Regular hexagon ABCDEF

[cschramke]

Regular hexagon ABCDEF has a perimeter of 36. O is the center of the hexagon and of circle O. Circles A, B, C, D, E, and F have centers at A, B, C, D, E, and F, respectively. If each circle is tangent to the two circles adjacent to it and to circle O, what is the area of the shaded region (inside the hexagon but outside the circles)?

In the explanation is states: "To find the portion of circles A, B, C, D, E, and F that is inside the hexagon, we must consider the angles of the regular hexagon. A regular hexagon has external angles of 360/6 = 60°, so it has internal angles of 180 - 60 = 120°. This means that each circle has 120/360 or 1/3 of its area inside the hexagon."

Why can't I, in order to determine the area of the circle that is part of the hexagon, find the area of the circle (9pie) and divide it by 6, since 60degrees = 1/6 of the circle? And then add all the the circles portions within the hexagon (total of 6) together, so it would be 9pie again (in addition the the whole circle in the middle which is also 9pie)? Why does the angle double to 120degrees? I don't understand the terms the explanations uses and calculates such as "internal angles" and "external angles". Please let me know, thanks!

[RonPurewal]

Why can't I, in order to determine the area of the circle that is part of the hexagon, find the area of the circle (9pie) and divide it by 6, since 60degrees = 1/6 of the circle? And then add all the the circles portions within the hexagon (total of 6) together, so it would be 9pie again (in addition the the whole circle in the middle which is also 9pie)?

Each of those portions of a circle is actually 1/3 of a circle, not 1/6 of a circle.

Reality check: If you think each of those 6 pieces is 1/6 of a circle, then you're not thinking about what stuff looks like.
* Imagine cutting a pizza into 6 equal slices. Will the slices look like that? No. They won't.
* Imagine cutting a pizza into 3 equal slices. Will the slices look like that? Yes.
Each is 1/3 of a circle.

Why does the angle double to 120degrees?

Each angle inside a regular hexagon is 120º.

The easiest way to see this is to draw lines between opposite corners, thus separating the hexagon into 6 equilateral triangles. Each equilateral triangle is 60º-60º-60º; each angle of the hexagon consists of two such angles stuck together, for a total of 2 x 60º = 120º.

[RonPurewal]

I don't understand the terms the explanations uses and calculates such as "internal angles" and "external angles".

I wouldn't worry about "external angles" -- it's not relevant here. (It actually took me a while to figure out what that was; it's not something that you need.)

"Internal angles", on the other hand, are what they sound like -- they're the angles inside the hexagon (the ones that are 120º apiece).

[kouranjelika]

Hi Guys,

So on this question, it seems the trap answer is D. And that is exactly the method I used, I just don't see why it is not acceptable here.

So the explanation and you guys here figure out the area of the Hexagon itself via finding the area of the Equilateral triangle and then multiplying that by 6. Which I get in retrospect, but when I first look at it, doesn't it look simpler to just find the area of the two trapezoids and multiply it by 2. Clearly there is a flaw in doing so (as D is the answer choice that does exactly that). Ron or anyone else, can you elaborate as to why we cannot do this?

Here is my process:
1) (n-2)*180 = 720 degrees total for 6-gon
2) 720/6 = 120 per angle
3) 360/120 = 3 Therefore, 1/3 of each outer circle is inside 6-gon
4) Perimeter - 36. 36/6 = 6 is each side. Half of that is radius of each circle. r=3
5) Area of Circle: r^2Pi = 9Pi 1/3*9Pi= 3Pi per sector; 6 such circles - 6*3Pi = 18Pi PLUS central circle 9Pi = 27Pi total (eliminate answer choices A-C)
6) Area of 6-gon:
Hexagon consists of two identical trapezoids.
Top bar 6, bottom bar 12; height 6
Area of Trapezoid: [(6+12)/2]*6 = 54
2 such Trapezoids: 54*2 = 108

Hence Answer choice D seemed correct: 108 - 27Pi

ANYONE READING THIS FROM THE END (D IS NOT THE CORRECT ANSWER, this is simply a layout of my reasoning for a WRONG CHOICE).

[RonPurewal]

The height of your trapezoids is not 6. It's 3√3.

If you don't see why this is so, then split the trapezoids into equilateral triangles, and label the sides of those triangles as 6 units. Then it will be obvious that the height is less than 6 units. (The 30º-60º-90º relationships let you find the actual vaule of 3√3.)

[kouranjelika]

I see it like that.
But why is it not just the diameter if one full circle is between the top and the midpoint? Or is it just an optical illusion that it's one circle?

[RonPurewal]

But why is it not just the diameter if one full circle is between the top and the midpoint?

I don't understand what you are asking. Can you draw it and post a link to the image?

[MamtaK378]

Hi, I did the calculations like Area of minor arc(shaded portion) = 60/360 * 2 * Pi * r = 60/360 * 2 * Pi * 3 = 3 Pi.
Now, the area of an equilateral triangle = (sqrt 3)/4 * s * s = (sqrt 3)/4 * 2r * 2r = (sqrt 3) * 9

Now if we subtract the area of three sectors from the area of equilateral triangle, it will give us the area of shaded portion = {(sqrt 3) * 9} - 3 Pi.
We have got 6 such shaded portions, so answer should be (sqrt*3)*54 - 18 Pi.

NOTE: I am unable to insert image. Hope you can understand my point.

[cgentry]

If I'm reading your post correctly, I believe you have mistakenly used the circumference formula instead of the area formula in your circle calculations.

Beyond that, I'm not sure what your question is?

[MamtaK378]

Apologies, I noticed the mistake.
Feeling awful at being so dumb.. ... I should have used Area of sector formula. I used Measure of Arc instead. That makes my calculations correct and is inline with the answer.

Thanks!!!

[RonPurewal]

if you have any other questions, please ask. thanks.