Question

Solve the following :
Find the value of $x$ in the trapezium ABCD given below.

Answer 1

Here ,

$\angle A$ $=(x-20{)}^{\circ },\angle D$ $=(x+40{)}^{\circ }$

$ABCD$ is trapezium,

$\Rightarrow AB||CD$, so $A$ and $D$ are supplementary , therefore-

$(x-20)+(x+40)={180}^{\circ }$

$2x+{20}^{\circ }={180}^{\circ }$

$2x=(180-20{)}^{\circ }$

$2x={160}^{\circ }$

$x={80}^{\circ }$