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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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