Question

Area of the rectangle is 78 square ft. The length is seven feet more than the width. Find the length and width.

Answer 1

L = 13 square ft and W = 6 square ft.

Given area of the rectangle is 78 square ft. Let L, W be length and width of the rectangle.

So, LxW = 78

W = 78 / L ->(Equation 1).

And length is seven feet more than the width.

So, L = 7 + W -> (Equation 2)

By substituting Equation 1 in equation 2, we get

L = 7 + (78 / L)

L = (7L + 78) / L

L^2 = 7L + 78

L^2 - 7L - 78 = 0

Factorizing the above eqaution,

(L - 13) x (L + 6) = 0

L = 13 or L = -6

Since, length cannot be negative, L = 13. Substitute L value in Equation 1

W = 78 / 13 = 6

So, length of rectangle = 13 square ft and width of rectangle = 6 square ft.

Answer 2

length = 13 feet , width = 6 feet

Explanation:

let w represent width then length is w + 7

area of rectangle is calculated as

A = wl = w(w + 7)

given A = 78 , then

w(w + 7) = 78

w² + 7w = 78 ( subtract 78 from both sides )

w² + 7w - 78 = 0 ← in standard form

consider the factors of the constant term (- 78) which sum to give the coefficient of the w- term (+ 7)

the factors are + 13 and - 6 , since

13 × - 6 = - 78 and + 13 - 6 = + 7 , so

(x + 13)(x - 6) = 0 ← in factored form

equate each factor to zero and solve for w

x + 13 = 0 ⇒ x = - 13

x - 6 = 0 ⇒ x = 6

however, w > 0 , so w = 6

then length = w + 7 = 6 + 7 = 13 feet and width = w = 6 feet