Question

Question 174
ABCD is a trapezium such that $AB\parallel CD$, $\angle A:\angle D=2:1,\angle B:\angle C=7:5$. Find the angles of the trapezium.

Answer 1

Let ABCD be a trapezium, where $AB\parallel CD$

Let the angles A and D be of measures 2x and x, respectively.
Then, $2x+x={180}^{\circ }$ [$\because$ in trapezium, the angles on either side of the base are supplementary]
$\Rightarrow 3x={180}^{\circ }\Rightarrow x={60}^{\circ }$
$\therefore \angle A=2\times {60}^{\circ }={120}^{\circ },\angle D={60}^{\circ }$
Again, let the angles B and C be 7x and 5x respectively. Then, $7x+5x={180}^{\circ }$
$\Rightarrow 12x={180}^{\circ }\Rightarrow x={15}^{\circ }$
Thus, $\angle B=7\times 15={105}^{\circ }$ and $\angle C=5\times 15={75}^{\circ }$