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missashleyn
Can someone help me, please? The sun roof in Troy’s house is shaped like a rectangle. He increases the length of the sides of the sun roof to five times the existing ones. How will this change affect the perimeter of the sun roof? It will be 20 times the original perimeter. It will be 5 times the original perimeter. It will be 10 times the original perimeter. It will be 15 times the original perimeter.
It depends on what the given perimeter is, my guess is that its 20
They didn't give me a perimeter :/
We don't need to know the original perimeter.
Oh then I have absolutely no idea how to figure this out :(
Well in some cases there should be. But if there isn't the best answer on the top of my head is like 20
Perimeter is just the sum of all of the sides. Let's make p be the original perimeter. Now he increases the sides to be 5 times the original ones. Well, for a rectangle, the perimeter is length + length + width + width, or 2*length + 2*width. Now we are making our perimeter be 5*length + 5*length + 5*width + 5*width = 10*length + 10*width. Compare the old perimeter (2*length + 2*width) with the new (10*length + 10*width). What is the ratio?
I am so confused, ratio of what?
If you like, take a roof that is square, and 1 unit on each side. (squares are rectangles). Perimeter is 1+1+1+1 = 4. Okay, now make it a square 5 units on a side. Perimeter is 5+5+5+5 = 20. 20/4 = 5. So, multiplying the sides by 5 makes the perimeter also be multiplied by 5.
The ratio of the new perimeter (bigger roof, with sides 5x as long) to the old perimeter (original roof).
So the perimeter will be 5 times the original perimeter?
Oh wow, that's confusing. I have another question like this.. could you help me with it too?
Fire away, let's make it less confusing :-)
Lol okay! A museum exhibit, ABCD, has infrared beams around it for security, as shown. If the length of the beams is doubled on each side, which statement is correct about the maximum area available for an exhibit to be displayed? It becomes six times the original area. It becomes double the original area. It becomes four times the original area. It becomes eight times the original area.
Okay, this is a little trickier, but not much :-) Area = length * width for a rectangle. If we double both length and width, what happens to area? New area = (2*old length)*(2*old width)
Think of it like: you have a piece of notebook paper that covers the original area exactly. Now you want to make a model that has each side twice as long. Could you do that by putting down more paper? How much?
So would it become 4 times the original area?
Yes! For the earlier problem, a similar mental model would be that the perimeter is a piece of string that goes around the roof. If you make each side 5x as long, you need 5x as much string to cover each side.
Why would it become 4 times the original area though?
Because the area is length * width, and we doubled each one, so there's 2 * 2 = 4. Think of the paper model: if you had one sheet of notebook paper that fit the area exactly, then putting down 3 more sheets in the same orientation would make a rectangle that was twice the length and twice the width, right? And you'd have 4 * the area...
But if there's only one of something, you'd only need to add one more of the same thing to be twice the length.. right? I'm so confused! :(
Basically, the idea is that you put the ratio you've changed the fundamental measurement by into the formula for the area, or perimeter, or volume, or shoe size, or whatever, and that gives you the ratio of the new to the old. Area = length * width, so when we put 2x length and 2x the width, we get 4x the area. Yes, only one more sheet of paper, but in each direction - both length and width.
Similarly, if we made each side be half the length they are now, we would up with (1/2)(1/2) = 1/4 the area.
Oh okay! Thank you very much :)