Question

The perimeter of a rectangular lot is 198 ft. The width of the lot is 57 ft less than twice length. Find the

length and width of the lot.

Answer 1

Explanation:

W = (2L -57)

P = 2×(L+W)

198 = 2× (L + 2L -57)

198 = 2× (3L - 57)

3L-57 = 198/2

3L-57 = 99

3L = 99+57

3L = 156

L = 156/3 = 52

W = 2(52) -57 = 104-57 = 47

so, the length = 52 ft

and the width = 47 ft

Answer 2

length = 52 ft

width = 47 ft

Explanation:

Perimeter of a rectangle

P = 2(W + L)

where:

• P = perimeter
• W = width
• L = length

Given:

• P = 198 ft
• W = 2L - 57

Substitute the given values into the formula and solve for L:

⇒ 2(2L - 57 + L) = 198

⇒ 2L - 57 + L = 99

⇒ 3L - 57 = 99

⇒ 3L = 156

⇒ L = 52

Substitute the found value of L into the expression for W and solve for W:

⇒ W = 2L - 57

⇒ W = 2(52) - 57

⇒ W = 104 - 57

⇒ W = 47

Therefore,

• length = 52 ft
• width = 47 ft