Triangle ABC is an acute angled triangle. D is the foot of perpendicular from A to BC. If AD=8,BC=11and (AB)^2-(AC)^2=66,find the area of triangle ACD.

Question

I hate these questions

Answer 1

Let DC = x

BD = 11-x

AB^2 = 66 + AC^2

AB^2 = 8^2 + (11-x)^2

AC^2 + 66 = 64 + x^2 - 22x + 121

AC^2 =64 + x^2

subtract last from 2nd to last to get

66 = -22x + 121

22x = 55

x = 5/2

Area of ADC = (5/2)(8)/2 = 10