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# Land for a pasture is enclosed in the shape of a 6-sided figure; all

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Land for a pasture is enclosed in the shape of a 6-sided figure; all [#permalink]
gmatbusters wrote:
Land for a pasture is enclosed in the shape of a 6-sided figure; all sides are the same length and all angles have the same measure. What is the area of the enclosed land?
(1) Each side is 8 meters long.
(2) The distance from the center of the land to the midpoint of one of the sides is 4$$\sqrt{3}$$ meters.

An additional solution to the one above, relying on the logic of regular polygons:
Since all sides and (internal) angles are equal, the polygon is regular.
The only thing you can do to a regular polygon is make it larger or smaller; all other ratios are fixed.
So, given any one of the polygons parameters (length of side, area, cross-section, whatever you like), you know EXACTLY how to draw the polygon.
This implies that you can calculate all of its parameters.

Since both (1) and (2) provide us with one of the parameters, they fix ALL the parameters of the polygon.
Then (D) is correct.­
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Land for a pasture is enclosed in the shape of a 6-sided figure; all [#permalink]
dave13 wrote:
gmatbusters wrote:
Land for a pasture is enclosed in the shape of a 6-sided figure; all sides are the same length and all angles have the same measure. What is the area of the enclosed land?

(1) Each side is 8 meters long.

(2) The distance from the center of the land to the midpoint of one of the sides is 4$$\sqrt{3}$$ meters.

STAMENT ONE: Hexagone consists of 6 equilateral triangles. if we know one side we can find area via this formula (i know eveyone knows but me )

$$A = \frac{s^2 \sqrt{3}}{ 4} * n$$ (where N is a number of sides of hexagone i.e. 6)

STAMENT TWO: Distance from center, means radius i.e one of the sides of equilateral triangle. So here we can find area here as well

D

It is not the side of an equilateral triangle but the altitude.­
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Re: Land for a pasture is enclosed in the shape of a 6-sided figure; all [#permalink]
Q . Land for a pasture is enclosed in the shape of a 6-sided figure; all sides are the same length and all angles have the same measure.

That Means the shape has to be Hexagon which consist of 6 equilateral triangle.
So If we could calculate the area of triangle we can easily calculate area hexagon.

In order to calculate area of equilateral triable, either height or base is enough to calculate area.
Since in both option base and height is given separately, both are sufficient to answer the question.
Hence Option D is correct.
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Land for a pasture is enclosed in the shape of a 6-sided figure; all [#permalink]
GMATBusters wrote:
Land for a pasture is enclosed in the shape of a 6-sided figure; all sides are the same length and all angles have the same measure. What is the area of the enclosed land?

(1) Each side is 8 meters long.

(2) The distance from the center of the land to the midpoint of one of the sides is 4$$\sqrt{3}$$ meters.

6-sided figure with sides of equal length and angles of equal measure is a regular hexagon. Area of a regular hexagon is $$\frac{3\sqrt{3}}{2}a^2$$ where a is the length of the side.

(1) a = 8m. Sufficient.

(2) Distance from center to the midpoint of a side of regular hexagon is $$a\frac{\sqrt{3}}{2}$$ where a is the side length. So we have $$a\frac{\sqrt{3}}{2}=4\sqrt{3}$$ or $$a=8.$$ Sufficient.

Hence, D.­
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Land for a pasture is enclosed in the shape of a 6-sided figure; all [#permalink]
GMATBusters wrote:
Land for a pasture is enclosed in the shape of a 6-sided figure; all sides are the same length and all angles have the same measure. What is the area of the enclosed land?

(1) Each side is 8 meters long.

(2) The distance from the center of the land to the midpoint of one of the sides is 4$$\sqrt{3}$$ meters.

In a 6-sided figure, hence all the angles are 60 degree

S1. each side 8 meters. Also the area of a regular hexagon = 6 X (area of an equilateral triangle formed by all the diagonals of a hexagon)
Sufficient

S2. also a\sqrt{3}/2 (a let's say be the side of the hexagon) is the distance from the centre to the midpoint of a side of a hexagon.
Hence Sufficient

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Re: Land for a pasture is enclosed in the shape of a 6-sided figure; all [#permalink]
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Re: Land for a pasture is enclosed in the shape of a 6-sided figure; all [#permalink]
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