the perimeter of a rectangle is 48 m. If the length is decreased by twice, the result is 12 m. Find the length and the width of the rectangle.

Question

The perimeter of a rectangle is 48 m. If the length is decreased by twice, the result is 12 m. Find the length and the width of the rectangle.

Answer 1

perimeter = P = 48

length = L

decrease by "twice" might mean L/2

or maybe L-2?

W

then P=12

LW = 48, and (L-2)W = 12

then L=48/W and (48/W-2)W = 12

48 -96W = 12

W = 36/96 = 3/8 m.

L= 48/(3/8) = 48(8)/3 = 16(8) = 128 m.

that's a little strange, lopsided

maybe you meant new L = L-2W?

then (L-2W)W= 12 where again L=48/W

(48/W -2W)W= 12

48-2W^2 = 12

2W^2 = 36

W^2 = 18

W = sqr18 = 3sqr2

L=48/W = 48/3sqr2 = 16/sqr2

or maybe you meant "decreased by twice" as halving L

(L/2)W = 48

and L=(48/W)(2) = 96W

(96W)W = 12

W^2 = 12/96 = 1/8

W= sqr(1/8) = 1/sqr8

L =48/W = 48/(1/sqr8) = 48sqr8= 96sqr2

That seems like a strange answer too

the question is an enigma wrapped in a riddle.

but maybe one of the above are the irrational solutions.

the 1st solution avoids irrational numbers, but it's a very lopsided rectangle

Sometimes the major problem is just figuring out what the problem is.