Question

The following figure shows a quadrilateral with sides AB and DC parallel to each other.

If $\angle A:\angle D=4:5$

$\angle B=(3x-15{)}^{\circ }and$

$\angle C=(4x+20{)}^{\circ }$

Which of the following options is/are correct ?

• A. $\angle A={100}^{\circ }$ $\angle B={60}^{\circ }$ $\angle C={120}^{\circ }$ $\angle D={80}^{\circ }$
• B. $\angle A={120}^{\circ }$ $\angle B={60}^{\circ }$ $\angle C={120}^{\circ }$ $\angle D={80}^{\circ }$
• C. $\angle A={80}^{\circ }$ $\angle B={60}^{\circ }$ $\angle C={120}^{\circ }$ $\angle D={100}^{\circ }$
• D. $\angle A={80}^{\circ }$ $\angle B={120}^{\circ }$ $\angle C={60}^{\circ }$ $\angle D={100}^{\circ }$

Answer 1

The correct option is C

$\angle A={80}^{\circ }$

$\angle B={60}^{\circ }$

$\angle C={120}^{\circ }$

$\angle D={100}^{\circ }$

The given figure is a trapezium as it has one pair of parallel sides.

$Given:\angle B=(3x-15{)}^{\circ }and\angle C=(4x+20{)}^{\circ }$

$\Rightarrow \angle B+\angle C={180}^{\circ }(Co-interiorangles)$

$\Rightarrow (3x-15)+(4x+20)={180}^{\circ }$

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

$\Rightarrow 7x={180}^{\circ }-5^{\circ }={175}^{\circ }$

$\Rightarrow x=\frac{{175}^{\circ }}{7}={25}^{\circ }$

$\therefore$

$\Rightarrow \angle B=3x-15=3\times 25-15=75-15$

$\angle B={60}^{\circ }$

$\Rightarrow \angle C=4x+20=4\times 25+20=100+20$

$\angle C={120}^{\circ }$

$Given:\angle A:\angle D=4:5$

Let $\angle A=4y,\angle D=5y$

$\Rightarrow \angle A+\angle D={180}^{\circ }(Co-interiorangles)$

$\Rightarrow 4y+5y={180}^{\circ }$

$\Rightarrow 9y={180}^{\circ }$

$\Rightarrow y={20}^{\circ }$

$\therefore$

$\Rightarrow \angle A=4y=4\times 20$

$\angle A={80}^{\circ }$

$\Rightarrow \angle D=5y=5\times 20$

$\angle D={100}^{\circ }$