A rectangular piece of land whose length is twice its width has a diagonal distance of 90 yards. How many yards, to the nearest tenth of a yard, does a person save by walking diagonally across the land instead of walking its length and its width?

the path a person walks diagonally creates a right triangle with the lenght and the width of the rectangular piece of land. The diagonal is 90 yards which becomes the hypotenuse.

let w be the width, then l becomes the length, which is twice the width. Therefore, leg l=2a and leg w=a.

By the Pythagorean Theorem:

[tex]c^{2}=l^{2}+w^{2}[/tex]

Replacing the values:

[tex]90^{2}=(2a)^{2}+a^{2}\\90^{2}=4a^{2}+a^{2}\\90^{2}=5a^{2}\\\\a=\sqrt{\frac{90^{2}}{5} } } =\frac{90}{\sqrt{5} }[/tex]

But that a is the value for the width, if a person walks the width and the length, it becomes:

the length plus the width=2a+a=3a.

So:

[tex]a=\frac{90}{\sqrt{5} } \\\\3a=\frac{3(90)}{\sqrt{5} }[/tex]

Now, the difference between walking across the land and surrounding it, is:

[tex]d=\frac{3(90)}{\sqrt{5} }-90=30.7476\\\\d=30.7[/tex]

We can conclude, a person saves 30.7 yards if walking across the land.

A rectangular piece of land whose length is twice its width has a diagonal distance of 90 yards. How many​ yards, to the nearest tenth of a​ yard, does a person save by walking diagonally across the land instead of walking its length and its​ width?

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A rectangular piece of land whose length is twice its width has a diagonal distance of 90 yards. How many yards, to the nearest tenth of a yard, does a person save by walking diagonally across the land instead of walking its length and its width?