Question

Construct a quadrilateral PQRS where PQ=3.5 cm, QR=6.5 cm, $\angle P=\angle R={105}^{\circ }$ and $\angle S={75}^{\circ }$

Answer 1

$\angle P={105}^{\circ },\angle R={105}^{\circ }$ and $\angle S={75}^{\circ }$
But $\angle P+\angle R+\angle S={360}^{\circ }$
(Sum of angles of a quadrilateral)
$\Rightarrow {105}^{\circ }+\angle Q+{105}^{\circ }+{75}^{\circ }={360}^{\circ }$
$\Rightarrow {285}^{\circ }+\angle Q={360}^{\circ }$
$\Rightarrow \angle Q={360}^{\circ }-{285}^{\circ }={75}^{\circ }$
steps of construction:
(i) Draw a line segments PQ =3.5 CM.

(ii) At P. draw a ray PX making an angle of ${105}^{\circ }$
(iii) At Q, draw another ray QY, making an angle of ${75}^{\circ }$ and cut off QR=6.5^\circ\)
(iv) At R, draw a ray RZ making an angle of ${105}^{\circ }$ which intersects PX at S.
Then PQRS is the required quadrilateral.