In the figure above, \(ABCD\) is a parallelogram, and \(E\) is the midpoint of side \(AD.\) The area of triangular regio

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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) \(\dfrac12\)

B) \(\dfrac13\)

C) \(\dfrac14\)

D) \(\dfrac15\)

E) \(\dfrac16\)

Answer: B

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If we take the horizontal side AD to be the base b of the parallelogram, then its area is bh, where h is its corresponding height. Using the horizontal side as the base of the triangle, the triangle then has the same height h as the parallelogram. The triangle's base is b/2, because E is the midpoint of AD. The area of the triangle is then half of its base times height, so is (1/2)(b/2)(h) = bh/4.

So the area of the triangle is 1/4 the area of the entire parallelogram. That is not, however, the ratio the question asks for. We want the ratio of the area in the triangle to the area outside of the triangle, and if the triangle's area is 1/4 of the parallelogram's area, the area of the rest, i.e. of quadrilateral BCDE, is 3/4 of the parallelogram's area, and those areas are thus in a 1 to 3 ratio, and the answer is 1/3.

Since this is a pure ratio question, it would also be fine just to imagine the parallelogram's area is 4, and proceed as above. Then the triangle's area is 1, and the remaining quadrilateral BCDE's area must be 3, and we get a 1 to 3 ratio, and an answer of 1/3.
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