Question

Construct a quadrilateral ABCD, where $\angle A={65}^{\circ },\angle B={105}^{\circ },\angle C={75}^{\circ },BC=5.7$ cm and CD=6.8 cm.

Answer 1

$\angle A={65}^{\circ },\angle B={105}^{\circ },\angle C={75}^{\circ }$
But $\angle A+\angle B+\angle C+\angle D={360}^{\circ }$
(Sum of angles of a quadrilateral) $\Rightarrow {65}^{\circ }+{105}^{\circ }+{75}^{\circ }+\angle D={360}^{\circ }$
$\Rightarrow {245}^{\circ }+\angle D={360}^{\circ }$
$\Rightarrow \angle D={360}^{\circ }-{245}^{\circ }={115}^{\circ }$
steps of construction :
(i) Draw a line segment BC=5.7 cm.
(ii) At B, draw a ray BX making an angle of${105}^{\circ }$

(iii) At C draw a another ray CY making an angle of ${75}^{\circ }$ and cut off CD=6.8 cm.
(iv At D, draw a ray DZ making an angle of ${115}^{\circ }$ which intersects BX at A.
Then ABCD is the required quadrilateral.