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In the figure below, isosceles \(\triangle ABC\) with base \(\overline{AB}\) has altitude \(CH = 24\) cm. , DE = GF, \(HF = 12\)  cm, and \(FB = 6\)  cm. What is the number of square centimeters in the area of pentagon \(CDEFG\)?

 

tertre  Feb 23, 2018
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 #1
avatar+12266 
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The area of the WHOLE figure is 1/2 b h    = 1/2 ( 36)(24) = 432 cm^2

Now subtract the small right triangles in the corners

-  2  x 1/2 x 6 x FG     

via similar triangles

CH/HB = FG/6    

24/18 = FG/6

FG = 8

 

- 2 x 1/2  x 6 x 8 = -48

 

Pentagon area = 432 - 48 = 384 cm^2

ElectricPavlov  Feb 23, 2018
 #2
avatar+2611 
+2

Great sol, EP! I understand it much better, now!

tertre  Feb 23, 2018

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