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Context/Question: Rachel bought a framed piece of artwork as a souvenir from her trip to Disney World. The diagonal of the frame is 20 inches. If the length of the frame is 4 inches greater than it’s width, find the dimensions of the frame.

Answer: The length is __ in. and the width is __ in.


Sagot :

The diagonal of a rectangular figure is an illustration of Pythagoras theorem

The length is 16 in, and the width is 12 in

The given parameters are:

[tex]\mathbf{Diagonal = 20}[/tex]

[tex]\mathbf{Length = 4 + Width}[/tex]

The diagonal is calculated using, the following Pythagoras theorem

[tex]\mathbf{Diagonal^2 = Length^2 + Width^2}[/tex]

Represent length with L and width with W

So, we have:

[tex]\mathbf{20^2 = L^2 + W^2}[/tex]

[tex]\mathbf{400 = L^2 + W^2}[/tex]

This gives

[tex]\mathbf{400 = (4 + W)^2 + W^2}[/tex]

Expand

[tex]\mathbf{400 = 16 + 8W + W^2 + W^2}[/tex]

[tex]\mathbf{400 = 16 + 8W + 2W^2}[/tex]

Divide through by 2

[tex]\mathbf{200 = 8 + 4W +W^2}[/tex]

Rewrite as:

[tex]\mathbf{W^2 + 4W +8 -200 = 0}[/tex]

[tex]\mathbf{W^2 + 4W - 192 = 0}[/tex]

Using a calculator, we have:

[tex]\mathbf{W = \{-16,12\}}[/tex]

The width cannot be negative.

So, we have:

[tex]\mathbf{W =12}[/tex]

Recall that:

[tex]\mathbf{Length = 4 + Width}[/tex]

This gives

[tex]\mathbf{L= 4 + 12}[/tex]

[tex]\mathbf{L= 16}[/tex]

Hence, the length is 16 in, and the width is 12 in

Read more about diagonals at:

https://brainly.com/question/18983839

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