Question

An equilateral triangle ABC has
AB=20√3. P is a point
placed in triangle ABC and D,E
and F are the foot of the
perpendiculars from P to AB, BC
and AC, respectively. If
PD=9 and PE=10, what is the value
of the length of PF? Please work
the complete solution.

Answer 1

= (1/2)*PD*AB + (1/2)*PE*BC + (1/2)*PF*AC
= (1/2)*AB*(PD + PE + PF)

Also, area of triangle ABC = (1/2)*height*AB
height = (√3/2)*20√3 = 30

So, PD + PE + PF = height = 30
=> PF = 11