Difference between revisions of "2013 AMC 10B Problems/Problem 16"
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==Problem== | ==Problem== | ||
| + | In triangle ABC, medians AD and CE intersect at P, PE=1.5, PD=2, and DE=2.5. What is the area of AEDC? | ||
| + | (A) 13 (B) 13.5 (C) 14 (D) 14.5 (E) 15 | ||
| + | ==Solution== | ||
| + | Let us use mass points: | ||
| + | Assign B mass 1. Thus, because E is the midpoint of AB, A also has a mass of 1. Similarly, C has a mass of 1. D and E each have a mass of 2 because they are between B and C and A and B respectively. Note that the mass of D is twice the mass of A, so AP must be twice as long as PD. PD has length 2, so AP has length 4 and AD has length 6. Similarly, CP is twice PE and PE=1.5, so CP=3 and CE=4.5. Now note that triangle PED is a 3-4-5 right triangle with the right angle DPE. This means that the quadrilateral AEDC is a kite. The area of a kite is half the product of the diagonals, AD and CE. Recall that they are 6 and 4.5 respectively, so the area of AEDC is 6*4.5/2=(B) 13.5. | ||
Revision as of 20:59, 21 February 2013
Problem
In triangle ABC, medians AD and CE intersect at P, PE=1.5, PD=2, and DE=2.5. What is the area of AEDC? (A) 13 (B) 13.5 (C) 14 (D) 14.5 (E) 15
Solution
Let us use mass points: Assign B mass 1. Thus, because E is the midpoint of AB, A also has a mass of 1. Similarly, C has a mass of 1. D and E each have a mass of 2 because they are between B and C and A and B respectively. Note that the mass of D is twice the mass of A, so AP must be twice as long as PD. PD has length 2, so AP has length 4 and AD has length 6. Similarly, CP is twice PE and PE=1.5, so CP=3 and CE=4.5. Now note that triangle PED is a 3-4-5 right triangle with the right angle DPE. This means that the quadrilateral AEDC is a kite. The area of a kite is half the product of the diagonals, AD and CE. Recall that they are 6 and 4.5 respectively, so the area of AEDC is 6*4.5/2=(B) 13.5.
