Question

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

Given,

$\angle A:\angle D=2:1$ and $\angle B:\angle C=7:5$

Let

$\angle A$ and $\angle B$ be $2x$ and $x$ and

$\angle C$ and $\angle D$ be $7y$ and $5y$

From the figure,

$\angle A+\angle D={180}^{\circ }$ (Alternate angles are supplementary)

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

$\Rightarrow 3x={180}^{\circ }$

$\Rightarrow x=\frac{{180}^{\circ }}{3}={60}^{\circ }$

And,

$\angle B+\angle C={180}^{\circ }$ (Alternate angles are supplementary)

$\Rightarrow 7y+5y={180}^{\circ }$

$\Rightarrow 12y={180}^{\circ }$

$\Rightarrow y=\frac{{18}^{\circ }}{12}={15}^{\circ }$

Hence,

$\angle A=2x=2\times {60}^{\circ }={120}^{\circ }$

$\angle D=x={60}^{\circ }$

$\angle B=7y=7\times {15}^{\circ }={105}^{\circ }$ and

$\angle C=5y=5\times {15}^{\circ }={75}^{\circ }$