ExamBro
ExamBro
JEE Advanced · Mathematics · 12. Circle

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

  1. A
    3
  2. B
    2
  3. C
    \(\frac{3}{2}\)
  4. D
    1
Verified Solution

Answer & Solution

Correct Answer

(B)
2

Step-by-step Solution

Detailed explanation

\[
\text { } 18=\frac{1}{2}(3 \alpha)(2 r) \Rightarrow \alpha r=6
\]


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
\[
r=2
\]
From JEE Advanced
Explore more questions on app