If ABCDEF is a regular hexagon and AB = 1cm, then what is the area of

Question

If ABCDEF is a regular hexagon and AB = 1cm, then what is the area of this regular hexagon ABCDEF?

A.  \frac{3\sqrt{3}}{4}

B. √3

C.  \frac{3\sqrt{3}}{2}

D. 2√3

E. 3√3

chetan2u · Answer

A regular hexagon has all sides equal. And when all vertices are joined at center we get SIX equilateral triangles with sides as 1 here. 
Area = 6*area of equilateral triangle of side 1. 
=  6*\frac{\sqrt{3}*1^2}{4}=3*\frac{\sqrt{3}}{2}

C

Karpov92 · Answer

There are 6 congruent triangles in a regular hexagon. The vertices of the triangles will have an angle of (360°/6) = 60°. So If we draw an apothem and a radius on any of the sides, it will create a 30-60-90 right triangle, where base = 1/2. So Apothem, a = √3/2.
Area of Hexagon = 1/2*asn, (where a = apothem, s = side length and n = num. of sides.) = 1/2 *√3/2 * 6 = 3√3/2. (C)

Abhishek009 · Answer

Area of a regular Hexagon is  a^2\frac{3\sqrt{3}}{2}

So, Area of a regular Hexagon is  1^2\frac{3\sqrt{3}}{2} = \frac{3\sqrt{3}}{2} , ANswer must be (C)

Karpov92 · Answer

Hi, I have a query. 
Instead of a hexagon, if we are asked to find the area of a regular heptagon/pentagon, can we find it using the formula of equilateral triangles?

chetan2u · Answer

No, because only regular hexagon has angles equal to 120, and when you join each vertex to center, it divides 120 into 60 each. On the other hand pentagon has each angle equal to (5-2)*180/5=108, so each triangle will be isosceles with 54-54-72

winterschool · Answer

O is center of hexagon, OAB equilateral triangle

OA = AB = OB =1, area = (3)^(1/2) /4 * 1^2 = (3)^(1/2)/4

area of total hexagon = 6 * (3)^(1/2)/4 = 3 (3)^(1/2) / 2

Answer C

If ABCDEF is a regular hexagon and AB = 1cm, then what is the area of

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GMAT Problem Solving (PS) Questions

If ABCDEF is a regular hexagon and AB = 1cm, then what is the area of

Prep Toolkit

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