Question

In the given quadrilateral ABCD, BC = 38 cm, QB = 27 cm, DC = 25 cm and AD $\perp$ DC find the radius of the circle.

Answer 1

Let $r$ be the radius of the circle.

In the given figure, $OP\perp DA,$ $OS\perp DC,$ $PDS={90}^0$ and $OP=OS$.

Therefore, $OPDS$ is a square and thus $DS=OS=r$.

It is given that $BC=38$ but $BC=BR+RC$ and $BR=QB$, also $CR=SC$

Now consider,

$BC=QB+SC\Rightarrow 38=27+SC\Rightarrow SC=38-27\Rightarrow SC=11$

It is also given that $DC=25$ cm, but $DC=DS+SC$, therefore,

$DS+SC=25\Rightarrow r+11=25\Rightarrow r=25-11\Rightarrow r=14$

Hence, the radius of the circle is $14$ cm.