Lösung 4.1:3c

Aus Online Mathematik Brückenkurs 1

(Unterschied zwischen Versionen)
Wechseln zu: Navigation, Suche
K (Lösning 4.1:3c moved to Solution 4.1:3c: Robot: moved page)
Zeile 1: Zeile 1:
-
{{NAVCONTENT_START}}
+
In this right-angled triangle, the side of length
-
<center> [[Image:4_1_3c.gif]] </center>
+
<math>\text{17}</math>
-
{{NAVCONTENT_STOP}}
+
is the hypotenuse (it is the side which is opposite the right angle). Pythagoras' theorem then gives
 +
 
 +
 
 +
<math>\text{17}^{2}=8^{2}+x^{2}</math>
 +
 
 +
 
 +
or
 +
 
 +
 
 +
<math>x^{2}=\text{17}^{2}-8^{2}</math>
 +
 
 +
 
 +
We get
 +
 
 +
 
 +
<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>

Version vom 09:36, 27. Sep. 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}

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