GMATinsight wrote:
In the figure above, ABCD is a parallelogram and E is the midpoint of side AD. The area of triangular region ABE is what fraction of the area of the quadrilateral region BCDE?
A) 1/2
B) 1/3
C) 1/4
D) 1/5
E) 1/6
PS49220.02
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The best part about the question is that it asks for a ratio and since no dimensions are given we are free to take any value. Refer figure (a) and figure (b):
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Area parallelogram ABCD.png [ 56.25 KiB | Viewed 17056 times ]
Consider figure (a).
Visual Solution:
Know that if we draw a parallel EF line to AB, the line we are dividing the area of parallelogram ABCD into two equal parts. Similarly, area of parallelogram AEFB is divided by AE into two equal parts.
Thus, if area of parallelogram ABCD = a THEN
Area of parallelogram AEFB = \(\frac{a}{2}\)
And Area of triangle AEB = 1/2 * \(\frac{a}{2}\) = \(\frac{a}{4}\)
Area of quadrilateral BCDE = a - \(\frac{a}{4}\) = \(\frac{3a}{4}\)
Hence, area of triangle ABE / area of quadrilateral BCDE = \(\frac{a}{4}\)/\(\frac{3a}{4}\) = \(\frac{1}{3}\)
ORConsider figure (b).
Parallelogram ABCD is remolded into rectangle ABCD to make things better understandable.
Let dimensions are AB = 3 and AD = 4. AE = 2.
Area of triangle ABE = \(\frac{1}{2} * 3 * 2 = 3\)
Area of quadrilateral BCDE = Area of rectangle ABCD - Area of triangle ABE
Area of quadrilateral BCDE = 4*3 - 3 = 9
Hence, , area of triangle ABE / area of quadrilateral BCDE = \(\frac{3}{9}\) = \(\frac{1}{3}\)
Answer B.
Note: A rectangle satisfies a parallelogram's conditions. _________________