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.
Answers
Answered by
32
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
Attachments:
Answered by
14
Answer:
hope it helps u well
Step-by-step explanation:
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tq........
Attachments:
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