Question

Let $A B C D$ be a quadrilateral with area $18$, side $A B$ parallel to the side $C D$, and $A B=2 C D$. Let $A D$ be perpendicular to $A B$ and $C D$. If a circle is drawn inside the quadrilateral $A B C D$ touching all the sides, then its radius is

• A. 3
• B. 2
• C. $\frac{3}{2}$
• D. 1

Answer 1

Answer: (B)

Given $A B || C D, C D=2 A B$.
Let $A B=a$. Then $C D$ $=2 a$.
Let the radius of the circle be $r$.
Let the circle touches $A B$ at $P, B C$ at $Q, A D$ at $R$, and $C D$ at $S$.
Then $A R=A P=r, B P=B Q=a-r, D R=D S=r$,
and $C Q=C S=2 a-r$.

In $\triangle B E C$,
$ B C^{2}=B E^{2}+E C^{2}$
$\Rightarrow (a-r+2 a-r)^{2}=(2 r)^{2}+(a)^{2}$
or $9 a^{2}+4 r^{2}-12 a r=4 r^{2}+a^{2}$
or $a=\frac{3}{2} r$ ...(i)
Also, area $(A B C D)=18\, sq$. unit
or Area $(A B E D)+$ Area $(\triangle B C E)=18\, sq$. unit
or $a \times 2 r+\frac{1}{2} \times a \times 2 r=18$
or $a r=6 $ or $\frac{3 r^{2}}{2}=6$ [Using Eq. (1)]
or $r^{2}=4$ or $r=2$