Question

The rectangle shown has a perimeter of 50 cm and the given area. Its length is 5 more than four times its width. Write and solve a system of equations to find the dimensions of the rectangle.

Answer 1

To solve for the dimensions of the rectangle with a perimeter of 50 cm and a length that is 5 more than four times its width, we set up a system of equations (2w + 2l = 50 and l = 4w + 5), and find that the width is 4 cm and the length is 21 cm.

Step-by-step explanation:

To find the dimensions of a rectangle given the perimeter and an equation for the length in terms of the width, we can set up a system of equations. Let's denote the width of the rectangle as w and the length as l. The problem states that the perimeter is 50 cm, which leads to one equation, 2w + 2l = 50. The other equation comes from the given relation between length and width, which is l = 4w + 5.

Firstly, we can simplify the perimeter equation to w + l = 25. Now we'll substitute the second equation into the first to solve for w:

1. w + (4w + 5) = 25
2. 5w + 5 = 25
3. 5w = 20
4. w = 4

With the width found to be 4 cm, we can now determine the length:

1. l = 4w + 5
2. l = 4(4) + 5
3. l = 16 + 5
4. l = 21

Therefore, the dimensions of the rectangle are 4 cm in width and 21 cm in length.