Question

A circle of unit radius touches the positive $x$ and $y$ axes. Two parallel tangents $AB$ and $DC$ having slopes $-1$ are drawn to the circle cutting $x$ -axis at $B$ and $C$ and $y$ -axis at $A$ and $D$ respectively. The square of the area of the trapezium $ABCD$ is equal to

Answer 1

Let the tangent line be $x+y=c$
Now, $\frac{\left|\right. 1 + 1 - \text{c} \left|\right.}{\sqrt{2}}=1\Rightarrow \text{c}=2\pm\sqrt{2}$
$\therefore A\left(0,2 + \sqrt{2}\right),B\left(2 + \sqrt{2} , 0\right)$
$C\left(2 - \sqrt{2} , 0\right),D\left(0,2 - \sqrt{2}\right)$
So, the area of the trapezium $ABCD$
$=\frac{1}{2}(2)[(\sqrt{2}(2+\sqrt{2}))+(\sqrt{2} \cdot(2-\sqrt{2}))]$
$=4\sqrt{2}$ sq. units
Hence, the square of the area $=\left(4 \sqrt{2}\right)^{2}=32$