Question

A rectangle has an area of 32 square feet and a

perimeter of 24 feet. What is the shortest of the side
lengths, in feet, of the rectangle?
F. 1
G. 2
H. 3
J. 4
K. 8

Answer 1

To find the shortest side of the rectangle with an area of 32 square feet and a perimeter of 24 feet, we solve the resulting quadratic equation from the given area and perimeter formulas. The shorter side length is determined to be 4 feet.

Step-by-step explanation:

The student's question deals with finding the shortest side lengths of a rectangle given its area and perimeter. To solve this, we can set up equations based on the properties of rectangles. Let l be the length and w be the width of the rectangle.

From the question, we have these two equations:

• The area of the rectangle is lw = 32 square feet.
• The perimeter of the rectangle is 2l + 2w = 24 feet.

Solving the perimeter equation for one variable:

1. w = 12 - l

Substituting this into the area equation gives:

1. l(12 - l) = 32
2. 12l - l2 = 32
3. l2 - 12l + 32 = 0
4. Factoring the quadratic equation gives: (l - 4)(l - 8) = 0
5. So, l = 4 or l = 8

Since the length cannot be shorter than the width in this case, the width w must therefore be the smaller value. If the length l is 8 feet, then the width w is 4 feet, and vice versa.

Therefore, the correct answer to the shortest side lengths, in feet, of the rectangle is: 4 feet (Option J)

Answer 2

Perimeter=2l+2w
24=2l+2w
2l=24-2w
l=12-w
Area=lw
32=(12-w)w
32=12w-w^2
w^2-12w+32=0
(w-4)(w-8)=0

w-4=0
w=4 ans. l=12-4=8 ans.
w-8=0
w=9 and. l=12-8=4 ans.
Proof:
24=2*8+2*4
24=16+8
24=24
32=8*4
32=32