Question

Let $ABCD$ be a trapezium with parallel sides $AB$ and $CD$ such that the circle $S$ with $AB$ as its diameter touches $CD$. Further, the circle $S$ passes through the midpoints of the diagonals $AC$ and $BD$ of the trapezium. The smallest angle of the trapezium is

• A. $\frac{\pi }{3}$
• B. $\frac{\pi }{4}$
• C. $\frac{\pi }{5}$
• D. $\frac{\pi }{6}$

Answer 1

Answer: (D)

$\frac{\pi }{6}$

In $\Delta ABC,BF\perp AC(\because AB$ is the diameter of the circle,angle subtended by the diameter is ${90}^{\circ })$

And $F$ is the midpoint of $AC$

Hence, $ABC$ is an isosceles $\Delta$ with $AB=BC$

Similaraly,$\Delta ADB$ is isosceles with $AD=AB$

$\Rightarrow AB=BC=AD\Rightarrow BC=AD$

Hence, $ABCD$ is an isosceles trapezium.

Now if $O$ is the centre of the circle and $OG,BH$ are $\perp CD$,

Then $AB\parallel CD$

$\Rightarrow \angle BOG=\angle OGH=\angle GHB={90}^{\circ }$

Hence, it is a rectangle.

$\Rightarrow OG=BH$

But $OG=OB$ ($\because$ radius in a circle is constant)

$\Rightarrow BH=OBBC=AB=2OB=2BH\Rightarrow \frac{BH}{BC}=\frac{1}{2}$

Now,in right-angled $\Delta CBH$,

$\sin \left(\angle BCH\right)=\frac{1}{2}$

$\Rightarrow \angle BCH=\frac{\pi }{6}$

Now since it is an isosceles trapezium,

$\angle BCH=\angle ADC=\frac{\pi }{6}$

$AB\parallel CD$

$\Rightarrow \angle ABC+\angle BCD={180}^{\circ }$

$\Rightarrow \angle ABC={150}^{\circ }=\angle BAD$

Therefore, smallest angle in a trapezium is $\frac{\pi }{6}$.