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Re: In the rectangle above, if the unshaded region is equilateral and equi [#permalink]
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Bunuel wrote:

In the rectangle above, if the unshaded region is equilateral and equiangular, what fractional part of the rectangular region is shaded?

(A) 1/6
(B) 2/9
(C) 1/4
(D) 1/3
(E) 4/9

Attachment:
2017-09-26_1117.png


Since the unshaded region is a hexagon and it is equilateral and equiangular, it must be a regular hexagon. We know that a regular hexagon is composed of six equilateral triangles. Note that the diagonal drawn inside the hexagon is equal to the length of the rectangle. Note, too, that twice the base of one of the six equilateral triangles is also twice a side of the hexagon. Furthermore, the width of the rectangle is twice the height of one of the six equilateral triangles.

Recall that the area of an equilateral triangle is (side^2*√3)/4 and the height is √3/2 of the base. Thus, if we let s = side of hexagon, then the area of the hexagon is 6 * (s^2*√3)/4 = 3s^2*√3/2 and the area of the rectangle is l * w = 2s * 2(√3s/2) = 2s^2*√3. Therefore, the area of the shaded region is 2s^2*√3 - 3s^2*√3/2 = s^2*√3(2 - 3/2) = s^2*√3(1/2) and the shaded region is:

(s^2*√3(1/2))/(2s^2*√3) = (1/2)/2 = ¼ of the rectangular region.

Answer: C
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Re: In the rectangle above, if the unshaded region is equilateral and equi [#permalink]
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Re: In the rectangle above, if the unshaded region is equilateral and equi [#permalink]
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