Answered

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An envelope measures 8 inches by 5 inches. A pencil is placed in the envelope at a diagonal. What is the maximum possible length of the pencil? If necessary, round to the nearest tenth.

Sagot :

ANSWER

[tex]9.4in[/tex]

EXPLANATION

First, let us make a sketch of the problem:

The envelope is shaped like a rectangle.

The maximum possible length of the pencil is the length of the diagonal of the envelope.

To find the length of the diagonal, apply the Pythagoras theorem:

[tex]\text{hyp}^2=a^2+b^2[/tex]

where hyp = hypotenuse

a, b = legs of the triangle formed by the diagonals and the sides of the envelope.

Therefore, for the question, we have that:

[tex]\begin{gathered} p^2=5^2+8^2 \\ p^2=25+64 \\ p^2=89 \\ p=\sqrt[]{89} \\ p\approx9.4in \end{gathered}[/tex]

The maximum possible length of the pencil is 9.4 inches.

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