See complete Problem 166 at:

www.gogeometry.com/problem/p166_parallelogram_triangle_area.htm

Parallelogram, Diagonal, Triangles, Areas. Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.

## Friday, August 29, 2008

### Elearn Geometry Problem 166

Labels:
area,
diagonal,
interior point,
parallelogram,
triangle

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draw diagonal AC which meets BD at G and BE at H.also join GE.

ReplyDeleteproof:we know that in a parallelogram diagonals bisect each other.so AG=GC and BG=GD.

also median of a triangle divides it into two triangles of equal area.

[tri BGE]=[tri GED]

[tri BGH]+[tri GHE]=[tri GEF]+[tri BFD]-----( 1 )

[tri AEG]= [tri GEC]

[tri AFG]+[tri GEF]=[tri GHE]+[tri HEC]------(2)

On adding 1 and 2 equations

[tri BGH]+[tri AFG]=[tri EFD]+[tri HEC]

On adding [tri AGB] on either side

[tri ABF]+[tri BGH]=[tri AGB]+[tri EFD]+[tri BHC]

but diagonals of a parallelogram divide it into four triangles of equal area.

[tri AGB]=[tri BGC] substitute in the above equation

and after simplifications we get the required result. S=S1+S2

Let A be (0,0), B;(b,h), C:(b+d,h) and D:(d,0) and let E be (p,q) Now we can determine the F as

ReplyDelete(dhp/(ph-qb+qd),dhq/(ph-qb+qd)). Now Tr. AFD =d*dhq/2(ph-qb+qd)= hqd^2/(ph-qb+qd) -----(1)

While, Tr. BFE =(1/2)(bdhq/(ph-qb+qd) -bq+dhpq/(ph-qb+qd)-pdh^2/(ph-qb+qd) +ph -pdhq/(ph-qb+qd))and Tr. ECD =(dh+bq-hp)/2

Now Tr. BFE + Tr. ECD = hqd^2/(ph-qb+qd) upon addition and simplification. In other words, Tr. AFD = Tr. BFE + Tr. ECD because of (1)

In parallelogram ABCD, Tr. ABD = Tr. BDC or S + Tr. AFD = S1 + S2 + Tr. BFE + Tr. ECD or S = S1 + S2 since Tr. AFD = Tr. BFE + Tr. ECD

Ajit: ajitathle@gmail.com

Es mucho más fácil: aplicación directa del teorema de la alfombra.

ReplyDeletePor un lado: [ABD]= S+[AFD]=0'5*[ABCD]

Y por otro: S1+S2+[AFD]=0'5*[ABCD]

Por tanto, S=S1+S2

Solution to problem 166.

ReplyDeleteLet be h1 and h2 the distances from E to lines BC and AD, respectively, and h3 the distance from F to line AD. We have BC = AD.

Then S1 = BC.h1/2 = AD.h1/2,

S2 = S(ADE) – S(ADF) = AD.h2/2 – AD.h3/2

and S = S(ABD) – S(ADF) = AD(h1+h2)/2 – AD.h3/2 = AD(h1+h2-h3)/2.

Hence S1 + S2 = AD.h1/2 + AD.h2/2 – AD.h3/2 = AD(h1+h2-h3)/2 = S.

To Antonio Ledesma:

ReplyDeleteWhy is it S1+S2+[AFD] = 0,5*[ABCD]?

(I know that "alfombra" means carpet - "tapete", in portuguese, but I don't know the "alfombra" theorem)

Thanks.