Find the length and width of a rectangle whose perimeter is 44feet and whose area is 112 suare feet

the dimensions are W = 8 ft and L = 14 ft

Step-by-step explanation:

Use the variables L and W.

Then P = 44 ft = 2L + 2W, and A = 112 ft² = L·W.

Reducing the first equation, we get 22 ft = L + W.  Solving for L, we get L = 22 ft - W.

Substituting 22 ft - W for L in the area equation, we obtain:

A = 112 ft² = L · W = (22 ft - W) · W, or 112 ft² = (22 ft)W - W²

Let's write this in standard form for a quadratic:

112 ft² = (22 ft)W - W² ↔ W² - (22 ft) · W + 112 ft² = 0

The coefficients of this quadratic equation are a = 1, b = -22 and c = 112.

The discriminant is thus  (-22)² - 4(1)(112), or 484 - 448 = 36.

Thus, the roots are:

        -(-22) ± √36

W =

                  2

          22 ± 6

     = , so that W = 14  and W = 8.

                2

Since L = 22 - W, L could be either 22 - 8 = 14   or   22 - 14 = 8

Thus, the dimensions are W = 8 and L = 14.

Check:  Does WL = 8(14) ft = 112 ft²?  YES

             Does P = 2W + 2L = 2(8 ft) + 2(14 ft) = 44 ft?  YES

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d

step-by-step explanation:

the correct answer is d

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