Question

The area of a rectangle is 66 ft^2 and the length of the rectangle is 7 feet less than three times the width find the dimensions of the rectangle

Answer 1

Length = 11

Width = 6

The area of a rectangle is 66 ft^2:

L * W = 66

Length of the rectangle is 7 feet less than three times the width:

L = 3W-7

Substitute L in terms of W:

(3W-7) * W = 66

Factorise the equation:

3W^2 -7W = 66

3W^2 - 7W - 66 = 0

Factors of 66 =

1 66, 2 33, 3 22, 6 11

(3W + 11) (W - 6) = 0

Solve for W:

3W + 11 = 0

3W = 11

W = 3/11

W - 6 = 0

W = 0 + 6

W = 6

Using the original equation, find L:

L = 3W-7

L = 3(6)-7

L = 18-7

L = 11

L * W = 66

11 * 6 = 66

Answer 2

Answer:

Step-by-step explanation:

Length * width =Area

3(x-7)*x=66

x^2-7x-22=0

Solve for 'x' using quadratic formula

Now width=9.35234995ft

Length=7.05704985ft