Question

AB is a chord of a circle with centre P and radius 13cm. If AB=24cm, find the distance of AB from centre P.​

Answer 1

Answer:

Perpendicular drawn from centre of a circle to chord bisects the chord.

AC=BC=12cm

And ∠OCB=90°

Thus, triangle OCB is a right angled triangle

OB=13cm.

Therefore by Pythagoras theorem

OB² =OC²+BC²

OC² = BC²- OB²

$\sqrt{13 {}^{2} - {12}^{2} }$

OC²= 169— 144 cm

OC²= 25cm

OC= 5cm

Answer 2

Answer:

1.8

Step-by-step explanation:

R=radius , C=chord

R/C

=1.8