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 ABC is x.
    The area of the triangle ABC can be calculated using Heron's formula, since all sides of the triangles are known.
    S=AB+BC+CA2=x+x+x2=3x2 cm.
     The area of the ABC=S(SAB)(SBC)(SCA)=3x2(3x2x)(3x2x)(3x2x)=3x2(x2)(x2)(x2)=3x2(x2)3=3(x2)4=3(x2)2=34(x)2  (1)
  3.  The area of the triangle AOB=AB×OP2=x×b2=bx2
  4. Similarly, the area of the triangle BOC=ax2
    and the area of the triangle AOC=cx2
  5.  The area of the triangle ABC=Area(AOB)+Area(BOC)+Area(AOC)=bx2+ax2+cx2=(a+b+c)x2  (2)
  6. By comparing the equations (1) and (2), we get:
    34x2=(a+b+c)x2x=2(a+b+c)3  (3)
  7. Now, Area(ABC)=34(x)2
  8. Area(ABC)=AB× Altitude of the triangle ABC234(x)2×2=x× Altitude of the triangle ABC32(x)= Altitude of the triangle ABC
    By putting the value of x from the equation (3), we get,
    Altitude of the triangle ABC=32(2(a+b+c)3)
    Altitude of the triangle ABC=a+b+c
  9. Hence, the altitude of the triangle is a+b+c.

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