From Förberedande kurs i matematik 1

Revision as of 09:34, 9 September 2008 (edit) Tekbot (Talk | contribs) m (Lösning 4.1:3b moved to Solution 4.1:3b: Robot: moved page) ← Previous diff		Revision as of 09:32, 27 September 2008 (edit) (undo) Ian (Talk | contribs) Next diff →
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-	{{NAVCONTENT_START}}	+	Because one of the angles in the triangle is
-	<center> [[Image:4_1_3b.gif]] </center>	+	<math>90^{\circ }</math>, we have a right-angled triangle and can use Pythagoras' theorem to set up a relation between the triangle's sides.
-	{{NAVCONTENT_STOP}}	+
		+	The side of length
		+	<math>\text{13}</math>
		+	is the hypotenuse in the triangle, and Pythagoras' theorem therefore gives us that
		+
		+
		+	<math>13^{2}=12^{2}+x^{2}</math>
		+
		+
		+	i.e.
		+
		+
		+	<math>x^{2}=13^{2}-12^{2}</math>
		+
		+
		+	This means that
		+
		+
		+	<math>x=\sqrt{13^{2}-12^{2}}=\sqrt{169-144}=\sqrt{25}=5</math>

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Revision as of 09:32, 27 September 2008

Because one of the angles in the triangle is \displaystyle 90^{\circ }, we have a right-angled triangle and can use Pythagoras' theorem to set up a relation between the triangle's sides.

The side of length \displaystyle \text{13} is the hypotenuse in the triangle, and Pythagoras' theorem therefore gives us that

\displaystyle 13^{2}=12^{2}+x^{2}

i.e.

\displaystyle x^{2}=13^{2}-12^{2}

This means that

\displaystyle x=\sqrt{13^{2}-12^{2}}=\sqrt{169-144}=\sqrt{25}=5