Question

A circle with
centre 0 in which diameter AB bisects the chord CD at point E. If CE = ED = 8 cm and
EB = 4 cm, find the radius of the circle.
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Answer 1

It is given that ⭐

CE = ED = 8cm and EB = 4cm Join the diagonal OC Consider OC = OB = r cm It can be written as OE = (r – 4) cm

Consider △ OEC By using the Pythagoras theorem OC2 = OE2 + EC2 By substituting the values we get r2 = (r – 4)2 + 82 On further calculation r2 = r2 – 8r + 16 + 64 So we get r2 = r2 – 8r + 80 It can be written as r2 – r2 + 8r = 80 We get 8r = 80 By division r = 10 cm Therefore, the radius of the circle is 10 cm.

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Answer 2

Step-by-step explanation:

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