Lösung 4.1:3c

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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
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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{{Displayed math||<math>17^2 = 8^2 + x^2</math>}}
+
{{Abgesetzte Formel||<math>17^2 = 8^2 + x^2</math>}}
or
or
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{{Displayed math||<math>x^2 = 17^2 - 8^2\,\textrm{.}</math>}}
+
{{Abgesetzte Formel||<math>x^2 = 17^2 - 8^2\,\textrm{.}</math>}}
We get
We get
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{{Displayed math||<math>\begin{align}
+
{{Abgesetzte Formel||<math>\begin{align}
x &= \sqrt{17^2-8^2} = \sqrt{289-64} = \sqrt{225}\\[5pt]
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\cdot 25} = \sqrt{3^2\cdot 5^2} = 3\cdot 5 = 15\,\textrm{.}
\end{align}</math>}}
\end{align}</math>}}

Version vom 08:47, 22. Okt. 2008

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}

Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-TU-Berlin/index.php/L%C3%B6sung_4.1:3c“