Question

A rectangle's length is 8.1 more than 10 times its width. Its perimeter is 515.6. Write and solve a system of equations to find the dimensions of the rectangle. Type the area of the rectangle, rounded to the nearest tenth.

Answer 1

A system of equations was set up to solve for the width and length of the rectangle, with the width being 22.7 and length 235 units. Multiplying these dimensions, the area of the rectangle is calculated and rounded to 5334.5.

Step-by-step explanation:

To find the dimensions of a rectangle where the length is 8.1 more than 10 times its width and its perimeter is 515.6, we can set up a system of equations. Let the width be w, then the length will be l = 10w + 8.1. The perimeter of a rectangle is given by the formula 2l + 2w. Using this, we can write our equations as:

• l = 10w + 8.1
• 2l + 2w = 515.6

Substitute the first equation into the second to find the value of w:

1. 2(10w + 8.1) + 2w = 515.6
2. 20w + 16.2 + 2w = 515.6
3. 22w + 16.2 = 515.6
4. 22w = 499.4
5. w = 22.7 (rounded to the nearest tenth)

Now, substitute w into the first equation to find l:

1. l = 10(22.7) + 8.1 = 235

Finally, compute the area by multiplying the length by the width (l × w):

• Area = l × w = 235 × 22.7 = 5334.5 square units

The area of the rectangle, rounded to the nearest tenth, is 5334.5.