Question

In the figure above a circle is inscribed in a parallelogram ABCD, PQRS is also a parallelogram and $\overline{AB}\perp \overline{PR}$ . If $\angle SRP={60}^{\circ }$ then find $\angle BPQ$

• A. ${60}^{\circ }$
• B. ${45}^{\circ }$
• C. ${30}^{\circ }$
• D. ${25}^{\circ }$

Answer 1

The correct option is C ${30}^{\circ }$

Given that:

$ABCD,PQRS$ are parallelograms.

$\overline{AB}\perp \overline{PR},\angle SRP={60}^{\circ }$

To find:

$\angle BPQ=?$

Solution:

$\angle SRP=\angle RPQ$ (Alternate interior angles)

or, $\angle SRP=\angle RPQ={60}^{\circ }$

$\angle BPR={90}^{\circ }$ $(\overline{AB}\perp \overline{PR})$

or, $\angle RPQ+\angle BPQ={90}^{\circ }$

or, $\angle BPQ={90}^{\circ }-{60}^{\circ }$

or, $\angle BPQ={30}^{\circ }$

Hence, C is the correct option.