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:
  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^@.

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