Question

A right angled trapezium is circumscribed about a circle.The radius of the circle.If the lengths of the bases(i.e., parallel sides) are equal to $a$ and $b$ is

• A. $a+b$
• B. $\frac{ab}{a+b}$
• C. $\frac{a+b}{ab}$
• D. $|a-b|$

Answer 1

Answer: (B)

$\frac{ab}{a+b}$

Let $ABCD$ be a right angled trapezium.Let radius of circle is $R$ and let $BD=BE=x$ and $CE=CF=y$
$\because AB=a$
$R+x=a\Rightarrow x=a-R$ ...............$\left(1\right)$
and $DC=b$
$R+y=b\Rightarrow y=b-R$ ............$\left(2\right)$
In $△ODB,\tan \alpha =\frac{R}{x}$ ............$\left(3\right)$
Now, in $△CHB$
$\sin 2\alpha =\frac{CH}{BC}$
$\Rightarrow \frac{2\tan \alpha }{1+{\tan }^2\alpha }=\frac{2R}{x+y}$
$\Rightarrow \frac{\frac{2R}{x}}{1+\frac{R^2}{x^2}}=\frac{2R}{x+y}$
$\Rightarrow \frac{2Rx}{x^2+R^2}=\frac{2R}{x+y}$
$\Rightarrow x^2+xy=x^2+R^2$
$\Rightarrow xy=R^2$
From $\left(1\right)$ and $\left(2\right)$
$\Rightarrow (a-R)(b-R)=R^2$
$\Rightarrow ab-(a+b)R+R^2=R^2$
$\Rightarrow ab=(a+b)R$
$\therefore R=\frac{ab}{a+b}$