in an equilateral triangle O is any point inside the triangle and perpendiculars are drawn from O.
Asked by | 10th Dec, 2009, 10:55: PM
the area of thetriangle ABC gets divided into threeparts i.e.
area of OAB,OBC and OCA.
let the lengths of the perpendiculars to AB,BC and CA be x ,y,z resply.
1/2[(AB)(x)]+1/2[(BC)(y)]+1/2[(CA)(z)]=area of ABC
but AB=BC=CA as the triangle is equilateral.
let AB=BC=CA=p (say)
x+y+z=2(area of ABC)/p
since area of the triangle ABC is fixed for given p, the right hand side is a constant number.
so the sum of the perpendiculars
x+y+z is also a constant.
Answered by | 1st Jan, 2010, 07:30: PM
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