Difference between revisions of "2009 UNCO Math Contest II Problems/Problem 8"
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== Solution == | == 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 <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 and as shown. What is the total area of the trapezoid?
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 .
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 |