Question

Question 163
Find the value of x in the trapezium ABCD given below

Answer 1

Given, a trapezium ABCD in which $\angle A=(x-20{)}^{\circ },\angle D=(x+40{)}^{\circ }$
Since, in a trapezium, the angles between the parallel lines are supplementary,
$(x-20)+(x+40)={180}^{\circ }\Rightarrow x-20+x+40={180}^{\circ }\Rightarrow 2x+{20}^{\circ }={180}^{\circ }\Rightarrow 2x=({180}^{\circ }-{20}^{\circ })={160}^{\circ }\Rightarrow x={80}^{\circ }$