Question

A circle touches the sides $AB$ and $AD$ of a rectangle $ABCD$ at $P$ and $Q$ respectively and passes through the vertex $C$. If the distance of $C$ from chord $PQ$ is $5$ units, then area of the rectangle is :

• A. $45$
• B. $25$
• C. $50$
• D. $75$

Answer 1

Answer: (B)

$25$

$\Rightarrow$ In given figure $ABCD$ is a rectangle.

$\Rightarrow$ Join $CQ$ and $CP$.

$\Rightarrow$ Let $\angle CPB=x$

$\Rightarrow$ $\angle CQP=x$ [Alternate segment theorem]

$\Rightarrow$ Let $\angle DQC=y$

$\Rightarrow$ $\angle QPC=y$ [Alternate segment theorem]

$\Rightarrow$ As $ABCD$ is a rectangle each angle is 90${}^{\circ }$

$\Rightarrow$ Now, $△CDQ\cong △CEP$ [By AA criteria]

$\Rightarrow$ $\frac{CD}{CE}=\frac{DQ}{EP}=\frac{CQ}{CP}$

$\Rightarrow$ Hence, $\frac{CD}{CE}=\frac{CQ}{CP}$ --- ( 1 )

And $△CBP\cong △CEQ$ [AA criteria]

$\Rightarrow$ $\frac{CB}{CE}=\frac{BP}{EQ}=\frac{CP}{CQ}$

$\Rightarrow$ $\frac{CB}{CE}=\frac{CP}{CQ}$

$\Rightarrow$ Hence, $\frac{CE}{CB}=\frac{CQ}{CP}$ --- ( 2 )

$\Rightarrow$ $\frac{CD}{CE}=\frac{CE}{CB}$ [From ( 1 ) and ( 2 )]

$\Rightarrow$ $CB\times CD=C{E}^2$

$\Rightarrow$ But, Area of rectangle $ABCD$ = $CB\times CD=C{E}^2=5^2=25sq.units$