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The length of a rectangle is nine inches less than twice the width. If the perimeter is 84 inches, find the dimensions of the rectangle

User Leventix
by
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1 Answer

18 votes
18 votes

Dimensions of the rectangle are 25 inches × 17 inches.

Explanation:

Let the length of the rectangle = l inches and width = w inches

The length of the rectangle is 9 inches less than twice the width.

Therefore, l = (2w - 9)

Perimeter of the rectangle is = 84 inches

And perimeter of the rectangle = 2(Length + width)

= 2(l + w)

= 2[(2w - 9) + w]

= 2[3w - 9]

Now 84 = 2{3w - 9)

3w - 9 = \frac{84}{2}

2

84

3w - 9 = 42

3w - 9 = 423w = 42 + 9

3w - 9 = 423w = 42 + 93w = 51

3w - 9 = 423w = 42 + 93w = 51w = \frac{51}{3}

3w - 9 = 423w = 42 + 93w = 51w = \frac{51}{3} 3

3w - 9 = 423w = 42 + 93w = 51w = \frac{51}{3} 351

3w - 9 = 423w = 42 + 93w = 51w = \frac{51}{3} 351

3w - 9 = 423w = 42 + 93w = 51w = \frac{51}{3} 351

w = 17 inches

Since l = 2w - 9

l = 2×17 - 9

= 34 - 9

= 25 inches

Therefore, dimensions of the rectangle are 25 inches × 17 inches.

User Bastek
by
2.9k points