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

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What is the area of the hexagonal region shown above ? [#permalink]  27 Jun 2018, 10:14
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What is the area of the hexagonal region shown above ?

A. $$54 \sqrt{3}$$

B. $$108$$

C. $$108 \sqrt{3}$$

D. $$216$$

E. It cannot be determined from the information given
[Reveal] Spoiler: OA

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Re: What is the area of the hexagonal region shown above ? [#permalink]  28 Jun 2018, 00:54
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Carcass wrote:
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The attachment ex.jpg is no longer available

What is the area of the hexagonal region shown above ?

A. $$54 \sqrt{3}$$

B. $$108$$

C. $$108 \sqrt{3}$$

D. $$216$$

E. It cannot be determined from the information given

There are no. of ways to solve it:

From the diagram we know it is a regular hexagon

and formula of the area of a regular hexagon = $$(3\sqrt{3} * side^2)$$ * $$\frac{1}{2}$$

= $$(3\sqrt{3} * 6^2)$$ * $$\frac{1}{2}$$ = $$54\sqrt{3}$$.

Now if we donot know the formula then, please refer to the diagram below:

Now we can write : y° + y° + y° + y° + y° + y° = 360°
or y° = 60°.

Now we can see the regular hexagon is divided into 6 triangle.
So each triangle will be equilateral triangle since it is inscribed in a regular hexagon.

Now area of Equilateral triangle = $$\sqrt{3} * (side^2)$$ * $$\frac{1}{4}$$ = $$\sqrt{3} * (6^2)$$ * $$\frac{1}{4}$$

=$$9\sqrt{3}$$

Therefore for 6 triangles we have = 6* $$9\sqrt{3}$$ = $$54\sqrt{3}$$.

Now if we donot know the area of equilateral triangle then we can always find out:

Take any one triangle inscribed in the regular hexagon:-
SO the area of th triangle = $$\frac{1}{2} * base * altitude$$

Now we need to find out the altitude, which can be found out by Pythagoras theorem, as we draw a line from the midpoint of the hexagon to the base of the triangle as shown in diagram,It divides the base at the midpoint as it is a equilateral triangle. (Also a 30 - 60 - 90 triangle will also have the same result)

So $$altitude^2 = 6^2 - 3^2$$ = 27

or altitude = $$3\sqrt{3}$$

Now area = $$\frac{1}{2} * 6 * 3\sqrt{3}$$ = $$9\sqrt{3}$$

Now area of all the six triangle = 6 * $$9\sqrt{3}$$ = $$54\sqrt{3}$$ = Area of the hexagon.

** Please note this applies only for a regular hexagon
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Re: What is the area of the hexagonal region shown above ?   [#permalink] 28 Jun 2018, 00:54
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