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30 Apr 2019, 00:46
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In the parallelogram ABCD above, AD = 12. If the area of triangle ABE is 3/8 the area of parallelogram ABCD, then what is the length of EC?
A. 3/2
B. 2
C. 8/3
D. 3
E. 4
Attachment:
2019-04-30_1144.png [ 19.37 KiB | Viewed 10040 times ]
30 Apr 2019, 05:22
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If we take AD as the base of both the parallelogram and of the triangle ADE, and take EC as the base of triangle ECD, notice that all three have the same height - the perpendicular (vertical) distance between the two sides of the parallelogram.
If h is that height, and b is the length of AD, then the area of the parallelogram is bh, and the area of triangle ADE is bh/2. So triangle ADE takes up half the area of the parallelogram, and we know ABE takes up another 3/8 of the area, which means triangle ECD takes up the rest, or 1/8 of the entire area. So the area of ADE is 4 times the area of ECD (one is 1/2 the parallelogram's area, the other 1/8), and since they have the same height, the base of ADE must be 4 times the base of ECD. Since the base of ADE is 12, the base of ECD must be 3, so that's the length of EC.
If h is that height, and b is the length of AD, then the area of the parallelogram is bh, and the area of triangle ADE is bh/2. So triangle ADE takes up half the area of the parallelogram, and we know ABE takes up another 3/8 of the area, which means triangle ECD takes up the rest, or 1/8 of the entire area. So the area of ADE is 4 times the area of ECD (one is 1/2 the parallelogram's area, the other 1/8), and since they have the same height, the base of ADE must be 4 times the base of ECD. Since the base of ADE is 12, the base of ECD must be 3, so that's the length of EC.
