Circle Ques 69
- 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
$(2007,3 M)$
(a) 3
(b) 2
(c) $\frac{3}{2}$
(d) 1
Show Answer
Answer:
Correct Answer: 69.(b)
Solution:
Formula:
- $18=\frac{1}{2}(3 \alpha)(2 r)$
$\Rightarrow \alpha r=6$
Line, $y=-\frac{2 r}{\alpha}(x-2 \alpha)$ is tangent to circle
$$ \begin{array}{r} (x-r)^{2}+(y-r)^{2}=r^{2} \\ 2 \alpha=3 r, \alpha r=6 \text { and } r=2 \end{array} $$
Alternate Solution
$$ \frac{1}{2}(x+2 x) \times 2 r=18 $$
$$ x r=6 $$
In $\triangle A O B, \tan \theta=\frac{x-r}{r}$ and in $\triangle D O C$,
$$ \begin{array}{rlrlrl} & & \tan \left(90^{\circ}-\theta\right) & =\frac{2 x-r}{r} \\ & \therefore & \frac{x-r}{r} & =\frac{r}{2 x-r} \\ & \Rightarrow & x(2 x-3 r) & =0 \\ & \Rightarrow & x & =\frac{3 r}{2} & \ldots \end{array} $$
From Eqs. (i) and (ii), we get
$$ r=2 $$
BETA
