Area problems Please Help!!

misheeru · 2010-04-30 09:55:44

I need to figure out how to find the area of a regular hexigon with the given measurement.

The only info they give you is: 2√3 apothem

please help idk what to do!

ZHero · 2010-04-30 12:50:43

Hi misheeru!

A Circle can be Inscribed in a Regular Hexagon and the Radius of such an Inscribed Circle is called an Apothem!

If u join all the Vertices of the Hexagon to the Centre of the Circle then 6 Equilateral Triangles will be formed, each with an Altitude equal to the Radius or the Apothem!

So, simply find the Area of 1 Equilateral Triangle using

and multiply it by 6 to find the Area of the Hexagon!

Last edited by ZHero (2010-04-30 15:06:01)

soroban · 2010-04-30 13:19:04

.

          * - - - *
         / \     / \
        /   \   /   \
       /     \ /     \
      * - - - * - - - *
       \     /:\     /  :   _
        \   / : \   /   : 2√3
         \ /  :  \ /    :
          * - - - *     -
              4

.

ZHero · 2010-04-30 15:08:41

Hi soroban!

I made a mistake in defining the Apothem and hence the Area of the Eq. Tri.
I've corrected it!!

misheeru · 2010-05-03 04:54:48

Okay, I see how you got that. I'm really having a hard time with these still, though. But I'll take a crack at a few more and let you know if I need any more help. thanks so much guys!!!

Last edited by misheeru (2010-05-03 04:55:13)

misheeru · 2010-05-03 05:19:37

"I'' ll try some on my own" wow that was short lived. Thank you every one for your help, unfortunaly I need some more!

*Find the area of an equilateral triangle (regular 3-gon) with the given measurement.

Given: 6-inch side

Please help!!

Last edited by misheeru (2010-05-03 05:19:57)

ZHero · 2010-05-03 05:43:57

plz look for the formula for Area in the other reply above...

here, ur Side is given!

bobbym · 2010-05-03 13:16:01

Hi misheeru;

Or try this:

Start with equilateral triangle ABC. Drop the perpendicular bisector BD to AC. It is easy to prove this can be done. Now AD = DC = 3. Just use pythagoras theorem to get the height BD.

Call h = BD, then find h.

h^2 + 3^2 + 6^2

a^2 = 27

h = √27 = 3√3

Now just use A = ( b * h ) / 2

A = ( 6 * 3√3 ) / 2 = 9 √3

misheeru · 2010-05-05 07:16:01

Okay so with triangles, I I was told to use : area (A) = 1/2 bh

So one of my problems is this: In a triangle the base is 4 inches and the height is 6 inches. Find the area.

I'd start with: A= 1/2bh

A= 1/2 (4*6)

A= 1/2 (24)

A= 12

So it would be: A= 12in² ? Is that right?

bobbym · 2010-05-05 07:27:45

Hi misheeru;

Correct! Formulas are great aren't they? You just plug in and whamo!

misheeru · 2010-05-05 09:07:35

LOL, I suppose. I just wish they were easier for me to figure out... oh well, I guess, like everything else, they require practice.

misheeru · 2010-05-06 19:14:00

Okay so my lessons have moved once again. Right now I'm in "Circles: Area Of Sectors".

I get a little bit of this, but not much. I found this video this lady made on youtube and it really helped me, but I get lost right after about 5:50 in it.

Her circle has a radius of 7cm and an angle of 120 degrees.

I used a problem from my work to follow her video. The circle I used had a radius of 8 inches and an angle of 40 degrees.

I worked with her up until the final equation wich was:

1/3 * 22/7 *7 * 7 (mine was 1/9 * 22/7 * 8 * 8)

at 5:50 the teacher announces that her answer is 51.3 cm. Can some one please help me, I can't seem to figure out how she came to that answer. That way I can figure out mine, and hopefully get this whole concept down. Thanks!

Heres the youtube link to the vid I used: http://www.youtube.com/watch?v=brErtLcvXvo

ZHero · 2010-05-06 19:18:25

hi

she is right and her answer is correct.
your's is also fine!!!

i guess somebody might explain to you in detail bout the Concept.
its got something to do with fractions.........

bobbym · 2010-05-06 20:44:51

Hi misheeru;

Where are you having a problem? Can you leave a link of the you tube vid?

misheeru · 2010-05-07 04:45:48

Oh shoot! I totally posted the wrong link! I should've checked that before I went to sleep. I'm sorry.

Okay, my bad I think its this one: http://www.youtube.com/watch?v=brErtLcvXvo

I just don't know how she got her answer from the final equation at about 5:50 or so. Like, how do I explain this, I don't know how to work the equation and come to the answer 51.3.

bobbym · 2010-05-07 04:51:20

Hi misheeru;

Do you understand her method of getting the area of a circle?

misheeru · 2010-05-07 05:09:13

Yeah, I think so. Here, one of the problems on my homework is an area of a circle problem:

Find the area of the circle with the given radius or diameter. Use π= 3.14. (my school prefers that we use the 3.14 instead of the 22/7)

A = πr^2 r =6

A= 3.14 * 6 * 6

A = 113.04

Hows that?

bobbym · 2010-05-07 05:13:44

That is correct. Now supposing I wanted the area of the sector formed by an angle of 60°

misheeru · 2010-05-07 05:24:23

Don't you have to know the radius, too?

bobbym · 2010-05-07 05:25:51

Same radius as she is using in the video, r = 7

misheeru · 2010-05-07 05:33:50

okay. So then we are left with an angle of 60° and a radius of 7.

Then we have 60/360 * πr^2 .

so thats.... 1/6 * 3.14 * 7 * 7?

bobbym · 2010-05-07 05:35:52

That's right. So what is it that you do not understand? Looks like you understand it fine, Just remember to include your units at the end.

misheeru · 2010-05-07 05:42:19

Um... okay. thanks then!

bobbym · 2010-05-07 05:45:16

Glad to help. Did you remember that this is an area problem? There is some unit of measure squared on the end. In this case cm²

misheeru · 2010-05-07 05:50:58

Okay so if I'm just supposed to multiply them together, would it be:

1/6 * 3.14 * 7 * 7 = 25.643

A= 25.643 cm²??

Math Is Fun Forum

#1 2010-04-30 09:55:44

Area problems Please Help!!

#2 2010-04-30 12:50:43

Re: Area problems Please Help!!

#3 2010-04-30 13:19:04

Re: Area problems Please Help!!

#4 2010-04-30 15:08:41

Re: Area problems Please Help!!

#5 2010-05-03 04:54:48

Re: Area problems Please Help!!

#6 2010-05-03 05:19:37

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#7 2010-05-03 05:43:57

Re: Area problems Please Help!!

#8 2010-05-03 13:16:01

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#9 2010-05-05 07:16:01

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#10 2010-05-05 07:27:45

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#11 2010-05-05 09:07:35

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#12 2010-05-06 19:14:00

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#13 2010-05-06 19:18:25

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#14 2010-05-06 20:44:51

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#15 2010-05-07 04:45:48

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#18 2010-05-07 05:13:44

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