Answer:
28 in²
Step-by-step explanation:
Without constraining the problem unduly, we can make the assumption that AB = 2 inches. Then the altitude from AB to D is h in ...
Area ABD = (1/2)(AB)h
16 in² = (1/2)(2 in)(h)
16 in = h . . . . . . . . . . . divide by 1 in
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The altitude D to AB is the sum of the heights from D to EC (h1) and from AB to EC (h2). That is ...
16 = h1 + h2
We also know that the height from FG to EC is 1/2 the height from D to EC, hence (1/2)h1. Likewise, the height to midsegment HI from either EC or AB is half the height from EC to AB, hence (1/2)h2. This means the total height of the quadrilateral HFGI is (1/2)h1 + (1/2)h2 = (1/2)(h1 +h2) = 8.
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We are given that FG is 50% longer than AB, so its length will be ...
FG = AB×(1 + .5) = (2 in)(1.5) = 3 in
Since FG is the mid-segment of triangle CDE, base EC is twice its length, or ...
EC = 2×FG = 2(3 in) = 6 in
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Mid-segment HI is the average of the base lengths of trapezoid ABCE, so is ...
HI = (EC +AB)/2 = (6 + 2)/2 = 4
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Now, we know the height and base lengths of trapezoid HFGI, so we can find its area as ...
A = (1/2)(b1 +b2)h = (1/2)(3 in + 4 in)(8 in) = 28 in²
The area of quadrilateral HFGI is 28 square inches.
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You can make any assumption you like about the dimension of AB, and the rest of the dimensions scale accordingly. The result is still the same.
