Question

The following figure shown a trapezium in which sides AB and DC are parallel. If $\angle A:\angle D=4:5,\angle B=(3x-15{)}^{\circ }$ and $\angle C=(4x+20{)}^{\circ }$, find each angle of the trapezium ABCD.

• A. $\angle A={80}^{\circ },\angle B={60}^{\circ },\angle C={120}^{\circ },$ and $\angle D={100}^{\circ }$
• B. $\angle A={60}^{\circ },\angle B={60}^{\circ },\angle C={120}^{\circ },$ and $\angle D={120}^{\circ }$
• C. $\angle A={70}^{\circ },\angle B={60}^{\circ },\angle C={120}^{\circ },$ and $\angle D={110}^{\circ }$
• D. None of the above

Answer 1

The correct option is A $\angle A={80}^{\circ },\angle B={60}^{\circ },\angle C={120}^{\circ },$ and $\angle D={100}^{\circ }$

In trapezium ABCD sides AB and DC are parallel and $\angle A:\angle D=4:5,\angle B=(3x-15{)}^o$ and $\angle C=(4x+20{)}^o,$

$\angle A+\angle D={180}^o$ (Sum of angles on the same sides of a transversal are supplementary.
$4y+5y={180}^o$

$y={20}^o$

$\angle A=4y={80}^o$ and $\angle D=5y={100}^o$

$\angle A+\angle B+\angle C+\angle D={360}^o$

${80}^o+(3x-15{)}^o+(4x+20{)}^o+{100}^o={360}^o$

$(7x+5{)}^o={180}^o$

$x={25}^o$

$\angle B=(3x-15{)}^o={60}^o$ and $\angle C=(4x+20{)}^o={120}^o$

$\angle A={80}^o,\angle B={60}^o,\angle C={120}^o,\angle D={100}^o$