A rectangle has an area of 32 square feet and a perimeter of 24 feet. What is the shortest side length, in feet, of the rectangle?

A. 1
B. 2
C. 3
D. 4
E. 8

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

Answered by Anonymous | 2024-06-10

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:

w = 12 - l

Substituting this into the area equation gives:

l(12 - l) = 32
12l - l__2 = 32
l__2 - 12l + 32 = 0
Factoring the quadratic equation gives: (l - 4)(l - 8) = 0
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)

Answered by JoaquinRafael | 2024-06-24

The shortest side length of the rectangle is 4 feet. This is found by solving the equations for area and perimeter. The dimensions satisfy both conditions given in the problem.
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Answered by Anonymous | 2024-09-04