Triangle ABC has AB = BC = 5 and AC = 6. Let E be the foot of the altitude from B to AC and let D be the foot of the altitude from A to BC. Compute the area of triangle DEC.
Triangle ABC has AB = BC = 5 and AC = 6. Let E be the foot of the altitude from B to AC and let D be the foot of the altitude from A to BC. Compute the area of triangle DEC.
AB = BC = 5
AC = 6
DC = 1/2 BC = 2.5
MD = 1/2 AC = 3
Let "F" be a midpoint of EC
DF = sqrt [(DC)² - (FC)²] = 2
The area of a triangle DEC = (DF) * (FC) = 3 u² 
+1486
AB=BC=5 AC=6
DC=? DC = cos(ACB) * AC = 3.6
Angle ACB = ? cos(ACB) = 3/5 ∠ ACB = 53.13°
Let F be the foot of the altitude from D to AC
DF = sin(ACB) * DC DF = 2.88
FC = cos(ACB) * DC FC = 2.16
EF = 3 - FC EF = 0.84
Area of Δ DFC = (DF * FC)/2 = 3.11 u²
Area of Δ DEF = (DF * EF)/2 = 1.21 u²
Area of Δ DEC = 3.11 + 1.21 = 4.32 u² 
