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

$\angle A={60}^{\circ },\angle B={105}^{\circ }\text{and}\angle D={90}^{\circ }$

But $\angle$ A +$\angle$ B + $\angle$ C+$\angle D={360}^{\circ }$ (Sum of angles of a quadrilateral)

$\Rightarrow {60}^{\circ }+{105}^{\circ }+\angle C+{90}^{\circ }={360}^{\circ }$

$\Rightarrow {255}^{\circ }+\angle C={360}^{\circ }$

$\Rightarrow \angle C={360}^{\circ }–{255}^{\circ }={105}^{\circ }$

Steps of construction :

Step 1 :Draw a line segment AB = 5.5 cm.

Step 2 : At A, draw a ray AX making an angle of

Step 3 : At B, draw another ray BY making an angle of ${105}^{\circ }$ and cut off BC = 3.7 cm.

Step 4 :At C, draw a ray CZ making an angle of ${105}^{\circ }$ which intersects AX at D.

Then ABCD is the required quadrilateral.