MCQ
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
  • A
    $3$
  • $2$
  • C
    $3 / 2$
  • D
     $1$

Answer

Correct option: B.
$2$
b
$18=\frac{1}{2}(3 \alpha)(2 \mathrm{r}) \Rightarrow \alpha \mathrm{r}=6$

Line $y=-\frac{2 r}{\alpha}(x-2 \alpha)$ is tangent to $(x-r)^2+$ $(y-r)^2=r^2$

$2 \alpha=3 \mathrm{r}$ and $\alpha \mathrm{r}=6$

$\mathrm{r}=2 \text {. }$

Alternate

$ \frac{1}{2}(x+2 x) \times 2 r=18 $

$ x r=6 $

$ \tan \theta=\frac{x-r}{r} \quad \tan (90-\theta)=\frac{2 x-r}{r} $

$ \frac{x-r}{r}=\frac{r}{2 x-r} $

$ x(2 x-3 r)=0 $

$ x=\frac{3 r}{2}$

From $(1)$ and $(2)$

$r=2 \text {. }$

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