
5. Let $A B C D$ be a quadrilateral in which $A B$ is parallel to $C D$ and perpendicular to $A D$; $A B=3 C D$; and the area of the quadrilateral is 4 . If a circle can be drawn touching all the sides of the quadrilateral, find its radius.


Solution: Let $P, Q, R, S$ be the points of contact of in-circle with the sides $A B, B C, C D$, $D A$ respectively. Since $A D$ is perpendicular to $A B$ and $A B$ is parallel to $D C$, we see that $A P=A S=S D=D R=r$, the radius of the inscribed circle. Let $B P=B Q=y$ and $C Q=C R=x$. Using $A B=3 C D$, we get $r+y=3(r+x)$.

![](https://cdn.mathpix.com/cropped/2024_06_05_b41d47c32aa98250392fg-3.jpg?height=328&width=526&top_left_y=-1&top_left_x=764)

Since the area of $A B C D$ is 4 , we also get

$$
4=\frac{1}{2} A D(A B+C D)=\frac{1}{2}(2 r)(4(r+x))
$$

Thus we obtain $r(r+x)=1$. Using Pythagoras theorem, we obtain $B C^{2}=B K^{2}+C K^{2}$. However $B C=y+x, B K=y-x$ and $C K=2 r$. Substituting these and simplifying, we get $x y=r^{2}$. But $r+y=3(r+x)$ gives $y=2 r+3 x$. Thus $r^{2}=x(2 r+3 x)$ and this simplifies to $(r-3 x)(r+x)=0$. We conclude that $r=3 x$. Now the relation $r(r+x)=1$ implies that $4 r^{2}=3$, giving $r=\sqrt{3} / 2$.
