From Förberedande kurs i matematik 1

Revision as of 09:34, 9 September 2008 (edit) Tekbot (Talk | contribs) m (Lösning 4.1:3c moved to Solution 4.1:3c: Robot: moved page) ← Previous diff		Current revision (10:21, 3 October 2008) (edit) (undo) Tek (Talk | contribs) m
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-	{{NAVCONTENT_START}}	+	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
-	<center> [[Image:4_1_3c.gif]] </center>	+
-	{{NAVCONTENT_STOP}}	+	{{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>}}

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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}