Question

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.

Answer 1

By solving a system of equations we will see that the dimensions are:

width = 16in

length = 76in

How to get the dimensions of the rectangle?

Here we have a rectangle of dimensions L = length and W = width.

We know that:

$L = 44in + 2W$

We also know that the diagonal is 2 inches more than the length, remember that the diagonal of a rectangle is:

$D = \sqrt{L^2 + W^2}$

Then we can write this as:

$\sqrt{L^2 + W^2} = L + 2in$

So we have the system of equations:

$\sqrt{L^2 + W^2} = L + 2in\\\\L = 44in + 2W$

First, we can rewrite the first equation as:

$L^2 + W^2 = (L + 2in)^2$

Now we can replace the second equation there, to get:

$(44in + 2W)^2 + W^2 = (44in + 2W + 2in)^2$

So we have a quadratic equation, expanding both sides we get:

$4W^2 + 176in*W + (44in)^2 + W^2 = 4W^2 + 184in*W + (46in)^2\\\\W^2 - 8in*W - 180 in^2 = 0$

Now we just need to solve that quadratic equation, using Bhaskara's, we get:

$W = \frac{8in \pm \sqrt{(-4in)^2 - 8*(-180in^2)} }{2} \\\\W = \frac{4in \pm 28in}{2}$

We only care for the positive solution, which is:

W = (4in + 28in)/2 = 16in

And the length is:

L = 44in + 2*W

L = 44in + 2*(16in) = 76in

If you want to learn more about systems of equations:

https://brainly.com/question/13729904

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Answer 2

Answer:

  18 inches by 80 inches

Step-by-step 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.