The area of a rectangular piece of cardboard is 90 square centimeters, and the perimeter is 46 centimeters. How do you find the dimensions of the rectangle?

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The area of a rectangular piece of cardboard is 90 square centimeters, and the perimeter is 46 centimeters. How do you find the dimensions of the rectangle?

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Answer 1 · 2016-11-14T03:45:40

Please see the explanation.

Explanation:

Let L = the length

Let W = the width

$L W = 90 {cm}^{2} [1]$ $LW = 90" cm"^2" [1]"$

$2 L + 2 W = 46 cm [2]$ $2L + 2W = 46" cm [2]"$

Divide equation [2] by 2:

$L + W = 23 cm$ $L + W = 23" cm"$

Subtract L from both sides:

$W = 23 cm - L$ $W = 23" cm" - L$

Substitute $23 cm - L$ $23" cm" - L$ for W in equation [1]:

$L (23 cm - L) = 90 {cm}^{2}$ $L(23" cm" - L) = 90" cm"^2$

Use the distributive property

$23 cm (L) - L^{2} = 90 {cm}^{2}$ $23" cm"(L) - L^2 = 90" cm"^2$

Subtract $90 {cm}^{2}$ $90" cm"^2$ from both sides:

$23 cm (L) - L^{2} - 90 {cm}^{2} = 0$ $23" cm"(L) - L^2 - 90" cm"^2 = 0$

Multiply both sides by -1:

$L^{2} - 23 cm (L) + 90 {cm}^{2} = 0$ $L^2 - 23" cm"(L) + 90" cm"^2 = 0$

Having solved this type of problem with the quadratic formula, many times, I know that the greater of the two solutions gives the length and the lesser, the width:

$L = \frac{23 cm + \sqrt{{(23 cm)}^{2} - 4 (1) (90 {cm}^{2})}}{2}$ $L = (23" cm" + sqrt((23" cm")^2 - 4(1)(90" cm"^2)))/2$

$L = 18 cm$ $L = 18" cm"$

$W = \frac{23 cm - \sqrt{{(23 cm)}^{2} - 4 (1) (90 {cm}^{2})}}{2}$ $W = (23" cm" - sqrt((23" cm")^2 - 4(1)(90" cm"^2)))/2$

$W = 5 cm$ $W = 5" cm"$

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The area of a rectangular piece of cardboard is 90 square centimeters, and the perimeter is 46 centimeters. How do you find the dimensions of the rectangle?

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Explanation:

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