Question

What is the sum of $\angle BAD+\angle BPR+\angle BCD+\angle BQR$ in the figure?

• A. ${540}^{\circ }$
• B. ${360}^{\circ }$
• C. ${240}^{\circ }$
• D. None of the above

Answer 1

The correct option is B ${360}^{\circ }$

In quadrilateral $ABCD$, since it is a cyclic quadrilateral, sum of opposite angles $={180}^o$.
Thus, $\angle BAD+\angle BCD={180}^o$.
In quadrilateral $BPRQ$, since it is a cyclic quadrilateral, sum of opposite angles $={180}^o$.
Thus,$\angle BPR+\angle BQR={180}^o$.

Therefore, $\angle BAD+\angle BCD+\angle BPR+\angle BQR={180}^o+{180}^o={360}^o$.

Hence, option $B$ is correct.