Circle 3 Question 7

7. 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

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Answer:

Correct Answer: 7. (b)

Solution:

  1. $18=\frac{1}{2}(3 \alpha)(2 r)$

$\Rightarrow \alpha r=6$

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

$$ r=2 $$



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