Area problems Please Help!!

bobbym · 2010-05-24 00:07:06

Using pythagoras again.

3^2 + b^2 = 6^2
b = √27

sides of the triangle are 3 , √27 , 6

misheeru · 2010-05-24 00:09:32

okay so we use that to find th area?

bobbym · 2010-05-24 00:14:17

I only see one triangle so I don't know what they are talking about when they say triangles.

A = ( 1/ 2 ) b * h

A = ( 1 / 2 ) 3 * √27 = ( 3 * √27 ) / 2 this is the area of the triangle.

misheeru · 2010-05-24 00:19:00

yeah im not folling this one

but its a multiple choice, so do you want to see the choices?

bobbym · 2010-05-24 00:24:35

Alright go ahead.

misheeru · 2010-05-24 00:30:05

just two:

A= 9π - (9/2) √3

and

A= 16

bobbym · 2010-05-24 00:36:16

I think the problem is that you told me that was a diameter. If they have pi in the answer then that is not true. Does the problem say anything about given conditions?

misheeru · 2010-05-24 00:37:28

no all all it says is: Find the difference in area between the circle and the triangle

and the picture

bobbym · 2010-05-24 00:38:41

That answer is not the area of the triangle. It is the area of the circle minus the triangle.

misheeru · 2010-05-24 00:41:10

yeah, okay. So since it need to e subtracted then its the first one! the 9π - (9/2) √3?

bobbym · 2010-05-24 00:44:42

Yes, but the wording of the problem says find the combined areas of the triangles. Did you copy the question correctly?

When I gave the area of the triangle.

( 3 * √27 ) / 2, I did not reduce. I expect you to do that. Do you see (9 / 2 )√3 in there?

misheeru · 2010-05-24 00:49:54

yeah. It says:Find the difference in area between the circle and the triangle

The answer is the area of the circle ( 9π) minus that of the triangle (9 / 2 )√3 ).

So A= 9π - (9/2) √3 is right, right?

bobbym · 2010-05-24 00:54:21

Ah Ha. Now it makes sense. You made it tougher by not copying the question right. Very tricky move. Yes that is the correct answer, you get the area of the circle as 9π and then you subtract the area of the triangle that I computed.

misheeru · 2010-05-24 00:56:05

I'm sorry i warned you that my brain wasn't in a good space! I'm sorry I complicated things! Geezus i feel dumb now....

bobbym · 2010-05-24 00:57:28

Well that's what you get for depriving yourself of much needed sleep... sleep... sleeeeeep...

misheeru · 2010-05-24 01:03:34

I am not denying it anylonger! Thank you for all you help! I am sorry to keep you and im sorry for being a nusence and im sorry i cant spell right now and im sorry for bugging you!

I'm glad I got to talk to you today

hope to see ya soon

thanks again!

Good night/good morning and have a terrific day!

Sayounara bobii sensei

bobbym · 2010-05-24 01:07:25

Sayounara bobii sensei

Thanks but the comment is not correct. Sensei, is not a title I am comfortable with. I understand there are many meanings to it. The one I adhere to does not apply to me. Let's stick with Forum man. CYA

misheeru · 2010-06-07 12:23:38

Finaly getting over being really sick, struggling in math (aren't I always?)... again

Guess what? More area crud! No matter how many times I go over it it just won't click!

Okay its lateral area, total area and volume of solid figures. In this case, pyramids.

Like this:

Find the total area for the regular pyramid.

Any help is really really appreciated, I am really behind and feeling really stupid. Thank you!

bobbym · 2010-06-07 13:22:39

Hi misheeru;

Incidentally mish, have you left out anything? Was there some givens that come with that diagram?

Look here for cool graphics, you appreciate art.

http://www.mathsisfun.com/geometry/pyramids.html

Total Surface Area = B + ( 1 / 2 )P s

B is the base area
P is the perimeter of the base
s = slant height

Everything has been given to you except the slant height, See that line segment from the 2 to the apex that's the slant height. Can you do that on your own? We have gone over this and I will cover it as many times as needed. Please give it a try on your own.

misheeru · 2010-06-07 21:50:46

Sure I'll give it a try. And to your question, no there is no given for this one, I checked.

Okay so total area or T.A. = 1/2pl + B

So to find the slant height I am basically finding the lenght of the long side of that triangle, right?

If that's what I'm supposed to do then I can use the Pythagorean Theorem...?

So: 1² + 3² = X²

1 + 9 = X²

= 10²

= √10

S = √10?

Am I good so far?

bobbym · 2010-06-07 22:37:08

Yep, that is good! Now what about the rest of it? Just plug in.

misheeru · 2010-06-09 08:04:21

Hi, Bobbym, thanks for responding, and how have you been?

Oh okay, I see how you got that, it fits all together nicely once you have all the information, doesn't it? Anyway, Sorry for the late response on my part, still not back on my feet...I hate being sick!

After this chapter I moved into total and lateral areas and volume for cylinders and cones. The formulas for those are a bit easier for me to understand, less steps too.

Just in case though, to make sure I am doing them right, would you mind too much if I asked you to check a few for me?
Its okay if you don't, and thank you for the help!

In case you do:

1) Find the lateral area for the cylinder with the given measurement.
r = 4, h = 5

L.A. = 2π(r)(h)

L.A. = 2π(4)(5)

L.A. = 2π(20)

L.A. = 40π

2) Find the total area for the cylinder with the given measurement.
r = 2, h = 4

T.A. = 2π(r)(h) + 2πr²
^
2π(2)(4)
^
2π(8)

L.A. = 16π

T.A. = 16π + 2πr²
^
2π(4)
^
8π

T.A.= 16π + 8π

T.A. = 24π

bobbym · 2010-06-09 08:24:07

Hi misheeru;

misheeru wrote:

Hi, Bobbym, thanks for responding, and how have you been?

Responding is what I live for. Is what I do. As long as there is a respo... I will respond. No thanks are necessary.

How have I been? Not dead yet.

Both math answers are correct. When will you test?

misheeru · 2010-06-09 11:40:52

In a few days, it will be a biggie. It will also cover the new chapter I just started which is all plotting stuff. You know, symmetry, ordered pairs, and graphs... more confusing stuff.

bobbym · 2010-06-09 12:50:48

Not confusing at all. That's the easy stuff.

Math Is Fun Forum

#126 2010-05-24 00:07:06

Re: Area problems Please Help!!

#127 2010-05-24 00:09:32

Re: Area problems Please Help!!

#128 2010-05-24 00:14:17

Re: Area problems Please Help!!

#129 2010-05-24 00:19:00

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#130 2010-05-24 00:24:35

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#131 2010-05-24 00:30:05

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#132 2010-05-24 00:36:16

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#133 2010-05-24 00:37:28

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#134 2010-05-24 00:38:41

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#135 2010-05-24 00:41:10

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#136 2010-05-24 00:44:42

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#137 2010-05-24 00:49:54

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#138 2010-05-24 00:54:21

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#139 2010-05-24 00:56:05

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#140 2010-05-24 00:57:28

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#141 2010-05-24 01:03:34

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#142 2010-05-24 01:07:25

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#143 2010-06-07 12:23:38

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#144 2010-06-07 13:22:39

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#145 2010-06-07 21:50:46

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#146 2010-06-07 22:37:08

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#147 2010-06-09 08:04:21

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#148 2010-06-09 08:24:07

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#149 2010-06-09 11:40:52

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#150 2010-06-09 12:50:48

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