Question

Construct a quadrilateral ABCD in which AB = 5.6 cm, BC = 4 cm, $\angle A={50}^{\circ }$, $\angle B={105}^{\circ }$ and $\angle D={80}^{\circ }$.

Answer 1

Steps of Construction :

$\because \angle A={50}^{\circ },\angle B={105}^{\circ }and\angle D={80}^{\circ }and\angle A+\angle B+\angle C+\angle D={360}^{\circ }\Rightarrow {50}^{\circ }+{105}^{\circ }+\angle C+{80}^{\circ }={360}^{\circ }\Rightarrow \angle C+{235}^{\circ }={360}^{\circ }\Rightarrow \angle C={360}^{\circ }-{235}^{\circ }\Rightarrow \angle C={125}^{\circ }$

(i) Draw a line segment AB 5.6 cm.

(ii) At B, draw a ray BY making an angle of ${105}^{\circ }$ and cut off BC = 4 cm.

(iii) At C, draw a ray CZ making an angle of ${125}^{\circ }$ and at A, a ray AX making an angle of ${50}^{\circ }$ intersecting each other at D.

then $\angle D={80}^{\circ }$

$\therefore$ ABCD is the required quadrilateral.