1997 AHSME Problems/Problem 15
Contents
Problem
Medians and of triangle are perpendicular, , and . The area of triangle is
Solution 1
One median divides a triangle into equal areas, so all three medians will divide a triangle into equal areas.
The median is divided into a ratio at centroid , so
Similarly,
The area of the right triangle is
The area of the whole figure is , and the correct answer is .
Solution 2
Notice that if you were to draw in line ED, you would get an orthodiagonal quadrilateral with diagonals 8 and 12. The area is going to be equal to 48. Now we need to examine the triangle AED. If the area we are trying to find is denoted as A, we can tell that the area of AEC is A/2. The area of AED is going to be half of that since AD = DC so it would be A/4. This means that 48 is 3/4 of A, so naturally A is going to be 64. Giving
See also
1997 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Problem 16 | |
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