### From a point in the interior of an equilateral triangle, perpendiculars are drawn on the three sides. If the lengths of the perpendiculars are $a, b,$ and $c$, find the altitude of the triangle.

$a + b + c$

Step by Step Explanation:
1. The following figure shows the required triangle:
2. Let's assume the side of the equilateral triangle $\triangle ABC$ is $x$.
The area of the triangle $\triangle ABC$ can be calculated using Heron's formula, since all sides of the triangles are known.
\begin{align} S & = \dfrac { AB + BC + CA }{2} \\ & = \dfrac { x + x + x }{2} \\ & = \dfrac { 3x }{2} \space cm. \end{align}
\begin{align} \text { The area of the } \triangle ABC & = \sqrt { S(S - AB)(S - BC)(S - CA) } \\ & = \sqrt{ \dfrac { 3x } { 2 }(\dfrac { 3x } { 2 } - x)(\dfrac { 3x } { 2 } -x)(\dfrac { 3x } { 2 } - x) } \\ & = \sqrt{ \dfrac { 3x } { 2 }(\dfrac { x } { 2 })(\dfrac { x } { 2 })(\dfrac { x } { 2 }) } \\ & = \sqrt{ \dfrac { 3x } { 2 }(\dfrac { x } { 2 })^3 } \\ & = \sqrt{ 3(\dfrac { x } { 2 })^4 } \\ & = \sqrt{ 3} {(\dfrac { x } { 2 })^2 } \\ & = \dfrac { \sqrt {3} }{4} (x)^2 \space \space ------(1) \end{align}
3. \begin{align} \text { The area of the triangle } AOB & = \dfrac { AB \times OP }{2}\\ & = \dfrac { x \times b }{2}\\ & = \dfrac { bx }{2} \end{align}
4. Similarly, the area of the triangle $\triangle BOC = \dfrac { ax }{2}$
and the area of the triangle $\triangle AOC = \dfrac { cx }{2}$
5. \begin{align} \text { The area of the triangle } \triangle ABC & = Area(\triangle AOB) + Area(\triangle BOC) + Area(\triangle AOC) \\ & = \dfrac { bx }{2} + \dfrac { ax }{2} + \dfrac { cx }{2} \\ & = \dfrac { (a + b + c)x }{2} \space \space -----(2) \end{align}
6. By comparing the equations $(1)$ and $(2),$ we get:
\begin{align} \dfrac { \sqrt { 3 } }{4}x^2 & = \dfrac { (a + b + c)x }{2}\\ \implies x & = \dfrac { 2(a + b + c) } { \sqrt { 3 } } \space \space------(3) \end{align}
7. Now, $Area(\triangle ABC) = \dfrac { \sqrt { 3 } } {4}(x)^2$
8. \begin{align}& Area(\triangle ABC) = \dfrac { AB \times \text { Altitude of the triangle } \triangle ABC }{2} \\ & \implies \dfrac { \sqrt { 3 } }{4}(x)^2 \times 2 = x \times \text { Altitude of the triangle } \triangle ABC \\ & \implies \dfrac { \sqrt { 3 } }{2}(x) = \text { Altitude of the triangle } \triangle ABC \\ \end{align}
By putting the value of $x$ from the equation $(3)$, we get,
Altitude of the triangle $\triangle ABC = \dfrac { \sqrt { 3 } }{2} (\dfrac { 2(a + b + c) } { \sqrt { 3 } })$
$\implies$ Altitude of the triangle $\triangle ABC = a + b + c$
9. Hence, the altitude of the triangle is $a + b + c$.