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The shorter sides of a rectangle measure 4 inches eachand one of its diagonals measures 8 inches. Which ofthe following is the measure of one of its longer sides?

Sagot :

Lets draw a picture of the rectangle:

From our figure, we can note that triangle ABC is a right triangle, so we can apply Pythagorean theorem, that is

[tex]4^2+x^2=8^2[/tex]

which gives

[tex]16+x^2=64[/tex]

If we move 16 to the right hand side, we get

[tex]\begin{gathered} x^2=64-16 \\ x^2=48 \end{gathered}[/tex]

Then, x is given by

[tex]x=\sqrt[]{48}[/tex]

since 48 = 16 x 3, we get

[tex]\begin{gathered} x=\sqrt[]{16\times3} \\ x=\sqrt[]{16}\times\sqrt[]{3} \\ x=4\sqrt[]{3} \end{gathered}[/tex]

therefore, the answer is

[tex]x=4\sqrt[]{3}[/tex]

which is the measure of the longer side.

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