Question

A circle touches two adjacent sides of a rectangle AB and AD at points P and Q respectively.Third vertex C of the rectangle lies on the circle. The length of perpendicular from vertex C to the chrod PQ is 5. Find the area of rectangle.

Answer 1

Let $C$ lie on the arc $C_1{C}_2$,Label the ${90}^{\circ }$ and ${45}^{\circ }$ angles.Since $PQ$ subtends a central angle of ${90}^{\circ }$, the inscribed angle $PCQ$ must be ${45}^{\circ }$

Let $\angle BCP=\alpha$.Label the remaining angles in terms of $\alpha$

Draw a perpendicular from$C$ to $PQ$.Let $E$ be the point of intersection

$\angle ECQ=\alpha$

$\Rightarrow \angle PCE=45-\alpha$

Since $EC=5$ and $\angle ECQ=\alpha ,CQ=\frac{5}{\cos \alpha }$

Since $\angle QCD45-\alpha ,CQ=\frac{CD}{\cos (45-\alpha )}$

$\Rightarrow \frac{5}{\cos \alpha }=\frac{CD}{\cos (45-\alpha )}$

$\Rightarrow CD=5\frac{\cos (45-\alpha )}{\cos \alpha }$

and $BC=\frac{5\cos \alpha }{\cos (45-\alpha )}$

Area of rectangle$ABCD=CD\times BC=5\frac{\cos (45-\alpha )}{\cos \alpha }\times \frac{5\cos \alpha }{\cos (45-\alpha )}=25$sq.units.