Geometry Problem

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

Click the figure below to see the complete problem 771.

## Friday, June 22, 2012

### Problem 771: Area of a Parallelogram, Star, Pentagon, Triangles

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http://img266.imageshack.us/img266/9572/problem771.png

ReplyDeleteAssign areas per attached sketch

We have Area(EBC)=S2+S3+P1= 1/2. BC*h1

Area(EAD)=S4+S5+P2= ½. AD * h2

So S2+S3+S4+S5+P1+P2=1/2. BC*(h1+h2)= ½.BC*h = ½ Area(ABCD)

Area(ABG)=1/2*h*BG=Area(AGF) => S1=S6+P1+P2

Combine above 2 expressions we get S1+S2+S3+S4+S5-S6= ½.Area(ABCD)

Let the area of "white" triangle with a vertex at F and at G be a and b respectively.

ReplyDeleteFirst, S(BCE) + S(ADE) = S/2,

(S2 + b + S3) + (S4 + a + S5) = S/2

(S2 + S3 + S4 + S5) + (a + b) = S/2.....(1)

Now consider trapezium ABGF,

let AG and BF intersect at H.

Then S(ABH) = S(FGH),

S1 = S6 + (a + b)

a + b = S1 - S6.....(2)

Substitute (2) into (1),

S1 + S2 + S3 + S4 + S5 - S6 = S/2