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Answer :
To find the rectangle's dimensions with a length twice the width and a perimeter of 120 cm, divide the perimeter by 6 to get the width (20 cm), then double that to get the length (40 cm).
The student is tasked with finding the dimensions of a rectangle where the length is twice as long as its width, and the perimeter is 120 centimeters.
First, let's establish the formulas we need:
The perimeter of a rectangle (P) is given by the formula P = 2l + 2w, where l represents the length and w represents the width of the rectangle.
According to the problem:
The length (l) is twice the width (w), so l = 2w.
The perimeter (P) is 120 cm.
Substitute l with 2w into the perimeter formula:
P = 2(2w) + 2w
P = 4w + 2w
P = 6w
Since P = 120 cm, we have:
120 = 6w
w = 120 / 6
w = 20 cm
Once we have the width, we can find the length:
l = 2w
l = 2(20 cm)
l = 40 cm
Thus, the dimensions of the rectangle are a width of 20 cm and a length of 40 cm.
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