Question

In a given figure,ABCD is a trapezium in which the parallel sides AB and CD are both perpendicular to BC.If $AB=16,AD=17$ and $CD=8$,then BC is

• A. $15$
• B. $25$
• C. $25.5$
• D. $39$

Answer 1

The correct option is A $15$

$AL=AB-BL=AB-CD=16-8=8$

$\therefore D{L}^2=A{D}^2-A{L}^2={17}^2-8^2=225$

Thus $DL=15$

As $DL=BC$

$\therefore$ $BC=15unit$