The rectangular banquet hall floor has a width that is 7/12 its length and a perimeter of 228 feet. What are the dimensions of the banquet hall floor?
(can someone solve and explain this answer)
2007-11-19 09:32:48 · 2 answers · asked by Kira 2 in Science & Mathematics ➔ Mathematics
let x be the length
7/12 x is the width
2(x + 7/12x) = 228
x + 7/12x = 114
12x + 7x = 1368
19x = 1368
x = 72
lenght = 72 width = 7*72/12 = 42
2007-11-19 09:38:43 · answer #1 · answered by norman 7 · 0⤊ 0⤋
The perimeter of a rectangle is width plus width plus length plus length. Basically, P=2w + 2l
Since w=7/12 l, you can plug that into the equation for w.
So, P= 2(7/12 l) +2l, which is the same as 7/6 l + 2l
Add them up and you get 3 and 1/6 lengths being the perimeter.
So, divide the perimeter by 3 and 1/6.
228 divided by 3 and 1/6 is the length (I'm afraid I don't have a calculator on hand).
Plug the length back into the original equation (P=2w +2l) and you'll get the width. Then you'll have the dimensions.
2007-11-19 17:49:13 · answer #2 · answered by Mariner Cat 2 · 0⤊ 0⤋