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unearthly09
What is the surface area of this three-dimensional, symmetrical, capital letter “I” block? Show your work.
All right, how familiar with surface area are you?
i know how to find the surface area of rectangular prisms just need a little help with the composite figurs
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.
Cool, can you determine the number of sides?|dw:1343144067033:dw|
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
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 ;)
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
You are given the lengths of all the sides. So, what would be the area of just the front ?
what do u mean the area of the I
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
Yeah, nice! So that takes care of the front, well, and the back areas.
area of front + area of back + area of left side + area of right side + area of top + area of bottom
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)
Now the side is more difficult, it is more complex than the front.
How many rectangles comprise a side of the surface?
is it 3 the 3 long ones and the one short one
i meant 2 long ones
|dw:1343144879045:dw| The black diagonal lines are for the left side, which we want to work with
i have a question is the answer 104???
Heh, cut the chase, eh? Just a second
No, that's not big enough
so where did i go wrong i have no clue what I did
Heh, I'm checking my math...
because I messed up, 104 is right
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
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
i really kind of did what u told me than i found the area of the 2 sides
my work is all jumbled up
Oh well, very cool though, that's a straightforward way to solve it, so nice work!
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.
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
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.
Let's see: 3 blocks: 2 long rectangles and 1 middle connector
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
|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
yes it does....thank u so much :)
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
omg thank u i get it now