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32
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avatar+819 

In the diagram, $AB\perp BC$ and $BC \perp CD$. $AB = 6'', BC = 4''$, and $CD = 10''$. $E$ is the midpoint of $AD$. How many square inches are in the area of $ABCD$?

 

 Sep 25, 2023
 #1
avatar+128732 
+1

Draw AC

 

Triangle BAC is right with angle ABC  = 90

 

AC  = sqrt [AB^2 + BC^2] =  sqrt [ 6^2 + 4^2 ]  = sqrt [ 52 ] 

 

sin [angle BCA ]  =  6/sqrt [52]

 

Angle BCD =  90

 

Angle BCD = angle BCA + angle ACD

 

sin [ angle BCA ]  = cos angle [ACD ]

 

cos [angle ACD ] =  6/sqrt 52

 

sin [angle ACD ] =  sqrt [  1  - (6/sqrt 52)^2 ] =  sqrt [ 1 - 36/52 ] = sqrt [ 16/52]  = 4 / sqrt 52

 

[ ABCD]  = 

 

[ABC] + [ACD]  =

 

(1/2)(BA)(BC)  + (1/2) (AC) (CD) ( 4 /sqrt 52)  = 

 

(1/2)(6) (4)  + (1/2) (sqrt 52) ( 10) (4 /sqrt 52)  =

 

12  +   20  = 

 

32

 

 

cool cool cool

 Sep 25, 2023

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