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  • If bisectors of A and B of a quadrilateral ABCD intersect each other at P, of B and C at Q, of C and D at R and of D and A at S, then PQRS is a (A) rectangle (B) rhombus (C) parallelogram (D) quadrilateral whose opposite angles are supplementary

If bisectors of \angle A and \angle B of a quadrilateral ABCD intersect each other at P, of \angle B and \angle C at Q, of \angle C and \angle D at R and of \angle D and \angle A at S, then PQRS is a
(A) rectangle
(B) rhombus
(C) parallelogram
(D) quadrilateral whose opposite angles are supplementary

Answers (1)

best_answer

Answer: [D] quadrilateral whose opposite angles are supplementary
Solution.
First of all, let us draw the figure according to the question.

From the above diagram
\angle QPS=\angle APB           …..(1)             (Vertically opposite angles)
In \triangle APB=\angle APB+\angle PAB+\angle ABP=180^{\circ}
\angle APB=\frac{1}{2}\angle A+\frac{1}{2}\angle B=180^{\circ}
\angle APB=180-\frac{1}{2}(\angle A+\angle B)            …..(2)
From equation 1 & 2
\angle QPS=180-\frac{1}{2}(\angle A+\angle B)            …..(3)
Similarly  \angle QRS=180-\frac{1}{2}(\angle C+\angle D)       …..(4)
Add equation 3 and 4
\angle QPS+\angle QRS=360^{\circ}-\frac{1}{2}(\angle A+\angle B+\angle C+\angle D)
=360^{\circ}-\frac{1}{2}(360^{\circ})
=360^{\circ}-180^{\circ}
=180^{\circ}
\therefore \angle QPS+\angle QRS= 180^{\circ}
Hence PQRS is quadrilateral whose opposite angles are supplementary
 

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