(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. About the author Elliana
[tex]∠ALB= \frac{1}{2}∠AOB= \frac{90}{2} = 45°[/tex] [tex] \tiny \: LQ \: is \: perpendicular \: bisection \: on \: AB.[/tex] [tex] \tiny \: Hence \: by \: isoceles \: triangle \: property[/tex] [tex]LA=LB[/tex] [tex]OB=10cm, and OA=10cm[/tex] [tex]AB= {10}^{2} + {10}^{2} =10 \: 2cm[/tex] [tex]QB= \frac{AB}{2} = 5 \: 2cm[/tex] [tex]OQ= \sqrt{ {OB}^{2} – {QB}^{2} } =100−50=50 = 5 \: \: 2cm[/tex] [tex]LQ=LO+OQ=(10+5 \: \: 2)cm[/tex] [tex]Area \: of ALBQA = \frac{\pi {r}^{2} }{4} −Ar.ofAOB[/tex] [tex] = \frac{\pi \times {10}^{2} }{4} – \frac{1}{2} \times {10}^{2} ⇒ {28.54cm}^{2} [/tex] Please mark me as the Brainliest. Reply
[tex]∠ALB= \frac{1}{2}∠AOB= \frac{90}{2} = 45°[/tex]
[tex] \tiny \: LQ \: is \: perpendicular \: bisection \: on \: AB.[/tex]
[tex] \tiny \: Hence \: by \: isoceles \: triangle \: property[/tex]
[tex]LA=LB[/tex]
[tex]OB=10cm, and OA=10cm[/tex]
[tex]AB= {10}^{2} + {10}^{2} =10 \: 2cm[/tex]
[tex]QB= \frac{AB}{2} = 5 \: 2cm[/tex]
[tex]OQ= \sqrt{ {OB}^{2} – {QB}^{2} } =100−50=50 = 5 \: \: 2cm[/tex]
[tex]LQ=LO+OQ=(10+5 \: \: 2)cm[/tex]
[tex]Area \: of ALBQA = \frac{\pi {r}^{2} }{4} −Ar.ofAOB[/tex]
[tex] = \frac{\pi \times {10}^{2} }{4} – \frac{1}{2} \times {10}^{2} ⇒ {28.54cm}^{2} [/tex]
Please mark me as the Brainliest.