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In the rectangle above, if the unshaded region is equilateral and equi

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

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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

[Reveal] Spoiler:
Attachment:
2017-09-26_1117.png
2017-09-26_1117.png [ 4.23 KiB | Viewed 515 times ]
[Reveal] Spoiler: OA

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

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Since the unshaded region is equilateral and equiangular,
the hexagon within the rectangle will have 6 equilateral triangles.
Each equilateral triangle will further contain two triangles of equal area.

These smaller triangles have an area equal to the shaded triangles,
because they have same base and height as the shaded triangles.

The hexagon itself will have 12 triangles(each of which has the same area as the shaded triangle)

There are a total of 16(12+4) smaller triangles of equal area within the rectangle and
the fractional part of the rectangular region which is shaded is \(\frac{4}{16} = \frac{1}{4}\)(Option C)
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Re: In the rectangle above, if the unshaded region is equilateral and equi [#permalink]

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New post 26 Sep 2017, 08:36
Bunuel wrote:
Image
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

[Reveal] Spoiler:
Attachment:
The attachment 2017-09-26_1117.png is no longer available

Attachment:
hexsquare.png
hexsquare.png [ 21.19 KiB | Viewed 237 times ]

Explaining this solution is a lot harder than deriving it.

1. SIX EQUILATERAL TRIANGLES IN THE HEXAGON

This regular hexagon (equal angles, equal sides) can be divided into 6 equilateral triangles

Diagram: each equilateral triangle has side 2x

2. EACH EQUILATERAL TRIANGLE DIVIDES INTO TWO CONGRUENT RIGHT TRIANGLES - see diagram, triangle on right

Equilateral triangles can be divided into 2 congruent right triangles 30-60-90 right triangles because:

--An equilateral triangle's median is also its angle bisector and perpendicular bisector

--One 60-degree vertex is bisected into a 30-degree angle in each right triangle

--The perpendicular bisector creates two 90-degree angles in each right triangle

-- Finally, the last vertex of the right triangle on both sides has measure of 60 degrees

SIDE RATIO OF 30-60-90 TRIANGLES

--30-60-90 triangles have sides in ratio \(x: x\sqrt{3}: 2x\)

3. HEXAGON HAS 12 CONGRUENT RIGHT TRIANGLES

There are 12 such identical triangles in the hexagon. SIX equilateral triangles * TWO right triangles each = 12 identical right triangles inside the hexagon

4. SHADED AREAS ARE IDENTICAL TO THE 12 RIGHT TRIANGLES INSIDE THE HEXAGON

The shaded areas (the corners of the rectangle) are congruent with all the right triangles inside the hexagon

Angles are congruent

--Each corner triangle is a 30-60-90 triangle - on diagram, see triangle ABC and the left side of the rectangle

-- 30- degree angle is on the straight line - see left side of rectangle

On that line, whose angles sum to 180, lies a 120-degree angle of a hexagon (or two 60-degree angles of the divided triangles)

180 - 60 - 60 = 60 degrees are left to be divided into two equal angles. On the line, each corner has a 30-degree angle

-- Each corner triangle has a 90-degree angle (by definition of rectangle)

-- Each corner triangle has a 60-degree angle - see left corner triangle, vertex C

Side lengths are equal

--Each corner triangle has exactly the same lengths as the 12 small triangles inside the hexagon: \(x, x\sqrt{3},\) and \(2x\) (see diagram)

5. FRACTION OF RECTANGLE SHADED?

12 triangles inside hexagon, 4 outside = 16 in rectangle. Four are shaded.

\(\frac{4}{16} = \frac{1}{4}\)

ANSWER C

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

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New post 29 Sep 2017, 10:17
Bunuel wrote:
Image
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

[Reveal] Spoiler:
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] 29 Sep 2017, 10:17
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