Question

$△\mathrm{ABC}$ is right angled at $\mathrm{A}$. $\mathrm{AD}$ is perpendicular to $\mathrm{BC}$. If $\mathrm{AB}=5\mathrm{cm}$, $\mathrm{BC}=13\mathrm{cm}$ and $\mathrm{AC}=12\mathrm{cm}$. Find the area of $△\mathrm{ABC}$. Also find the length $\mathrm{AD}.$

Answer 1

Step 1: Find the area of the given triangle

Given: $△\mathrm{ABC}$ is right angled at $\mathrm{A}$, $\mathrm{AB}=5\mathrm{cm}$, $\mathrm{BC}=13\mathrm{cm}$ and $\mathrm{AC}=12\mathrm{cm}$

As area of a right-angled triangle $=\frac{1}{2}\mathrm{bh}$, where $\mathrm{b}$ is base and $\mathrm{h}$ is height.

So, area of $△\mathrm{ABC}$$=\frac{1}{2}\left(\mathrm{AB}\right)\left(\mathrm{AC}\right)$

$=\frac{1}{2}\left(5\right)\left(12\right)$

$=30{\mathrm{cm}}^2$

Step 2: Find the length of $\mathrm{AD}.$

From the figure, Area of $△\mathrm{ABC}$$=\frac{1}{2}\left(\mathrm{BC}\right)\left(\mathrm{AD}\right)$

$\Rightarrow$ $30=\frac{1}{2}\left(13\right)\left(\mathrm{AD}\right)$

$\Rightarrow$ $\mathrm{AD}=\frac{30\times 2}{13}$

$\Rightarrow$ $\mathrm{AD}=4.6\mathrm{cm}$

Hence, area of $△\mathrm{ABC}$ is $30{\mathrm{cm}}^2$ and length of $\mathrm{AD}$ is $4.6\mathrm{cm}$.