A) rhombus
B) parallelogram
C) cyclic quadrilateral
D) trapezium
Correct Answer: C
Solution :
\[\angle APB+\angle PAB+PBA={{180}^{o}}\] and \[\angle CRD+\angle RCD+\angle RDC={{180}^{o}}\] \[\Rightarrow \] \[\angle APB+\frac{1}{2}\angle A+\frac{1}{2}\angle B={{180}^{o}}\] and \[\angle CRD+\frac{1}{2}\angle C+\frac{1}{2}\angle D={{180}^{o}}\] \[\Rightarrow \]\[\angle APB+\angle CRD+\frac{1}{2}(\angle A+\angle B+\angle C+\angle D)\] \[={{360}^{o}}\] \[\Rightarrow \] \[\angle APB+\angle CRD={{180}^{o}}\] Hence \[PQRS\] is a cyclic quadrilateral.You need to login to perform this action.
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