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From the figure below,  ABC and DEF are right triangles, and AF and DC are the their respective altitudes. Point G is the intersection of AC and DF and GH is drawn from it such that it is perpendicular to BC. If  AF = 6  and GH = 4 and FC = 9, what is the area of polygon AGDEB?

 

 Jan 4, 2021
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Triangles AFC  and  GHC  are similar

 

So  AF /FC  = GH/HC

AF/ GH  = FC /HC

6/4 =  FC /HC

3/2  = FC /HC

3/2= 9/HC

(3/2)HC  =FC

(3/2) HC  = 9

HC  = 6

So FH  = 3

 

And triangle GFH is similar to triangle DFC

So

FC/ FH  = DC/ GH

9/3 =  DC/ 4

3 = DC/ 4

DC  = 12

 

And we have  these relationships

BF/ AF = AF/ FC                               FC /DC = DC / CE

FC * BF = AF^2                                 CE * FC =  DC^2

9 BF  =  6^2                                       CE * 9 = 12^2                                        

BF  = 36/9  = 4                                  CE  =144/9   =  16

 

Area of the polygon is  =  area of ABC  +  area of FDE  - area of FGC  =

 

(1/2)(BF + FC) (AF)  +  (1/2) (FC + CE) (DC)  - (1/2) (FC)(GH)   =

 

(1/2) (4 + 9)(6) + (1/2) (9 + 16)(12)   -  (1/2)(9)(4)  =

 

       39       +  150    -   18  =

 

171 units^2

 

 

cool cool cool

 Jan 4, 2021
edited by CPhill  Jan 4, 2021

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