Question

In the figure given below, $CD$ is the diameter of the circle which meets the chord $AB$ at $P$ such that $AP=BP=12cm$. If $DP=8cm$, find the radius of the circle.

Answer 1

REF.Image

Given : AP = BP = 12 cm, DP = 8 cm

Let OP be x, OA = OD = [Radius]

$OA=OD=x+8cm$

$\because AP=BP$

$\therefore \angle OPA=\angle OPB={90}^{\circ }$

In right $△OPA$

$O{A}^2=O{P}^2+A{P}^2$

$(8+x{)}^2=x^2+(12{)}^2$

$64+16x+x^2=x^2+144$

$16x=144-64$

$16x=80$

$x=\frac{80}{16}$

$x=5$

Now, $OD=OP+PD$

$=5+8$

$=13cm$

Radius of circle is $13cm$