Final answer:
To find the possible dimensions of the rectangle with a perimeter of 24 inches and an area of 32 square inches, we set up a system of equations and solve to get the width as either 4 inches or 8 inches. This would make the length the other value since length and width are interchangeable, giving dimensions of the rectangle as 4 inches by 8 inches or 8 inches by 4 inches.
Step-by-step explanation:
A student needs to find the dimensions of a rectangle when given the perimeter and the area. Given the perimeter of the rectangle is 24 inches and the area is 32 square inches, we can write two equations based on what we know about rectangles: perimeter (P) is 2 times the length (L) plus 2 times the width (W), and area (A) is the length times the width. So, we have P = 2L + 2W and A = LW.
To solve this, set up the system of equations:
- 2L + 2W = 24 (perimeter equation).
- LW = 32 (area equation).
From the perimeter equation, we can express L as L = 12 - W, and then substitute into the area equation: (12 - W)W = 32.
Solving this quadratic equation by expanding and rearranging gives us W^2 - 12W + 32 = 0. Factoring yields (W - 4)(W - 8) = 0, which gives us W = 4 or W = 8.
Therefore, the possible dimensions of the rectangle, since length and width are interchangeable, are 4 inches by 8 inches or 8 inches by 4 inches. Both of these will provide the correct area and perimeter as stated in the problem.