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 $\mathrm{AD}.$

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Solution

## Step 1: Find the area of the given triangleGiven: $△\mathrm{ABC}$ 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)$$⇒$ $30=\frac{1}{2}\left(13\right)\left(\mathrm{AD}\right)$$⇒$ $\mathrm{AD}=\frac{30×2}{13}$$⇒$ $\mathrm{AD}=4.6\mathrm{cm}$Hence, area of $△\mathrm{ABC}$ is $30{\mathrm{cm}}^{2}$ and length of $\mathrm{AD}$ is $4.6\mathrm{cm}$.

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