Solution 4.1:3c
From Förberedande kurs i matematik 1
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- | In this right-angled triangle, the side of length | + | 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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- | is the hypotenuse (it is the side which | + | |
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+ | {{Displayed math||<math>17^2 = 8^2 + x^2</math>}} | ||
or | or | ||
- | + | {{Displayed math||<math>x^2 = 17^2 - 8^2\,\textrm{.}</math>}} | |
- | <math>x^ | + | |
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We get | We get | ||
- | + | {{Displayed math||<math>\begin{align} | |
- | <math>\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{.} |
- | & =\sqrt{9\ | + | \end{align}</math>}} |
- | \end{align}</math> | + |
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} |