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The rectangle shown has a perimeter of 38 cm and the given area. Its length is 4 more than twice its width. Write and solve a system of equations to find the dimensions of the rectangle.

The length of the rectangle is ___cm and the width of the rectangle is
___ cm.

2 Answers

1 vote

Final answer:

To find the dimensions of the rectangle, a system of equations is created: 2w + 2l = 38 and l = 2w + 4. By substituting the second equation into the first and solving, the width w is found to be 5 cm and the length l is 14 cm.

Step-by-step explanation:

The student is asked to find the dimensions of a rectangle with a specific area and with the length being 4 more than twice its width. To find these dimensions, we will set up and solve a system of equations.

Let's denote the width of the rectangle as w and the length as l. The problem gives us two key pieces of information:

  • The perimeter of the rectangle is 38 cm, which gives us the equation: 2w + 2l = 38.
  • The length is 4 more than twice the width, giving us another equation: l = 2w + 4.

We can solve this system of equations by substitution or elimination. Here, substitution is more straightforward. Substituting the expression for l from the second equation into the first gives us:

2w + 2(2w + 4) = 38

Simplifying, we have:

2w + 4w + 8 = 38

6w + 8 = 38

Subtracting 8 from both sides, we get:

6w = 30

Dividing both sides by 6, we find that:

w = 5 cm

Substituting w back into the second equation, we get:

l = 2(5) + 4 = 14 cm

Thus, the dimensions of the rectangle are:

Length = 14 cm

Width = 5 cm

User David Van Rijn
by
6.7k points
0 votes

Answer:

L: 19, W: 5 Area: 70

Step-by-step explanation:

2 * (L+W) = 38

L + W = 19

L = 2 * W + 4

2 * W + 4 + W = 19

3W=15

W=5

L=14

Area= 5 x 14 =70

User Meze
by
8.0k points

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