Question

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

• A. 19.5 sq. cm
• B. 20.5 sq. cm
• C. 21 sq. cm
• D. 16 sq. cm

Answer 1

The correct option is A 19.5 sq. cm

In triangle ABC, using Pythagoras theorem
$A{C}^2=A{B}^2+B{C}^2$
$A{C}^2={12}^2+(6.5{)}^2$
$A{C}^2=144+42.5$
$A{C}^2=186.25$
AC = 13.6 cm
Also,
AD = DC = $\frac{1}{2}AC=\frac{1}{2}\times 13.6=6.8cm$
Falling a perpendicular line from D to AB, such that ED is parallel to BC.

$\angle AED=\angle ABC={90}^{\circ }$
Since D is the mid point of AC and ED is parallel to BC, so applying mid pint theorem,
$ED=\frac{1}{2}BC=\frac{1}{2}\times 6.5=3.25cm$
Now in triangle ABD, ED is the altitude and AB is the base
So the area of triangle ABD $=\frac{1}{2}(AB\times ED)$
ar(ABD) $=\frac{1}{2}(12\times 3.25)$
$=\frac{39}{2}$
= 19.5 sq. cm