Question

In ®A shown below, radius AB is perpendicular to chord XY at point C. If XY=24 and AC=5 cm, what is the radius of the circle?

Answer 1

AB and AC are two equal chord of a circle, therefore the centre of the circle lies on the bisector of ∠BAC.

OA is the bisector of ∠BAC.

Again, the internal bisector of an angle divides the opposite sides in the ratio of the sides containing the angle.

P divides BC in the ratio 6:6=1:1.

P is mid-point of BC.

OP ⊥ BC.

In △ ABP, by pythagoras theorem,

AB2=AP2+BP2

BP2=36−AP2 ....(1)

In △ OBP, we have

OB2=OP2+BP2

52=(5−AP)2+BP2

BP2=25−(5−AP)2 .....(2)

From 1 & 2, we get,

36−AP2=25−(5−AP)2

36=10AP

AP=3.6cm

Substitute in equation 1,

BP2=36−(3.6)2=23.04

BP=4.8cm

BC=2×4.8=9.6cm

Answer 2

B. 13cm

Step-by-step explanation

The radius of the circle becomes the hypotenuse of the right triangle formed.

We can use the Pythagoras Theorem to obtain,

AC²+CY²=r²

This implies that,

r²=5²+12²

r²=25+144

r²=169

Take positive square root to get;

r=√169

r=13