Geometry Problem. Post your solution in the comments box below.

Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to see the complete problem 880.

## Thursday, May 23, 2013

### Problem 880: Triangle, Midpoints, Sides, Perpendiculars, Hexagon, Area

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Let S(PQR) be the area of ΔPQR.

ReplyDeleteForm ΔDEF.

Since ΔBFD, ΔFAE and ΔDEC are congruent,

and H, G, M are their orthocenters.

Thus

S(HFD)+S(FGE)+S(DEM) = S(DEF)

So

Area of hexagon = 2×S(DEF) = 12

Connect EF, FD and DE

ReplyDeleteNote that area(BFD)=Area(AFE)=Area(CED)=Area(EFD)=6

Triangles BFD, DEC and FAE are congruence (case SSS)

Triangle EDM congruence to triangle FBH ( case ASA)

Triangle FGE congruence to triangle BHD ( case ASA)

So Area(EGFHDM)= Area(EFD)+Area(FBD)= 12

join F to D, D to E; A to G. Tr FHD = Tr AGE, Tr EMD = tr AGF

ReplyDelete=> Sx = S (AFDE) = 1/2 S (ABC) = 12

A beautiful solution I have :consider the triangle FDE WE have FD=1/2AC. and DE=1/2AB and FE=1/2BC

ReplyDeletethen the area of triangle FDE is FD×DE×sin^D/2(1) and WE know that ^d=^A(FDEA is parallellogram)

Consider the triangle ABC WE have AB×AC×sinA=2×24=48 so sinA=48/AB×AC(2)

then WE get two equations have same sin

(1)and(2)

So the area of triangle FDE 1/2AC×1/2AB×48/AB×AC=6

the other triangles is similar (∆FDE~DEC)and(∆AFE~FDB)

They are same area 6

Let S'area of traingle FHD and S" the are of traingle EMD and S'''thé Area of triangle FGE

triangle BHD is similar to DMC

and BHF is similar to EMD

and FHD is similar to EMC

SO S'+S"+S'''=6

And the are of hexagon is the total area of four triangle FHD.FGE.EMD and FED

SO S=12