Question

In the given figure, ABCD is a rectangle, PQRS is a parallelogram and TUVR is a rhombus, DQM is a straight line. Then find (i) $\angle QPS$, (ii) $\angle PQD$, (iii) $\angle MQN$

• A. (i) ${45}^0$, (ii) ${98}^0$, (iii) ${105}^0$
• B. (i) ${36}^0$, (ii) ${42}^0$, (iii) ${121}^0$
• C. (i) ${75}^0$, (ii) ${46}^0$, (iii) ${121}^0$
• D. (i) ${70}^0$, (ii) ${43}^0$, (iii) ${105}^0$

Answer 1

The correct option is C (i) ${75}^0$, (ii) ${46}^0$, (iii) ${121}^0$

$\left(i\right)\angle TRV={180}^{\circ }-\angle UVR={180}^{\circ }-{114}^{\circ }={66}^{\circ }$ (adjacent angles of a rhombus)

$\angle SRU=\angle SRT+\frac{1}{2}\angle TRV={42}^{\circ }+{33}^{\circ }={75}^{\circ }$

$\angle QPS=\angle SRU={75}^{\circ }$ (opposite angles of a parallelogram)

$\left(ii\right)\angle PDQ=\angle PDN-\angle QDN={90}^{\circ }-{31}^{\circ }={59}^{\circ }$

$\angle PQD={180}^{\circ }-\angle QPS-\angle PDQ={180}^{\circ }-{75}^{\circ }-{59}^{\circ }={46}^{\circ }$ (angles of a triangle)

$\left(iii\right)\angle DQN=\angle PDQ={59}^{\circ }$ (alternate angles of transversal)

$\angle MQN={180}^{\circ }-\angle DQN={180}^{\circ }-{59}^{\circ }={121}^{\circ }$ (angles on a straight line)

$\therefore \angle QPS={75}^{\circ },\angle PQD={46}^{\circ }and\angle MQN={121}^{\circ }\left[C\right]$