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M70-17

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Joined: 02 Sep 2009
Posts: 52285

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03 Sep 2018, 03:37
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In rectangle ABCD above, $$AE = \frac{1}{3}AD$$, $$4×AG = GB$$. The area of triangle AEF is what fraction of the area of the area of rectangle FHCI?

A. $$\frac{1}{30}$$
B. $$\frac{1}{16}$$
C. $$\frac{1}{24}$$
D. $$\frac{5}{64}$$
E. $$\frac{1}{32}$$

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Math Expert
Joined: 02 Sep 2009
Posts: 52285

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03 Sep 2018, 03:37
Official Solution:

In rectangle ABCD above, $$AE = \frac{1}{3}AD$$, $$4×AG = GB$$. The area of triangle AEF is what fraction of the area of the area of rectangle FHCI?

A. $$\frac{1}{30}$$
B. $$\frac{1}{16}$$
C. $$\frac{1}{24}$$
D. $$\frac{5}{64}$$
E. $$\frac{1}{32}$$

We’ll go for ALTERNATIVE because all we have is ratio.

We’ll assign easy numbers to AE and AG: AE = 1, AG = 2. This makes AEF’s area $$\frac{1×2}{2}=1$$. This means GB = 8 and ED = AD – AE = 3 – 1 = 2. Through symmetry, FI = ED = 2 and FH = GB = 4, so that FHCI = 8 × 2 = 16. Thus, $$\frac{AEF}{FHCI}=\frac{1}{16}$$.

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Re M70-17 &nbs [#permalink] 03 Sep 2018, 03:37
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