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hexagon problem

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hexagon problem

by anjaligeorge1 » Sun Jun 08, 2008 9:52 am
If the perimeter of a regular hexagon is P, in terms of P, what is the length of the diagonals?

(A) P*root3/2
(B) P/3
(C) P*root3/3
(D) P/4
(E) P/6

ans :B
how do i solve this ?
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by Ian Stewart » Sun Jun 08, 2008 10:57 am
In a regular hexagon ABCDEF, all angles are 120 degrees. Connect AD, BE and CF: you've divided the hexagon into six equilateral triangles (they must be equilateral, because you've divided each 120 degree angle exactly in half, so two angles in each triangle are certainly 60 degrees). Look at the line AD; from A to the centre of the hexagon is one side of one equilateral triangle, and from the centre to D is another side. Each side of each equilateral is as long as one side of the hexagon, and one side in the hexagon is P/6 in length, so AD, BE and CF are each 2*P/6 = P/3 in length.

Much easier to see if you draw the picture!
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by chidcguy » Sun Jun 08, 2008 5:19 pm
Jeez! Ian

Are you a math major or what?

Excellent work!
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by netigen » Sun Jun 08, 2008 6:48 pm
There is another way to solve this, though not as efficient as pointed out in a previous post

Connect a diagonal, you get a isosceles trapezoid

draw perpendiculars connecting the two bases of the trapezoid to get two right angle triangle and one rectangle.

length of rectangle = side of hexagon = p/6

triangle is a 30-60-90 right angle triangle so the base will be (p/6) x 1/2 = p/12

add up to get = p/6+p/12+p/12 = p/3

I recently got a similar question in one of my PR tests. See the attachment.
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Untitled 1.png
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by llewellyn27 » Tue Jun 10, 2008 3:20 pm
Easy way to solve this is by plugging in numbers

We know that a hexagon is made up of 6 Equilateral triangles

Assume s = length of 1 side
So perimeter = 6s

Let P= 36 (divisible by 6)
So s = 6

Diagonal is formed by the sides of 2 of the 6 Eq triangles in the figure

so Diagonal is 12

Input p =36 in the answer choices and you will find that p/3 = 12
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