Best Answer

Assuming this question is asking about a rectangle:

Perimeter = 2l + 2w and according to this problem Perimeter = 96, so

96 = 2l + 2w

If l = 5w, then we can replace each l in the statement above with 5w, so

96 = 2(5w) + 2w

96 = 10w + 2w

96 = 12w

8 = w

If 8 = w, and l = 5w, then l = 5 times 8 = 40

The area of a rectangle is length (l) times width (w), so

A = lw

A = (40) (8)

A = 320 square units

Q: What is the area if perimeter is 96 and l equals 5w?

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48 times 96 equals 4,608.

The perimeter of the square is 96.

8 by 12

Any number less than 576 cm2. A square, with sides of 24 cm will have a perimeter of 96 cm and an area of 576 cm2. Suppose the sides of the rectangle are 24+x cm and 24-x cm, where 0<x<24. Then the perimeter will be 96 cm while the area will be (24+x)*(24-x) = 576 - x2 cm2 The area can have any value in the range (0, 576) depending on the value of x.

Area = 3*45 = 135 square feet. Perimeter = 2*(3 + 45) = 2*48 = 96 feet.

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96 feet.

What are the dimensions of a rectangle that has a perimeter of 56 units and an area of 96 square units

48 times 96 equals 4,608.

4X24

The perimeter of the square is 96.

4 x 24

4 x 24

8 by 12

Any number less than 576 cm2. A square, with sides of 24 cm will have a perimeter of 96 cm and an area of 576 cm2. Suppose the sides of the rectangle are 24+x cm and 24-x cm, where 0<x<24. Then the perimeter will be 96 cm while the area will be (24+x)*(24-x) = 576 - x2 cm2 The area can have any value in the range (0, 576) depending on the value of x.

Area = 3*45 = 135 square feet. Perimeter = 2*(3 + 45) = 2*48 = 96 feet.

The area is 96*sqrt(3) = 166.3 sq inches, approx.

Assuming 96 refers to the area of therectangle, the answer is: infinite. Consider the following sequence of rectangles with breadh B units and length L units.: Breadth = 1 Length = 96. Area = 96, Perimeter = 194 B= 0.1, L = 960. A = 96, P = 1920.2 B = 0.01, L =9600. A = 96, P = 19200.02 B = 0.001, L = 96000. A = 96, P = 192000.002 There is no limit to how small B can get and therefore, how large P can get.