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?
+128407 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
