Difference between revisions of "2006 UNCO Math Contest II Problems/Problem 5"
Mathisfun04 (talk | contribs) (→Solution) |
Hashtagmath (talk | contribs) (Added solution to the problem) |
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==Solution== | ==Solution== | ||
− | { | + | Since we know <math>BF</math> is equal to the short side of both triangles and that <math>DB</math> is equal to the long sides of the hypotenuse of the triangle. Thus we know that if we make a line <math>DF</math> (as seen below), all four triangles in <math>\triangle ACE</math> 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 <math>128</math>, we can solve the easy division problem <math>128/4</math> and see that the area of <math>\triangle ABF = \boxed{32}</math> | ||
==See Also== | ==See Also== |
Revision as of 16:07, 18 December 2018
Problem
In the figure is parallel to and also is parallel to . The area of the larger triangle is . The area of the trapezoid is . Determine the area of triangle .
Solution
Since we know is equal to the short side of both triangles and that is equal to the long sides of the hypotenuse of the triangle. Thus we know that if we make a line (as seen below), all four triangles in are congruent.
Then, since we know the sum of all four triangles is , we can solve the easy division problem and see that the area of
See Also
2006 UNCO Math Contest II (Problems • Answer Key • Resources) | ||
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 |