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
- The following figure shows the required triangle:
- 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(S−AB)(S−BC)(S−CA)=√3x2(3x2−x)(3x2−x)(3x2−x)=√3x2(x2)(x2)(x2)=√3x2(x2)3=√3(x2)4=√3(x2)2=√34(x)2 −−−−−−(1) - The area of the triangle AOB=AB×OP2=x×b2=bx2
- Similarly, the area of the triangle △BOC=ax2
and the area of the triangle △AOC=cx2 - The area of the triangle △ABC=Area(△AOB)+Area(△BOC)+Area(△AOC)=bx2+ax2+cx2=(a+b+c)x2 −−−−−(2)
- By comparing the equations (1) and (2), we get:
√34x2=(a+b+c)x2⟹x=2(a+b+c)√3 −−−−−−(3) - Now, Area(△ABC)=√34(x)2
- Area(△ABC)=AB× Altitude of the triangle △ABC2⟹√34(x)2×2=x× Altitude of the triangle △ABC⟹√32(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 - Hence, the altitude of the triangle is a+b+c.