Solution 4.1:3c

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In this right-angled triangle, the side of length
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In this right-angled triangle, the side of length 17 is the hypotenuse (it is the side which is opposite the right angle). The Pythagorean theorem then gives
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<math>\text{17}</math>
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is the hypotenuse (it is the side which is opposite the right angle). Pythagoras' theorem then gives
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<math>\text{17}^{2}=8^{2}+x^{2}</math>
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{{Displayed math||<math>17^2 = 8^2 + x^2</math>}}
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or
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{{Displayed math||<math>x^2 = 17^2 - 8^2\,\textrm{.}</math>}}
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<math>x^{2}=\text{17}^{2}-8^{2}</math>
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We get
We get
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{{Displayed math||<math>\begin{align}
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<math>\begin{align}
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x &= \sqrt{17^2-8^2} = \sqrt{289-64} = \sqrt{225}\\[5pt]
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& x=\sqrt{\text{17}^{2}-8^{2}}=\sqrt{289-64}=\sqrt{225} \\
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&= \sqrt{9\cdot 25} = \sqrt{3^2\cdot 5^2} = 3\cdot 5 = 15\,\textrm{.}
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& =\sqrt{9\centerdot 25}=\sqrt{3^{2}\centerdot 5^{2}}=3\centerdot 5=15. \\
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\end{align}</math>}}
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\end{align}</math>
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Current revision

In this right-angled triangle, the side of length 17 is the hypotenuse (it is the side which is opposite the right angle). The Pythagorean theorem then gives

\displaystyle 17^2 = 8^2 + x^2

or

\displaystyle x^2 = 17^2 - 8^2\,\textrm{.}

We get

\displaystyle \begin{align}

x &= \sqrt{17^2-8^2} = \sqrt{289-64} = \sqrt{225}\\[5pt] &= \sqrt{9\cdot 25} = \sqrt{3^2\cdot 5^2} = 3\cdot 5 = 15\,\textrm{.} \end{align}

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