A) A rectangle
B) A rhombus
C) A square
D) Any other parallelogram
Correct Answer: A
Solution :
\[\angle APQ+\angle CQP={{180}^{o}}\] \[\Rightarrow \]\[\frac{\angle APQ}{2}+\frac{\angle CQP}{2}{{90}^{o}}\] \[\Rightarrow \]\[\angle RPQ+\angle PQR={{90}^{o}}\] \[\Rightarrow \]\[\Delta PQR\]is a right angled triangle Similarly, \[\Delta PQR\]is a right angled triangle \[\Rightarrow \]\[PQRS\]is rectangle.You need to login to perform this action.
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