Question

From a point in the interior of an equilateral triangle,perpendiculars are drawn on th three sides.the lenths of the prependiculars are 14cm,10cm and 6cm.find the area of the triangle.

Answer 1

Let ABC be a equilateral triangle
Let the side of equilateral triangle be a cm
P is a point in the interior of the triangle. PQ is perpendicular to BC and PR is perpendicular to CA and PS is perpendicular to AB.
Let PS= 14 cm, PQ=10cm, and PR=6 cm 
Area of triangle ABC= area of APB+ area of BPC+ area of CPA
√3/4*a²=1/2*AB*PS+1/2*BC*PQ +1/2*CA*PR
√3/4*a²=1/2*a*14+1/2*a*10+1/2*a*6
√3/4*a²=7a+5a+3a
√3/4*a²=15a
a=60/√3
Therefore side of equilateral side are 60/√3
Area =√3/4*(60/√3)²=√3/4*(60*60/3)=300√3

Hope I helped!!

Answer 2

$300\sqrt{3}cm^{2}$ is the answer

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