2009 UNCO Math Contest II Problems/Problem 8
Problem
Two diagonals are drawn in the trapezoid
forming four triangles. The areas of two of the
triangles are 
and 
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]](http://latex.artofproblemsolving.com/4/0/2/402684b2f1013e5842db62e6b3d685aa5698aec9.png)
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 64.
See also
| 2009 UNCO Math Contest II (Problems • Answer Key • Resources) | ||
| 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 | ||
