From a point in the interior of an equilateral triangle, perpendiculars are drawn on the three sides. The lengths of the perpendiculars are 4 cm, 6 cm, and 2 cm. Find the area of the triangle.


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Answer:

48√3 cm2

Step by Step Explanation:
  1. The following figure shows the required triangle,

    Let's assume the sides of the equilateral triangle ΔABC be x.
    The area of the triangle ΔABC can be calculated using Heron's formula since all sides of the triangles are known.
    S =  
    AB + BC + CA
    2
     
    =  
    x + x + x
    2
     
    =  
    3x
    2
      cm.
    The area of the ΔABC = √S(S - AB)(S - BC)(S - CA)
    = ^@ \sqrt{ \dfrac {3x} { 2 } \left(\dfrac {3x} { 2 } - x \right) \left(\dfrac {3x} { 2 } -x \right) \left(\dfrac {3x} { 2 } - x \right) } ^@
    = ^@ \sqrt{ \dfrac {3x} { 2 } \left(\dfrac {x} { 2 }\right)\left(\dfrac {x} { 2 }\right)\left(\dfrac {x} { 2 }\right) } ^@
    = ^@ \sqrt{ \dfrac {3x} { 2 }\left(\dfrac {x} { 2 }\right)^3 } ^@
    = ^@ \sqrt{ 3\left(\dfrac {x} { 2 }\right)^4 } ^@
    = ^@ \sqrt{ 3} {\left(\dfrac {x} { 2 }\right)^2 } ^@
    =  
    √3
    4
     (x)2 ------(1)
  2. The area of the triangle AOB =  
    AB × OP
    2
     
    =  
    'x' × 6
    2
     
    =  
    6x
    2
     
  3. Similarly, the area of the triangle ΔBOC =  
    4x
    2
     
    and the area of the triangle ΔAOC =  
    2x
    2
     .
  4. The the area of the triangle ΔABC = Area(ΔAOB) + Area(ΔBOC) + Area(ΔAOC)
    =  
    6x
    2
      +  
    4x
    2
      +  
    2x
    2
     
    =  
    12x
    2
      -----(2)
  5. By comparing equation (1) and (2), we get,
     
    √3
    4
     (x)2 =  
    12x
    2
     
    ⇒ x =  
    24
    √3
     
  6. Now, Area(ΔABC) =  
    √3
    4
     (x)2
    =  
    √3
    4
     ( 
    24
    √3
     ) 2
    =  
    144
    √3
      = 48 √3 cm2
  7. Hence, the area of the triangle is 48√3 cm2.

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