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What is the area, in square units, of triangle $ABC$ in the figure shown if points $A$, $B$, $C$ and $D$ are coplanar, angle $D$ is a right angle, $AC = 13$, $AB = 15$ and $DC = 5$?

 Apr 4, 2020
 #1
avatar+111321 
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Right  triangle  ACD  is a  5 -12 -13  Pythagorean Triple

 

So AD  = 12

 

Since AB  = 15.....then   BD  =sqrt  [ AB^2 - AD^2] = sqrt ( 15^2 - 12^2 ] = sqrt  [225 - 144] = sqrt [81] = 9

 

So....the  area  of triangle  ADB  = (1/2)(BD)(AD)  = (1/2)(9)(12)  = 54    (1)

 

And  the area of triangle ACD = (1/2)(5)(12) = 30    (2)

 

So the area of triangle ACB = (1)  - (2)  =  54  - 30  =  24

 

 

cool cool cool

 Apr 4, 2020

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