What is the surface area of this three-dimensional, symmetrical, capital letter “I” block? Show your work.

- anonymous

- katieb

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- anonymous

##### 1 Attachment

- anonymous

All right, how familiar with surface area are you?

- anonymous

i know how to find the surface area of rectangular prisms just need a little help with the composite figurs

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- anonymous

We need to determine the number of sides, or specific planes, that the 'I' has. Like a cube, a cube has 6 sides that are in the shape of squares.
This 'I' has quite a 'sides'. So, we'll want to find the area of all these different sides and sum them.

- anonymous

why the area?

- anonymous

Cool, can you determine the number of sides?|dw:1343144067033:dw|

- anonymous

Surface area is just the sum of all the area of the outside of this 3D object. That is the surface, the part you can touch if you were actually holding the 'I', so surface area just means you find the area of the surface, or the outside skin if you will of the object

- anonymous

oh i get it

- anonymous

So perhaps the best way to go about this would be to look at the 'I' straight on. If you can imagine that you have this object and can rotate, this will be simple ;)

- anonymous

|dw:1343144255358:dw|

- anonymous

ok

- anonymous

The best part is that this 'I' is symmetric, so it it has the same properties from different view points. so if we view only the front of the 'I'|dw:1343144361875:dw| this is what the back looks like too

- anonymous

yeah

- anonymous

You are given the lengths of all the sides. So, what would be the area of just the front ?

- anonymous

what do u mean the area of the I

- anonymous

Find the area of the three rectangles that make up the front. Like how I drew the front, even though the attached picture has the front being one large shape of a capital I, you could break that shape down into smaller shapes: like 2 big rectangles at the top and a square in the middle

- anonymous

would it be 22cm

- anonymous

Yeah, nice! So that takes care of the front, well, and the back areas.

- anonymous

area of front + area of back + area of left side + area of right side + area of top + area of bottom

- anonymous

Due to the symmetry, we can just find the area of the front (=area of back), area of right side (=area of left), and the area of the top (=area of bottom)

- anonymous

Now the side is more difficult, it is more complex than the front.

- anonymous

How many rectangles comprise a side of the surface?

- anonymous

is it 3 the 3 long ones and the one short one

- anonymous

i meant 2 long ones

- anonymous

|dw:1343144879045:dw| The black diagonal lines are for the left side, which we want to work with

- anonymous

i have a question is the answer 104???

- anonymous

Heh, cut the chase, eh? Just a second

- anonymous

ha:D

- anonymous

No, that's not big enough

- anonymous

...

- anonymous

so where did i go wrong i have no clue what I did

- anonymous

Heh, I'm checking my math...

- anonymous

because I messed up, 104 is right

- anonymous

really!!!!

- anonymous

5 squares for the left side surface, each 2x2 = 20
2 long rectangles, 5x2, and 1 small square, 2x1 for the front = 22
1 long rectangle, 5x2, for the top = 10
That is 52, so due to symmetry, 52x2 is the total surface area

- anonymous

Yes, how did you get it? If you wrote down your work,you can take a picture of it with maybe your cell phone and upload your work on here

- anonymous

i really kind of did what u told me than i found the area of the 2 sides

- anonymous

my work is all jumbled up

- anonymous

Oh well, very cool though, that's a straightforward way to solve it, so nice work!

- anonymous

does this make sense symmetrical, capital letter i
I would find the surface area of one of the "bars" 5 by 2 by 2 (ignoring the fact that the middle connector hides part of the surface).
So one bar has 4 sides with 5 by 2 or 4*10= 40 sq cm
the 2 ends have area 2*2=4 or 8 total.
So the bar has a surface area of 48
Double this for the other bar: 96
Now the middle connector has 4 sides. I used 2 of them to "fill in" the hidden part of the bar We have 2 sides of 2x2 or 2 sides of 4 or 8 sq cm to add in
96 8 104
Therefore the surface area of this figure is 104 centimeters squared.

- anonymous

i meant 96+8= 104

- anonymous

The middle part however has 2 2x1 sides, and 2 2x2 sides that are visible, and 2 2x1 sides that not visible, if you break the 'I' in 3 3D objects: 2 prisms on the top, 1 in the middle

- anonymous

yeah

- anonymous

right...???

- anonymous

Usually, if you are going to include a part, such as the under side of the top block, which is not part of the 'I''s surface area, that means you need to actually subtract that amount of 'hidden surface area' from the final total.

- anonymous

Let's see:
3 blocks: 2 long rectangles and 1 middle connector

- anonymous

the long rectangles are similar, so let's just focus on one: We are going to use your concept of building the 'I' from these 3 objects and accounting for hidden area.
So, the surface area of one long block is:
4 - 5x2 rectangles
2 - 2x2 rectangles
SA = 4(10) + 2(4) = 48
So surface area of both blocks is 48x2 = 96
Now, the middle connector:
2 - 2x2 rectangles
4 - 2x1 rectangles
SA = 2(4) + 4(2) = 16
But 2 of those 2x1 rectangles are hidden once all the blocks are connected
And, in each long block, an area of 2x1 is hidden too.
96 + 16 - 4(2x1) = 112 - 8 = 104

- anonymous

|dw:1343146032157:dw| In the 2 circled parts, we have to account for the hidden parts of the surface when the blocks are all connected. So we you quite close, and I think you're idea of working with the 3 blocks is a little more intuitive

- anonymous

Does this make sense?

- anonymous

yes it does....thank u so much :)

- anonymous

No problem, I think you can do your other questions if they're similar, you seem to have the right intuition about how to solve the problem

- anonymous

:)thanks again

- anonymous

bye

- anonymous

omg thank u i get it now

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