Question

8. 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:

i) CD= 24 cm

ii) AD= 26.83 cm

iii) CB= 13.4cm

Step-by-step explanation:

Here as u can see, CD is the chord and the diameter AB will divide it equally.

i) OB= 15CM and OP= 9CM.

In triangle OCP OC = 15CM and OP = 9CM

therefore CP = root over 15^2 - 9^2=root over 144= 12 CM

hence CD = 2×12 CM = 24 CM

ii) Similarly from triangle ADP one can find out AD.

Here AP = OA+ OP= 15 +9= 24CM and PD = CP= 12CM

iii) Same as ii) one can find it also...

Here CP we know and PB = OB - OP = 15 -9 = 6CM

HERE WE CAN FIND ii and iii by pythagoras theorem.

Thank you.