2006 UNCO Math Contest II Problems/Problem 5

Revision as of 16:07, 18 December 2018 by Hashtagmath (talk | contribs) (Added solution to the problem)


In the figure $BD$ is parallel to $AE$ and also $BF$ is parallel to $DE$. The area of the larger triangle $ACE$ is $128$. The area of the trapezoid $BDEA$ is $78$. Determine the area of triangle $ABF$.

[asy] draw((0,0)--(1,2)--(4,0)--cycle,black); draw((1/2,1)--(2.5,1)--(2,0),black); MP("A",(4,0),SE);MP("C",(1,2),N);MP("E",(0,0),SW); MP("D",(.5,1),W);MP("B",(2.5,1),NE);MP("F",(2,0),S); [/asy]


Since we know $BF$ is equal to the short side of both triangles and that $DB$ is equal to the long sides of the hypotenuse of the triangle. Thus we know that if we make a line $DF$ (as seen below), all four triangles in $\triangle ACE$ are congruent.

[asy] draw((0,0)--(1,2)--(4,0)--cycle,black); draw((1/2,1)--(2.5,1)--(2,0),black); draw((1/2,1)--(2,0),black); MP("A",(4,0),SE);MP("C",(1,2),N);MP("E",(0,0),SW); MP("D",(.5,1),W);MP("B",(2.5,1),NE);MP("F",(2,0),S); [/asy]

Then, since we know the sum of all four triangles is $128$, we can solve the easy division problem $128/4$ and see that the area of $\triangle ABF = \boxed{32}$

See Also

2006 UNCO Math Contest II (ProblemsAnswer KeyResources)
Preceded by
Problem 4
Followed by
Problem 6
1 2 3 4 5 6 7 8 9 10
All UNCO Math Contest Problems and Solutions
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