Question

A chord CD of a circle, whose centre is O is bisected at P by a diameter AB.
Given OA = OB = 15 cm and OP = 9 cm
Calculate the lengths of :
(i) CD (ii) AD (iii) CB.​

Answer 1

Answer:

a) 24 cm

b) 12√5 cm

c) 10.81 cm

Step-by-step explanation:

From the  the figure in the Attachment we can say that

OA = OB = 15 = radius of the circle

and OP bisect chord CD in into halves

⇒CP =PD

In ΔOCP

CO^2 =OP^2+CP

15^2 =9^2 +CP^2

⇒CP =12 cm

therefore PD= 12 cm and CD =2CP = 2*12= 24 cm

In ΔAPD

we can write

AP^2+ PD^2 =Ad^2

AD^2= 24^2+ 12^2

AD= 12√5 cm

now in ΔCPB

we can write

CP^2+ PB^2 = CB^2

CB^2 = 9^2 + 6^2= 10.81 cm

Answer 2

Answer:

hope it helps u well

Step-by-step explanation:

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