SOLUTION: Suppose that the width of a certain rectangle is three-fourths of its length, and the area of that same rectangle is 48 square meters. What is the length and the width of this rect

Algebra ->  Rectangles -> SOLUTION: Suppose that the width of a certain rectangle is three-fourths of its length, and the area of that same rectangle is 48 square meters. What is the length and the width of this rect      Log On


   

Question 349249: Suppose that the width of a certain rectangle is three-fourths of its length, and the area of that same rectangle is 48 square meters. What is the length and the width of this rectangle?
Found 2 solutions by nyc_function, ewatrrr:
Answer by nyc_function(2741) About Me  (Show Source):
You can put this solution on YOUR website!
Length = x

Width = (3/4)x = (3x/4)

Area = 48m^2

Use A = L*W

48 = x(3x/4)

After solving for, we get x = -8 and x = 8.

Disregard x = -8 since length and width are distances and distance is positive.

The length is 8 meters.

The width is (3*8)/4 or 6 meters.

Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!
Hi,
*Note: l*w = the area of a rectangle.
.
the question states the following to be true:
w = (3/4)*l
.
l%2A%283%2F4%29%2Al=+48m%5E2
.
%283%2F4%29%2Al%5E2=+48m%5E2
.
Multiply both sides of the equation by (4/3)
l%5E2=+%284%2F3%29%2A48
l%5E2=+64m%5E2
l = +-8
plus value is what we can use to represent length.
l = 8cm
w = 6m
.
check you answer
8m+%2A6m+=+48m%5E2