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Question

From a point in the interior of an equilateral triangle, perpendiculars are drawn on the three sides. The lengths of the perpendiculars are 14 cm, 10 cm, and 6 cm.
The area of the triangle is 300mcm2 where m is _______

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Solution

LetPbeanypointontheinteriorofanequilateraltriangleABC.JoinPtoall3verticesofthetriangle.Ondoingso,thetrianglegetsdividedintothreesmallertriangleswithheights6cm,10cm&14cmrespectively.Letthesideofthetrianglebey.Theareaofa=12×base×heightAreaofPAB=12×y×14=7yAreaofPBC=12×y×10=5yAreaofPCA=12×y×6=3ySo,thetotalareaoftheABC=PAB+PBC+PCA=7y+5y+3y=15y(1)Nowweknow,theareaofanequilateraltriangle=34×y2(whereyisthesideofthetriangle).(2)Henceequatingtheareas1&2weget,15y=34×y215=34×y[cancellingyfrombothsides]15×4=3×yy=603Theareaofthetriangle=15y=15×603=15×20×33=3003
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