m e n t = ?$
$2) A r e a o f m a j o r s e g m e n t = ?$
$(i) A r e a o f m i n o r s e g m e n t$ $O A P B$
$= \frac{θ}{360} \times π r^{2}$
$= \frac{90}{360} \times \frac{22}{7} \times (10)^{2}$
$= \frac{1}{4} \times \frac{22}{7} \times 100$
$= \frac{550}{7}$
$= 78.5 s q . c m$ .
$i i) A r e a o f M i n o r S e g m e n t,$ $A P B$ :
$= A r e a o f$ $O A P B -$ $A r e a o f$ $A O A B$
$= 78.5 c m^{2} - \frac{1}{2} \times O A \times O B$
$= 78.5 - \frac{1}{2} \times 10 \times 10$
$= 78.5 - 50$
$= 28.5 s q . c m$ .

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