Question

AB is a chord of a circle having centre O. If angle AOB = 90°, prove that the chord AB is of length r√ 2

where r is the radius of the circle.​

Answer 1

Answer:

isosceles Δ BOA,

so, ∠ BAO = ∠ABO,

now, by ASP of a Δ,

∠BAO+∠ABO+∠AOB=180°

⇒2∠BAO=180°-60°

⇒2∠BAO=120°

⇒∠BAO=60°

∴ all angles in ΔBOA are 60°

⇒ΔBOA is equilateral

∴ OA=OB=AB

⇒AB= radius.                              (∵ OA and OB are radii)

thank you