Question

(1) AB is a chord of a circle with centre O and radius 10 cm. makes a right angle at the centre of a circle AB divides a circle in to two segments. Find the area of the minor segment.

Answer 1

$∠ALB= \frac{1}{2}∠AOB= \frac{90}{2} = 45°$

$\tiny \: LQ \:  is \: perpendicular \: bisection \: on \:  AB.$

$\tiny \: Hence \: by \: isoceles \: triangle \: property$

$LA=LB$

$OB=10cm, and OA=10cm$

$AB= {10}^{2} + {10}^{2} =10 \: 2cm$

$QB= \frac{AB}{2} = 5 \: 2cm$

$OQ= \sqrt{ {OB}^{2} - {QB}^{2} } =100−50=50 = 5 \: \: 2cm$

$LQ=LO+OQ=(10+5 \: \: 2)cm$

$Area \: of ALBQA = \frac{\pi {r}^{2} }{4} −Ar.ofAOB$

$= \frac{\pi \times {10}^{2} }{4} - \frac{1}{2} \times {10}^{2} ⇒ {28.54cm}^{2}$

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