Question

$△\mathrm{ABC}$is an equilateral triangle with each side of length $2\mathrm{p}$. If $\mathrm{AD}\perp \mathrm{BC}$, find the value of $\mathrm{AD}.$

Answer 1

Finding the value of $\mathrm{AD}$:
Given,
$△\mathrm{ABC}$ is an equilateral triangle with each side of length $2\mathrm{p}.$
That means,$\mathrm{AB}=\mathrm{BC}=\mathrm{CA}=2\mathrm{p}$
Also, $\mathrm{AD}\perp \mathrm{BC}$
As we know, in an equilateral triangle, the perpendicular drawn from a vertex to the opposite side bisects the opposite side.
So,$\mathrm{BD}=\mathrm{DC}=\frac{1}{2}\mathrm{BC}=\mathrm{p}$
Now, triangle $△\mathrm{ADB}$ is a right triangle.
By Pythagoras theorem,
$\begin{array}{rcl}{\mathrm{AD}}^2+{\mathrm{BD}}^2& =& {\mathrm{AB}}^2\\ {\mathrm{AD}}^2+{\mathrm{p}}^2& =& {\left(2\mathrm{p}\right)}^2\\ {\mathrm{AD}}^2& =& 4{\mathrm{p}}^2–{\mathrm{p}}^2\\ {\mathrm{AD}}^2& =& 3{\mathrm{p}}^2\\ \mathrm{AD}& =& \mathrm{p}\sqrt{3}\end{array}$

Therefore, the value of $\mathrm{AD}=\sqrt{3}\mathrm{p}$.