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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.

 Feb 8, 2020
 #1
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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  indecision

 Feb 8, 2020
 #2
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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   smiley

Dragan  Feb 8, 2020
edited by Dragan  Feb 8, 2020

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