A) Kite
B) Rhombus
C) Rectangle
D) Trapezium
Correct Answer: C
Solution :
Given, APB and CQD are two parallel lines. Let the bisectors of angles APQ and CQP meet at a point M and bisectors of angles BPQ and PQD meet at a point N. Since, \[APB||CQD\] \[\therefore \]\[\angle APQ=\angle PQD\] [Alternate interior angles] \[\Rightarrow \]\[\frac{1}{2}\angle APQ=\frac{1}{2}\angle PQD\] \[\Rightarrow \]\[\angle MPQ=\angle NQP\] This shows that alternate interior angles are equal. \[\therefore \] \[PM||QN\] Similarly,\[\angle NPQ=\angle MQP,\]which shows that alternate interior angles are equal. \[\therefore \] \[PN||QM\] So, quadrilateral PMQN is a parallelogram. Also, \[\angle CQP+\angle DQP={{180}^{o}}\] [Linear pair] \[\Rightarrow \]\[2\angle MQP+2\angle NQP={{180}^{o}}\] \[\Rightarrow \]\[2(\angle MQP+\angle NQP)={{180}^{o}}\] \[\Rightarrow \]\[\angle MQN={{90}^{o}}\] Thus, PMQN is a rectangle.You need to login to perform this action.
You will be redirected in
3 sec