Question

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

Answer 1

In triangle $ABC$, using Pythagoras theorem

$A{C}^2=A{B}^2+B{C}^2$

$A{C}^2={12}^2+(6.5{)}^2$

$A{C}^2=144+42.5$

$A{C}^2=186.25$

$AC=13.6cms$

Also,

$AD=DC=\frac{AC}{2}=\frac{13.6}{2}=6.8cms$

Falling a perpendicular line from $D$ to $AB$, such that $ED$ is parallel to $BC$. $\angle AED=\angle ABC={90}^o$

Since $D$ is the mid point of $AC$ and $ED$ is parallel to $BC$, so applying mid pint theorem,

$ED=\frac{BC}{2}=\frac{6.5}{2}=3.25cms$

Now in triangle $ABD$, $ED$ is the altitude and $AB$ is the base

So the area of triangle $ABD=\frac{(AB\times ED)}{2}$

$Ar\left(ABD\right)=\frac{12\times 3.25}{2}$

$=\frac{39}{2}$

$=19.5sq.cms$