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Help with 5th grade math ECR? don't really understand my bro's hw.?
Mr. harrison has a very large backyard. He contacted Clover Pool Company(CPC) to have a pool installed in the shape of a rectangle. The CPC would build a pool that has a perimeter of 72 yards. Mr Harrsion would be allowed to choose the dimensions (length and width) of the pool. He chose the dimensions that would give him the greatest area possible.
A.) What are the dimensions of the pool tat would give Mr. harrison the greatest area?
B.) How did u get that answer(if u can-pls do so, but it's not a must)
B2) Suppose Mr. Harrison decides he wants to save space and now decides to make the pool with the smallest area possible remembering that the perimeter must still be 72 yards. What are the dimensions that would give Mr. Harrison the greatest area.
^^^(if u could help us find A, we should be able to do B2 by ourselves)
PLS &THANK YOU!!!
3 Answers
- 1 decade agoFavorite Answer
This is a question about the relationship between area and perimeter. It's purpose is to reinforce the idea that even if rectangles have the same perimeter, they can have different areas.
For example:
Rectangle 1
Perimeter: 72 yards
Length: 1 yard
Width: 35 yards
Area: 35 square yards
Rectangle 2
Perimeter: 72 yards
Length: 2 yard
Width: 34 yards
Area: 68 square yards
Rectangle 3
Perimeter: 72 yards
Length: 10 yard
Width: 26 yards
Area: 260 square yards
Rectangle 4
Perimeter: 72 yards
Length: 18 yard
Width: 18 yards
Area: 324 square yards
Rectangle 4 is the answer to queston A.
Rectangle 1 is the answer to question B2.
You can show your work (question B) by setting up a table with the following headings:
Perimeter Length Width Area
Start with a length of 1 and a width of 35, and work your way down to a pool that is 18 X 18.
Source(s): I taught fifth grade for years. - 1 decade ago
OK, the perimeter is 72 yards, so you divide that in half to get the what the length and width are when added together--36. Divide that in half again, and you have both the width and length--18. You multiply the length (18) and the width (also 18) and that's the greatest possible area with the given information.
For the smalles, you would start out the same, dividing the total perimeter in half to get 36. Now, assuming you are still working with whole units (no fractions of yards and such), the smallest width is 1, and subtract that from 36 to get the length, 35. That would give you the smallest area.
