Question

Construct a quadrilateral ABCD where AB=5.5 cm, BC=3.7 cm, $\angle A={60}^{\circ }$ $\angle B={105}^{\circ }$ and $\angle D={90}^{\circ }$.

Answer 1

$\angle A={60}^{\circ },\angle B={105}^{\circ }$ and $\angle D={90}^{\circ }$
But$\angle A+\angle B+\angle C+\angle D={360}^{\circ }$
(Sum of angles of a quadrilateral) $\Rightarrow {60}^{\circ }+{105}^{\circ }+\angle C+{90}^{\circ }={360}^{\circ }$
$\Rightarrow {255}^{\circ }+\angle C={360}^{\circ }$
$\Rightarrow \angle C={360}^{\circ }-{255}^{\circ }={105}^{\circ }$
(i) Draw a line segment AB=5.5 cm.
(ii) At A, draw a ray AX making an angle of ${60}^{\circ }$

(iii) At B, draw another ray BY making an angle of ${105}^{\circ }$ and cut off BC=3.7 cm.
(iv) At C, draw a ray CZ making an angle of ${105}^{\circ }$ which intersect AX at D.
Then ABCD is the reqtired quadrilateral.