Q:

The length of a rectangle is 3 meters less than twice the width. if the area of the rectangle is 527 square​ meters, find the dimensions.

Accepted Solution

A:
The rectangle is 17 meters wide and 31 meters long. To solve this problem, first work out the equation for the length of the rectangle. Using the information given, we find that length (L) is given by L=2W-3. Next we setup the formula for the area of a rectangle, A=(W)(L). We are told that the area is 527 sq. meters, so our equation is 527=(W)(L). We also know that L=2W-3, so we can plug that into the equation for area as well which gives us 527=W(2W-3), which can be written 0=2(W^2)-3W-527. We now need to solve this equation for W. Factoring the left side we get, 0=(W-17)(2W+31). Thus, either (W-17)=0 or (2W+31)=0 must equal zero. Solving these two equations, we get W=17 or W=-15.5. Since we are looking for the width (which is a length), we know it cannot be negative and therefore the width must be 17 meters. Now that we know the width, we can plug W=17 back into the equation for length (L=2W-3). Doing this give us L=2(17)-3 which is equal to 31. Thus the length is 31 meters.