Question

Let ABCD be a quadrilateral in which $AB\parallel CD,AB\perp AD$ and AB=3CD. The area of quadrilateral ABCD is 4. The radius of a circle touching all the sides of quadrilateral is:

• A. $sin\frac{\pi }{12}$
• B. $sin\frac{\pi }{6}$
• C. $sin\frac{\pi }{4}$
• D. $sin\frac{\pi }{3}$

Answer 1

The correct option is D $sin\frac{\pi }{3}$

Consider the problem

According to the given information we have draw the diagram

Here,

$CD=X$, $SO$ $AB$ $=3X$ (As given $AB=3CD$)

And the points $P,Q,RandS$ are points where circle meets line $AB,BC,CDandDA$

So,

Radius of circle$=OP=OQ=OR=OS=r$

Here we construct $CM\perp AB$ and given $CD\perp AD$ So, $AMCD$ is a rectangle

$AM=CD=XandCM=DA=2r........\left(A\right)$

As given $CD\perp AD$ and $OP\perp AB,OR\perp CD$ (we know the radius is perpendicular to tangent at the point of tangency)

So $APOS$ and $DROS$ are square,Then

$AP=AS=OP=OS=DR=DS=OR=OS=r........\left(B\right)$

And

$CR=CD-DR=X-r$ (From equation $B$ and we assume $CD=X$)

We know length of tangents drawn from external point to the circle are equal

So,

$CR=CQ=X-r.......\left(1\right)$

And

$BP=AB-AP=3X-r$ (From equation $B$ and we get $AB=3X$)

So,

$BP=BQ=3X-r.......\left(2\right)$

And

$BC=BQ+CQ=3X-r+X-r=4X-2r(Fromequation\left(1\right)and\left(2\right)$

And

$\begin{array}{c}BM=AB-AM\\ =3X-X\\ =2X\end{array}$ (From equation $A$ and we get $AB=3X$)

Now, apply Pythagoras theorem in triangle $CMB$

$B{C}^2=C{M}^2+B{M}^2$

$\begin{array}{c}{\left(4X-2r\right)}^2={\left(2r\right)}^2+{\left(2x\right)}^2\\ \Rightarrow 16X^2+4r^2-16Xr=4r^2+4X^2\\ \Rightarrow 12X^2=16Xr\\ \Rightarrow 12X=16r\\ \Rightarrow 3X=4r\\ \Rightarrow X=\frac{4r}{3}.......\left(1\right)\end{array}$

Here, in quadrilateral $ABCD,AB\parallel CD$ So that $ABCD$ is a trapezium.

And, we know area of trapezium $=\frac{sumofparallelsides}{2}\times height$

Given area of ABCD$=4sq.unit$

$\begin{array}{c}\frac{AB+CD}{2}\times DA=4\\ \Rightarrow \frac{3X+X}{2}\times 2r=4\\ \Rightarrow \frac{4X}{2}\times 2r=4\\ \Rightarrow 4Xr=4\\ \Rightarrow X=\frac{1}{r}\end{array}$

Substitute value from equation $3$ and we get

$\begin{array}{c}\frac{4r}{3}=\frac{1}{r}\\ \Rightarrow 4r^2=3\\ \Rightarrow r^2=\frac{3}{4}\\ \Rightarrow r=\sqrt{\frac{3}{4}}\\ \Rightarrow r=\frac{\sqrt{3}}{2}\end{array}$

Therefore, Radius of given circle $=\frac{\sqrt{3}}{2}$ unit

Hence, Option $D$ is the correct answer which is $\sin \frac{\pi }{3}$