Question

find the length of a chord which is the outer diameter of 3 cm from centre of a circle whose radius is 5 cm​

Answer 1

Radius of a circle is 5cm

therefore OA= 5cm

perpendicular drawn to the chord from the centre bisects the chord

therefore AM=MB

triangle OMA is right angled triangle at M

∠OMA=90°

by applying Pythagoras theorum

OA²= OM² + MA²

MA= √OA² − OM²

MA = √5² -3²

MA = √16

MA = 4

THEREFORE THE CORD OF AB = 2MA

AB = 8cm

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