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anonymous
 3 years ago
What is the surface area of this threedimensional, symmetrical, capital letter “I” block? Show your work.
anonymous
 3 years ago
What is the surface area of this threedimensional, symmetrical, capital letter “I” block? Show your work.

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0All right, how familiar with surface area are you?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i know how to find the surface area of rectangular prisms just need a little help with the composite figurs

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0We 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
 3 years ago
Best ResponseYou've already chosen the best response.0Cool, can you determine the number of sides?dw:1343144067033:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Surface 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
 3 years ago
Best ResponseYou've already chosen the best response.0So 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
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1343144255358:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0The 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
 3 years ago
Best ResponseYou've already chosen the best response.0You are given the lengths of all the sides. So, what would be the area of just the front ?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0what do u mean the area of the I

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Find 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
 3 years ago
Best ResponseYou've already chosen the best response.0Yeah, nice! So that takes care of the front, well, and the back areas.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0area of front + area of back + area of left side + area of right side + area of top + area of bottom

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Due 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
 3 years ago
Best ResponseYou've already chosen the best response.0Now the side is more difficult, it is more complex than the front.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0How many rectangles comprise a side of the surface?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0is it 3 the 3 long ones and the one short one

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1343144879045:dw The black diagonal lines are for the left side, which we want to work with

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i have a question is the answer 104???

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Heh, cut the chase, eh? Just a second

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0No, that's not big enough

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so where did i go wrong i have no clue what I did

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Heh, I'm checking my math...

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0because I messed up, 104 is right

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.05 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
 3 years ago
Best ResponseYou've already chosen the best response.0Yes, 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
 3 years ago
Best ResponseYou've already chosen the best response.0i really kind of did what u told me than i found the area of the 2 sides

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0my work is all jumbled up

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Oh well, very cool though, that's a straightforward way to solve it, so nice work!

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0does 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
 3 years ago
Best ResponseYou've already chosen the best response.0The 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
 3 years ago
Best ResponseYou've already chosen the best response.0Usually, 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
 3 years ago
Best ResponseYou've already chosen the best response.0Let's see: 3 blocks: 2 long rectangles and 1 middle connector

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0the 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
 3 years ago
Best ResponseYou've already chosen the best response.0dw: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
 3 years ago
Best ResponseYou've already chosen the best response.0Does this make sense?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yes it does....thank u so much :)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0No 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
 one year ago
Best ResponseYou've already chosen the best response.0omg thank u i get it now
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