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The length of a rectangle is 44 inches greater than twice the width. If the diagonal is 2 inches more than the​ length, find the dimensions of the rectangle.

User Selmaril
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1 Answer

4 votes

Answer:

18 inches by 80 inches

Explanation:

The given relations can be used in conjunction with the Pythagorean theorem to find the rectangle dimensions.

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setup

Let x and y represent the width and length of the rectangle, respectively. One of the relations is that between length and width:

y = 2x +44 . . . . . length is 44 more than twice the width

The other relation is described by the Pythagorean theorem. The square of the diagonal is the sum of the squares of the length and width:

x² +y² = (y +2)²

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solution

Expanding the second equation, and subtracting y², we find ...

x² +y² = y² +4y +4

x² = 4y +4

Substituting for y using the first equation gives the quadratic ...

x² = 4(2x +44) +4

x² = 8x +180 . . . . . . . eliminate parentheses

x² -8x +16 = 196 . . . . add 16 -8x to make perfect squares

(x -4)² = 14²

We're only interested in the positive solution, so ...

x = 4 +14 = 18 . . . . . . . . square root, add 4

y = 2(18) +44 = 80

The dimensions of the rectangle are 18 inches wide by 80 inches long.

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Check

The diagonal is √(18² +80²) = √6724 = 82.

User TryPyPy
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