A chord of a circle of radius 10 centimetre substance a right angle at centre find area of minor segment
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Answer:
28.57
Step-by-step explanation:
length of chord = 2rsinA/2
= 2× 10 sin(90/2) = 20 × 1/√2
= 10√2cm
3rd side of OAB = √[(10√2)^2 - (10}^2]
= 10 cm
are of right ∆OAB = 1/2 * 10 *10 = 50
area of minor segment
= 90/360πr^2 - area of ∆OAB
=1/4 *22/7*(10)^2 -50= 28.57cm^2
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