Difference between revisions of "2009 UNCO Math Contest II Problems/Problem 8"

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== Solution ==
 
== Solution ==
{{solution}}
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By definition of a trapezoid, the two bases are parallel. Because the intersecting lines are transversals of the two parallel lines, we know that the two triangles with known areas are similar by angle-angle-angle similarity. Noticing that the areas of the triangles are square numbers, we know that the base of the triangle must be 2 times an integer and that that integer squared is the area of the triangle. We find that the bases are 6 and 10 and the heights are 3 and 5. Using that formula for the area of a trapezoid, we find that the area of the trapezoid is ((6+10)/2)*(5+3) and this is 8*8, which is <math>\boxed{64}</math>.
  
 
== See also ==
 
== See also ==

Latest revision as of 02:20, 29 January 2019

Problem

Two diagonals are drawn in the trapezoid forming four triangles. The areas of two of the triangles are $9$ and $25$ as shown. What is the total area of the trapezoid?

[asy] draw((0,0)--(20,0)--(2,4)--(14,4)--(0,0),black); draw((0,0)--(2,4)--(14,4)--(20,0),black); MP("25",(9,.25),N);MP("9",(9,2.25),N); [/asy]


Solution

By definition of a trapezoid, the two bases are parallel. Because the intersecting lines are transversals of the two parallel lines, we know that the two triangles with known areas are similar by angle-angle-angle similarity. Noticing that the areas of the triangles are square numbers, we know that the base of the triangle must be 2 times an integer and that that integer squared is the area of the triangle. We find that the bases are 6 and 10 and the heights are 3 and 5. Using that formula for the area of a trapezoid, we find that the area of the trapezoid is ((6+10)/2)*(5+3) and this is 8*8, which is $\boxed{64}$.

See also

2009 UNCO Math Contest II (ProblemsAnswer KeyResources)
Preceded by
Problem 7
Followed by
Problem 9
1 2 3 4 5 6 7 8 9 10
All UNCO Math Contest Problems and Solutions
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