Hexagon

[vampirewitchreine]

I totally thought that I had this one...... Then I realized that I'd done the problem completely wrong.


I'm trying to find the area.

I do know that half of the given line is the apothem.... other than that I'm really not sure (I also know that half of the diagonals in a hexagon is congruent to one side)


(According to the back of the textbook, my answer should come out to be $\displaystyle 98\sqrt {3} ~\approx 170$ somehow)

 

[vampirewitchreine]

Your hexagon is made up of 6 equilateral triangles of height 7; see that?
Find the area of one of them, then multiply by 6; give it a shot!
We KNOWWWWWWW you can do it

Can you explain to me how to find it with the only given information being the height? Pretending that I know nothing about special right triangles....


(Sorry, I'm really tired)

 

[vampirewitchreine]

Staying up all night looking for victims to feed on will make you sleepy.

(I'm home schooled and my course expires soon, so dad's making me stay up till I finish everything)

For reasons not known to me, few take Morris Kline's point to use the history of mathematics to figure out what is an intuitive way for people to learn mathematics. The REASON that Euclidean geometry spends so much time on triangles is that ANY RECTILINEAR figure can be decomposed into triangles. A regular hexagon can be decomposed into six equilateral triangles. You do not need to know anything about angles or special triangles to find the area of an equilateral triangle.

The formula for the area of any triangle is what?

The height of any equilateral triangle can be determined how?

Okay, a triangle's area is determined by the formula $\displaystyle \frac {1}{2} bh$

To find the height, use the Pythagorean Theorem, as the following $\displaystyle c^2 - a^2= b^2$ with a being half of the base and b being the height of the triangle.


Since I already have the height both of those formulas become
$\displaystyle \frac {1}{2} b(7)$ and $\displaystyle c^2 - a^2= 7^2$

 

[vampirewitchreine]

that "c" of yours equals 2a : ain't that evident?

Should have been, since I even said that it would be when I posted originally (As it is half of a diagonal, making it congruent to a side)....


So then I have:
$\displaystyle 2a^2 - a^2 = 7^2$ right?

 

[vampirewitchreine]

Couple of tips

(1) Draw your figure and label it with your variable names. Have you drawn the equilateral triangles? If not you are missing clues.

(2) Write down the definitions of your variable names. You have mixed up your formulas by not doing so.

c = length of hypotenuse of given right triangle

a = length of leg of given right triangle that constitutes a segment of the given hexagon.

b = height of equilateral triangle = third leg of given right triangle.

x = length of base of equilateral triangle.

M = area of equilateral triangle.

N = area of hexagon.

SIX variables. So you need six equations to solve.

N = 6M

b = 7

M = (1/2)bx = 7x/2

c² = a² + b² = a² + 49

You need two more equations. Look at that equilateral triangle for clues.

I know that c should = x, as half of a diagonal is congruent to one side, so a should be half of c/x
(I have a feeling that I'm still missing the other formula....)



Given the new information.... here is the new diagram:

 

[pka]

This is a painful to watch thread for me.
Here are two websites that could have shorten all of this.

HEXGON & EQUILATERAL TRIANGLE

 

[Mrspi]

to PKA....as a retired geometry teacher, this thread is painful to me as well.

I will NOT buy "pretend I don't know anything about special right triangles."

35+ years of teaching geometry tell me that the kinds of problems I'm seeing from this student ONLY come after the special right triangle relationships have been presented and (presumably) mastered by the student.

If you all want to re-invent the wheel, have at it.

Vampirewitch, I'll suggest again that you commit those special right triangles to MEMORY and save yourself (and us, too) a lot of grief.

 

[vampirewitchreine]

I will NOT buy "pretend I don't know anything about special right triangles."

35+ years of teaching geometry tell me that the kinds of problems I'm seeing from this student ONLY come after the special right triangle relationships have been presented and (presumably) mastered by the student.

This is in my section of Areas of Regular Polygons and Circles (3 chapters before I get to Special Right Triangles)

If you all want to re-invent the wheel, have at it.

Vampirewitch, I'll suggest again that you commit those special right triangles to MEMORY and save yourself (and us, too) a lot of grief.

I do I fact know how to do the special right triangles fairly well now..... had I been able to use them then I probably wouldn't have posted this problem other than to check my work for it.

This is a painful to watch thread for me.
Here are two websites that could have shorten all of this.

HEXGON & EQUILATERAL TRIANGLE

I followed the links that you provided...... They're still banking on special right triangles... which are not suppose to be known in this section.

Nooooooo....4a^2 - a^2 = 7^2

Anyway, forget all that....keep it simple:
formula for area of equilateral triangle is h^2 / sqrt(3) ; where h is height.
We have height of 7, and 6 of them equilateral critters; so:
AREA = 6(7^2) / sqrt(3) = ~169.74 ; that's it, over and out!

:roll: Had I thought about looking that up, I could have made this so much simpler..... and probably would have never had need to post the thread..... Plus, it's much faster than working though all the variable that JeffM had me using to solve...... Thank you.