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An envelope measures $15$ inches by $8$ inches. A pencil is placed in the envelope at a diagonal. What is the maximum possible length of the pencil? $\newline$ $\_\_\_\_\_$ inches

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Pythagorean theorem: word problems

Full solution

Q. An envelope measures

15

inches by

8

inches. A pencil is placed in the envelope at a diagonal. What is the maximum possible length of the pencil?

\newline

\_\_\_\_\_

inches

Define Diagonal Length: We know the envelope's length is $15$ inches and width is $8$ inches. Let's call the diagonal $d$ inches, which is also the maximum length of the pencil. $\newline$ Using the Pythagorean Theorem: $(\text{Leg})^2 + (\text{Leg})^2 = (\text{Hypotenuse})^2$ $\newline$ $15^2 + 8^2 = d^2$
Calculate Squares: Calculate the squares: $\newline$ $15^2 = 225$ $\newline$ $8^2 = 64$ $\newline$ $225 + 64 = d^2$
Add Squares: Add the squares: $\newline$ $225 + 64 = 289$ $\newline$ $289 = d^2$
Find Diagonal Length: Find $d$ by taking the square root of $289$ : $\sqrt{289} = \sqrt{d^2}$ $17 = d$

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