### 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.

**Answer:**

^@a + b + c^@

**Step by Step Explanation:**

- The following figure shows the required triangle:
- 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}^@ - ^@\begin{align} \text { The area of the triangle } AOB & = \dfrac { AB \times OP }{2}\\ & = \dfrac { x \times b }{2}\\ & = \dfrac { bx }{2} \end{align}^@
- Similarly, the area of the triangle ^@\triangle BOC = \dfrac { ax }{2}^@

and the area of the triangle ^@\triangle AOC = \dfrac { cx }{2}^@ - ^@\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}^@
- 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}^@ - Now, ^@Area(\triangle ABC) = \dfrac { \sqrt { 3 } } {4}(x)^2 ^@
- ^@\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^@ - Hence, the altitude of the triangle is ^@a + b + c^@.