In triangle ABC, AB = 17, AC = 8, and BC = 20. Let D be the foot of the altitude from C to AB. Find the area of triangle ACD.
+23246 Notice that CD falls outside the triangle; BA + AD = BD.
In triangle(BDC), BD2 + CD2 = BC2 In triangle(ADC), AD2 + CD2 = AC2
---> (BA + AD)2 + CD2 = BC2 AD2 + CD2 = 82
(17 + AD)2 + CD2 = 202 AD2 + CD2 = 64
289 + 34AD + AD2 + CD2 = 400
34AD + AD2 + CD2 = 111
Combining these two: 34AD + AD2 + CD2 = 111
AD2 + CD2 = 64
Subtracting: 34AD = 47 ---> AD = 47/34
In triangle(ADC), AD2 + CD2 = AC2
(47/34)2 + CD = 82
CD = 7.8796632...
Area of triangle(ACD) = ½·AD·DC = ...
