From a given point in the interior of an equilateral triangle,perpendiculars are drawn on the three sides.the lengths of the perpendicularare 14cm,10cm & 6cm. Find the area of triangle.

Asked by araima2001 | 14th Sep, 2014, 05:02: PM

Expert Answer:

∆ABC is an equilateral triangle.

Let ABC be equilateral triangle of side 'a' cm.
 
Let P be a point in the interior of the ∆ABC. so PQ⊥BC, PR⊥CA and PS⊥AB
 
So PS = 14 cm, PQ = 10 cm and PR = 6 cm
 
Area of ∆ABC = Area of ∆APB + Area of ∆BPC + Area of ∆CPA
 
fraction numerator square root of 3 over denominator 4 end fraction a squared equals 1 half open square brackets P S cross times A B plus P Q cross times B C plus P R cross times A C close square brackets rightwards double arrow fraction numerator square root of 3 over denominator 4 end fraction a squared equals 1 half cross times a cross times open square brackets 14 plus 10 plus 6 close square brackets rightwards double arrow fraction numerator square root of 3 over denominator 4 end fraction a squared equals 1 half cross times a cross times 30 rightwards double arrow fraction numerator square root of 3 over denominator 4 end fraction a squared equals 15 a rightwards double arrow a equals fraction numerator 15 cross times 4 over denominator square root of 3 end fraction equals fraction numerator 60 over denominator square root of 3 end fraction space c m  A r e a space o f space t r i a n g l e space equals space fraction numerator square root of 3 over denominator 4 end fraction cross times fraction numerator 60 over denominator square root of 3 end fraction cross times fraction numerator 60 over denominator square root of 3 end fraction equals 300 square root of 3 space s q. c m

Answered by Dharma Teja | 14th Sep, 2014, 11:25: PM

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