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Triangles $BAD$ and $BDC$ are right triangles with $AB = 12$ units, $BD = 15$ units, and $BC = 17$ units. What is the area, in square units, of quadrilateral $ABCD$?

 Apr 4, 2020
edited by helpppp  Apr 4, 2020
 #1
avatar+111329 
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In right triangle  ABD, AD  = sqrt [ BD^2  - BA^2] =  sqrt [ 15^2  -  12^2]  =sqrt [ 225 -144] = sqrt [81  = 9

 

So...its area = (1/2)product of the legs  =(1/2) BA * AD =  (1/2)12 * 9  =  54

 

And  in right triangle BDC, DC = sqrt [ BC^2  - BD^2] = sqrt  [ 17^2  - 15^2] = sqrt [ 289 - 225] = sqrt [ 64] = 8

 

So....the  area of triangle  BDC  =(1/2) (8)(15)  = (1/2) (120) = 60

 

So....the  area of ABCD =   54 + 60 =   114

 

cool cool cool

 Apr 4, 2020
 #2
 #3
avatar+111329 
+1

Thanks, Cal   !!!!

 

cool cool cool

CPhill  Apr 4, 2020

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