How to find the volume and surface area of right prisms in the attachments ?Attachment 18585Attachment 18584Attachment 18583

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- Aug 15th 2010, 01:49 AMhaftakhanFinding volume and surface area of right prism
How to find the volume and surface area of right prisms in the attachments ?Attachment 18585Attachment 18584Attachment 18583

- Aug 15th 2010, 01:55 AMEducated
$\displaystyle \mbox{Volume} = \mbox{Base} * \mbox{Height} * \mbox{Depth} $

$\displaystyle \mbox{Surface area (Of each face)} = \mbox{Base} * \mbox{Height}$

Divide the prisms into many simpler shapes (eg. rectangular cubes for volume, or rectangles/squares for surface area) so it is easier to work with and find the surface area and volume seperately. Afterwards add them together to get the final surface area or volume of each prism.

For the wedge, use:

$\displaystyle \mbox{Volume} = \dfrac{\mbox{Base} * \mbox{Height} * \mbox{Depth}}{2} $ for the volume

$\displaystyle \mbox{Surface area (Of each triangular face)} = \dfrac{\mbox{Base} * \mbox{Height}}{2}$ for the surface area of the triangular faces - Aug 15th 2010, 02:00 AMUnknown008
I think I'll put more detail for the volume of a prism.

$\displaystyle Volume\ =\ Area\ of\ Cross\ Section\ \times\ Depth$

You need to first identify the cross section of the solid, the area which 'repeats' itself through the solid. In the first one, the cross section is a cross, in the second, it's this 'L' shape and in the last, it's a right angled triangle. Once you get their cross sectiona; area, you'll be able to find the volume by multiplying it by the depth of the solid. - Aug 15th 2010, 02:03 AMhaftakhan
For attachment 1 Volume=12*16*20=3840 ?

and area=12*16=192

Is this the right answer ? - Aug 15th 2010, 02:04 AMhaftakhan
Please solve the 1st one with ur details so that i can understand.

- Aug 15th 2010, 02:09 AMEducated
First one:

$\displaystyle \mbox{Total Volume} = (4 * 5 * 3) + (4 * 5 * 3) + (14 * 3 * 4)$

$\displaystyle \mbox{Total Volume} = 288~ \mbox{units}^3$

$\displaystyle \mbox{Surface Area} = (5*4*(8)) + (4*4*(2)) + (4*3*(4)) + (5*3*(8))$ (The single numbers in brackets () are how many faces of each there are.)

$\displaystyle \mbox{Surface Area} = 360~\mbox{units}^2$ - Aug 15th 2010, 02:11 AMUnknown008
The area of the cross can be obtained by dividing the cross into 5 parts; 4 rectangles and 1 central square.

Each rectangle has area 5*4 = 20.

Area of 4 rectangles = 20*4 = 80.

Area of central square = 4*4 = 16.

Total area of cross section = 16 + 80 = 96.

Volume of Prism = 96 x 3 = 288 units^3 - Aug 15th 2010, 02:16 AMhaftakhan
Very difficult question to understand.

- Aug 15th 2010, 02:23 AMhaftakhan
Unit^2 or unit^3 ?

- Aug 15th 2010, 02:32 AMUnknown008
Oh, sorry, a typo from my part. It's units^3

http://i38.tinypic.com/17etf9.png

Here is a picture to help you understand. The rectangles are reddish and the central square is grey.