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# What is the area of the hexagonal region shown in the figure above?

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Math Expert
Joined: 02 Sep 2009
Posts: 47037
What is the area of the hexagonal region shown in the figure above? [#permalink]

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20 Sep 2017, 22:50
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79% (01:23) correct 21% (02:15) wrong based on 27 sessions

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What is the area of the hexagonal region shown in the figure above?

(A) 54√3
(B) 108
(C) 108√3
(D) 216
(E) cannot be determined

Attachment:

2017-09-20_1021_002.png [ 8.74 KiB | Viewed 870 times ]

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Joined: 02 Jul 2017
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GMAT 1: 730 Q50 V38
Re: What is the area of the hexagonal region shown in the figure above? [#permalink]

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20 Sep 2017, 23:43
Given all angles= x and all sides = 6. So given hexagon is regular hexagon.
=> x= 120

To find area, hexagon an be divided into 1 rectangle and 2 isosceles triangles. [Attached figure]

To find sides of isoselec trinagle 1 isoselces triangle => drop perpendicular on base => 2 right triangles of 30-60-90

Sides as base = $$3\sqrt{3}$$ and height = 3

=> Sides of rectangles = 6 and ($$3\sqrt{3} + 3\sqrt{3}$$) = 6 and $$6\sqrt{3}$$
=> Height and base of triangle = 3 and $$6\sqrt{3}$$

Area = 6 * $$6\sqrt{3} + 2 * \frac{1}{2} *3* 6\sqrt{3}$$ = $$54\sqrt{3}$$

Attachments

hexegon.png [ 15.54 KiB | Viewed 716 times ]

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What is the area of the hexagonal region shown in the figure above? [#permalink]

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21 Sep 2017, 12:48
Bunuel wrote:
What is the area of the hexagonal region shown in the figure above?

(A) 54√3
(B) 108
(C) 108√3
(D) 216
(E) cannot be determined

Attachment:

2017-09-20_1021_002ed.png [ 21.16 KiB | Viewed 686 times ]

The hexagon can be divided into triangles. (Ignore the right corner of the diagram if you know the formula for area of an equilateral triangle.)

This hexagon is regular: side lengths and angles are all equal.

1) Draw diagonals. A regular hexagon can be divided into six identical equilateral triangles*

2) Find the area of one equilateral triangle and multiply by 6 to get total area of hexagon

Area of equilateral triangle** with side = s (side here = 6) is

$$\frac{s^2\sqrt{3}}{4}$$

$$\frac{6^2\sqrt{3}}{4}$$

$$\frac{36\sqrt{3}}{4}$$ = $$9\sqrt{3}$$

$$(9\sqrt{3}) * 6$$ =

$$54√3$$

*Regular hexagon can be divided into six identical equilateral triangles

Each angle measure is 120 degrees
The sum of interior angles of n-sided polygon, where n is the number of sides, is (n-2)* 180
(4 * 180) = 720 degrees total for all interior angles
There are six interior angles
720/6 = 120 per angle

Diagonals bisect the 120 degree angles, so at each vertex there are two 60 degree angles (diagram: a and b)

At the center, where the diagonals intersect, there are 6 angles whose sum is 360

The angle nearest the center of each identical triangle also equals 60: 360/6 = 60 degrees (diagram: c)

Three 60 degree angles (a, b, c ) = equilateral triangle, and there are six of them

**If formula for equilateral triangle is not known, see lower right corner of diagram

Derive area by dropping an altitude from a vertex

The altitude of an equilateral triangle
---is a perpendicular bisector of the opposite side
---and bisects the vertex opposite that side, so

There are two angles with measure 30 at the top vertex, two right angles where altitude meets base, and two angles with measure 60

Thus there are two 30: 60: 90 triangles with side ratios x: $$x\sqrt{3}: 2x$$

Calculate the area of one such triangle and multiply by two to get the area of one equilateral triangle (then multiply by 6 for the hexagonal area)
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What is the area of the hexagonal region shown in the figure above?   [#permalink] 21 Sep 2017, 12:48
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