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Let $A B C D$ be a quadrilateral with area 18 , with 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
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The correct answer is:
2
2
$$
\text { } 18=\frac{1}{2}(3 \alpha)(2 r) \Rightarrow \alpha r=6
$$
Line, $y=-\frac{2 r}{\alpha}(x-2 \alpha)$ is tangent to circle
$$
\begin{aligned}
(x-r)^2+(y-r)^2 & =r^2 \\
2 \alpha & =3 r \text { and } \alpha=6 \\
r & =2
\end{aligned}
$$
$$
\begin{aligned}
\frac{1}{2}(x+2 x) \times 2 r & =18 \\
x r & =6 \\
\tan \theta & =\frac{x-r}{r} \\
\tan \left(90^{\circ}-\theta\right) & =\frac{2 x-r}{r} \\
\frac{x-r}{r} & =\frac{r}{2 x-r} \times(2 x-3 r)=0
\end{aligned}
$$
$$
x=\frac{3 r}{2}
$$
From Eqs. (i) and (ii), we get
$$
r=2
$$
\text { } 18=\frac{1}{2}(3 \alpha)(2 r) \Rightarrow \alpha r=6
$$
Line, $y=-\frac{2 r}{\alpha}(x-2 \alpha)$ is tangent to circle
$$
\begin{aligned}
(x-r)^2+(y-r)^2 & =r^2 \\
2 \alpha & =3 r \text { and } \alpha=6 \\
r & =2
\end{aligned}
$$
$$
\begin{aligned}
\frac{1}{2}(x+2 x) \times 2 r & =18 \\
x r & =6 \\
\tan \theta & =\frac{x-r}{r} \\
\tan \left(90^{\circ}-\theta\right) & =\frac{2 x-r}{r} \\
\frac{x-r}{r} & =\frac{r}{2 x-r} \times(2 x-3 r)=0
\end{aligned}
$$
$$
x=\frac{3 r}{2}
$$
From Eqs. (i) and (ii), we get
$$
r=2
$$
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