Question

Construct a quadrilateral PQRS in which PQ = 5 cm, QR = 6.5 cm, $\angle P=\angle R={100}^{\circ }$ and $\angle S={75}^{\circ }$.

Answer 1

$\because \angle P+\angle Q+\angle R+\angle S={360}^{\circ }\therefore {100}^{\circ }+\angle Q+{100}^{\circ }+{75}^{\circ }={360}^{\circ }\Rightarrow \angle Q+{275}^{\circ }={360}^{\circ }\Rightarrow \angle Q={360}^{\circ }-{275}^{\circ }\therefore \angle Q={85}^{\circ }$

Steps of Construction :

(i) Draw a line segment PQ = 5 cm.

(ii) At Q, draw a ray QX making an angle of ${85}^{\circ }$ and cut off QR = 6.5 cm.

(iii) At R, draw a ray making an angle of ${100}^{\circ }$ and at, P, another ray making an angle of ${100}^{\circ }$ which intersect each other at S.

then $\angle S={75}^{\circ }$.

PQRS is the required quadrilateral.