Solution 4.1:3c
From Förberedande kurs i matematik 1
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| - | {{  | + | In this right-angled triangle, the side of length   | 
| - | <  | + | <math>\text{17}</math>  | 
| - | {{  | + | 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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| + | or  | ||
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| + | <math>x^{2}=\text{17}^{2}-8^{2}</math>  | ||
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| + | We get  | ||
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| + | <math>\begin{align}  | ||
| + | & x=\sqrt{\text{17}^{2}-8^{2}}=\sqrt{289-64}=\sqrt{225} \\   | ||
| + | & =\sqrt{9\centerdot 25}=\sqrt{3^{2}\centerdot 5^{2}}=3\centerdot 5=15. \\   | ||
| + | \end{align}</math>  | ||
Revision as of 09:36, 27 September 2008
In this right-angled triangle, the side of length \displaystyle \text{17} is the hypotenuse (it is the side which is opposite the right angle). Pythagoras' theorem then gives
\displaystyle \text{17}^{2}=8^{2}+x^{2}
or
\displaystyle x^{2}=\text{17}^{2}-8^{2}
We get
\displaystyle \begin{align}
& x=\sqrt{\text{17}^{2}-8^{2}}=\sqrt{289-64}=\sqrt{225} \\ 
& =\sqrt{9\centerdot 25}=\sqrt{3^{2}\centerdot 5^{2}}=3\centerdot 5=15. \\ 
\end{align}
