Question

In $△$ABC, $\angle$ABC $={90}^{\circ }$; AD = DC; AB =12 cm, BC = 6.5 cm. Find the area of $△$ADB.

Answer 1

$△ADB=\frac{1}{2}△ABC[\because BD\text{is a median of}△ABC]$
$=\frac{1}{2}\times [\frac{1}{2}\times AB\times AC]$
$=\frac{1}{4}\times 12\times 6.5$
$=19.5{\text{cm}}^2$