Question

Let $ABCD$ be a quadrilateral with area $18$, side $AB$ parallel to the side $CD$, and $AB=2CD$. Let $AD$ be perpendicular to $AB$ and $CD$. If a circle is drawn inside the quadrilateral $ABCD$ touching all the sides, then its radius is

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

Answer 1

Answer: (B)

Given $AB||CD,CD=2AB$.
Let $AB=a$. Then $CD$ $=2a$.
Let the radius of the circle be $r$.
Let the circle touches $AB$ at $P,BC$ at $Q,AD$ at $R$, and $CD$ at $S$.
Then $AR=AP=r,BP=BQ=a-r,DR=DS=r$,
and $CQ=CS=2a-r$.

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