Solution 4.1:3c
From Förberedande kurs i matematik 1
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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 |
- | < | + | |
- | {{ | + | {{Displayed math||<math>17^2 = 8^2 + x^2</math>}} |
+ | |||
+ | or | ||
+ | |||
+ | {{Displayed math||<math>x^2 = 17^2 - 8^2\,\textrm{.}</math>}} | ||
+ | |||
+ | We get | ||
+ | |||
+ | {{Displayed math||<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{.} | ||
+ | \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} |