A right angled trapezium is circumscribed about circle.What is the radius of the circle,if the lengths of the bases(i.e. parallel sides ) are $a$ and $b.$
By using the property that the length of tangents from the external point to the circle are equal.We conclude that in the right angled trapezium $ABCD$,
$AB+CD=BC+AD$
where $AD=a$ and $BC=b$ are parallel sides.
So $AB+CD=a+b$ but i do not know how to find the radius of the circle.
On
Let $r$ be the radius of circle, with the center $O$, circumscribed in the trapezium $ABCD$, having $\angle ABC=\angle BCD=90^\circ$, $AB=a$ & $CD=b$ & drop a perpendicular $AE$ from vertex $A$ to the side $CD$, drop perpendiculars $OM$ & $ON$ from $O$ to the sides $AB$ & $CD$ respectively. Then apply Pythagorean theorem
in right $\triangle OMA$, $$\color{red}{OA^2}=AM^2+OM^2=(a-r)^2+r^2$$ in right $\triangle OND$, $$\color{blue}{OD^2}=DN^2+ON^2=(b-r)^2+r^2$$ in right $\triangle AOD$, $$AD^2=\color{red}{OA^2}+\color{blue}{OD^2}=(a-r)^2+r^2+(b-r)^2+r^2$$ $$AD^2=4r^2+a^2+b^2-2r(a+b)\tag 1$$ in right $\triangle AED$, $$AD^2=AE^2+DE^2=(2r)^2+(b-a)^2$$ $$AD^2=4r^2+a^2+b^2-2ab\tag 2$$ equating the values of $AD^2$ from (1) & (2), $$4r^2+a^2+b^2-2r(a+b)=4r^2+a^2+b^2-2ab$$ $$r(a+b)=ab$$
$$\bbox[5px, border:2px solid #C0A000]{\color{red}{r=\frac{ab}{a+b}}}$$
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