Question

The figure shows a circle of radius one unit with centre O and

Answer 1

AB bisects chord PQ at point C.

Therefore,

PR=QR=8cm

RB=4cm

Let, radius OA=OB=r

OB=OR+RB

=>r=OR+4

=>OR=r−4 (i)

Now,

OP² = OR² + PR²

=>r = ( r - 4)² + 8² ( using i)

=>r² = r² −8r+16+64

= >8r=80

=>r=10cm

the radius of the circle is 10cm

Answer 2

Answer:

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