# total surface area of prism

• Apr 5th 2011, 08:11 AM
andyboy179
total surface area of prism
hi,

i need to work out the total surface area of this prism:

Attachment 21372

i know its not draw well but it gets the point across. i was going to divide it up into 3 parts and times the final answer by 3 because there are 3 parts.

however i don't know how to work out the surface area of a cuboid but i know it would contain the length of 8cm, width 3cm and height 3cm.

so if somone could help i would be very very grateful!!

thanks
• Apr 5th 2011, 09:11 AM
Quote:

Originally Posted by andyboy179
hi,

i need to work out the total surface area of this prism:

Attachment 21372

i know its not draw well but it gets the point across. i was going to divide it up into 3 parts and times the final answer by 3 because there are 3 parts.

however i don't know how to work out the surface area of a cuboid but i know it would contain the length of 8cm, width 3cm and height 3cm.

so if somone could help i would be very very grateful!!

thanks

Hi andyboy179,

if the 3 blocks have identical dimensions,
then their volumes are equal,
so multiplying by 3 would give total volume
(assuming the side with no dimension given is 3 units).

However, one block is on top of another, removing 2 sides from the surface area
and the 2 base blocks are side-by-side removing another 2 sides from the total surface
of the shape.
• Apr 5th 2011, 01:42 PM
andyboy179
Hi,
I understand but how would I actually work out the total surface area of the prism?
• Apr 5th 2011, 03:16 PM
Quote:

Originally Posted by andyboy179
Hi,
I understand but how would I actually work out the total surface area of the prism?

Have a look at your drawing again.

It seems that all "long" faces are 8X3, according to your dimensions.
We can consider your figure as 3 identical blocks joined together.

3 long faces are exposed on the bottom right block.
2 long faces are exposed on the bottom left block.
3 long faces are exposed on the top block.

All six 3X3 edges are exposed.

If you add all those independent parts of the surface area,
you will have the total surface area.

You should get the exact same answer if you calculate the surface area of all six flat sides of a block,
then multiply the answer by 3,
then subtract the areas of the 4 hidden flat sides.

Try experimenting with that.
• Apr 6th 2011, 06:19 AM
andyboy179
8 long faces are exposed and 6 3x3 ones. so would i do 8x8=64 and 3x6=18 64+18=82 82x3=246 ??
• Apr 6th 2011, 06:41 AM
Quote:

Originally Posted by andyboy179
8 long faces are exposed and 6 3x3 ones. so would i do 8x8=64 and 3x6=18 64+18=82 82x3=246 ??

It's a roundabout way of getting there.
I wouldn't recommend those calculations by any means though.

To get the surface area of an 8X3 rectangle, how many 1X1 "unit" squares would fit inside ?
Also, for a 3X3 square, how many 1X1 "unit" squares fit inside ?
Those calculations tell you how many "unit" squares are in those sides.
The total surface area is the number of "unit" squares that cover the entire surface
(imagine the surface being tiled with 1X1 squares).
• Apr 6th 2011, 06:44 AM
andyboy179
would there be 11 and 6 1x1 squares?
• Apr 6th 2011, 06:53 AM
Quote:

Originally Posted by andyboy179
would there be 11 and 6 1x1 squares?

If you had a 3X1 rectangle, then you could say it's made of 1 row of 3 "unit" 1X1 squares.
Hence it's area is 3 "square units".
If you had a 2X3 rectangle, then you could say it's made of 2 rows of 3 "unit" 1X1 squares.
Hence it's area is 2X3=6 "square units".
We are counting the "unit squares" within the rectangles.

If you continue in this way, how many "unit" squares fit inside a 3X3 square and an 8X3 rectangle ?

It looks like you were guessing above.
Guessing gets you nowhere in maths.
You should try to get the ideas instead.
• Apr 6th 2011, 07:00 AM
andyboy179
oh i understand so it would be 24 and 9 square units
• Apr 6th 2011, 07:02 AM
Quote:

Originally Posted by andyboy179
oh i understand so it would be 24 and 9 square units

Perfect!

So you only need discover how many of the 24 "square unit" sides are exposed
and how many 9 "square unit" squares are also exposed.

You will get the result you had earlier, which was correct,
but you needed a much clearer way to calculate it.
• Apr 6th 2011, 07:14 AM
andyboy179
how do i actually find out if they are exposed or not?
• Apr 6th 2011, 07:32 AM
Quote:

Originally Posted by andyboy179
how do i actually find out if they are exposed or not?

What I mean by "exposed" is....

Look at the block on the bottom right.
It's upper face is exposed.
So is it's rightmost face.
It's underneath face is exposed also
(think of the entire object suspended on a piece of string
attached to the celing or something like that).
It's leftmost face is not exposed since it is against another block.

Therefore, those 3 faces contribute to the overall surface area.
That block also has two 9 "square unit" faces exposed.

Notice that the bottom-left block has only 2 of the long faces exposed
and two 9 "square unit" square sides exposed.

What about the block at the top ?
• Apr 6th 2011, 07:46 AM
andyboy179
the block at the top has two 9 square unit sides exposed and 3 long faces exposed.
so in total the amount of 9 square unit sides exposed are 6 and 24 square unit has 8 so the total surface area is 14cm^2?
• Apr 6th 2011, 10:10 AM