Question

In the given figure, $ABCD$ is a trapezium where $AB||DC$ and $AD=BC$. If $\angle CBA=82°$, then the measure of $\angle ADC$ is equal to

• A. $85°$
• B. $82°$
• C. $98°$
• D. $88°$

Answer 1

The correct option is C $98°$


Here, AD = BC.
Thus, ABCD is an isosceles trapezium.
$\therefore \angle DAB=\angle CBA$
$\Rightarrow \angle DAB=82°$ …..(i)
Since $AB||CD$ and AD is transversal.
$\therefore \angle ADC+\angle DAB=180°$ (Pair of interior angles on the same side of transversal)
$\Rightarrow \angle ADC=180°–82°$ [From (i)]
$\Rightarrow \angle ADC=98°$
Hence, the correct answer is option (c).