Solution 2.2:9a - Revision history

Tek at 13:58, 24 September 2008

Tek — Wed, 24 Sep 2008 13:58:51 GMT

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-	We can start by drawing the points	+	We can start by drawing the points (1,4), (3,3) and (1,0) in a coordinate system and draw lines between them, so that we get a picture of how the triangle looks like.
-	~~<math>\left~~( 1 ~~\right.~~,~~\left.~~ 4 ~~\right~~)~~</math>~~,	+
-	~~<math>\left~~( 3 ~~\right.~~,~~\left.~~ 3 ~~\right~~)~~</math>~~	+
-	and	+
-	~~<math>\left~~( 1 ~~\right.~~,~~\left.~~ 0 ~~\right~~)~~</math>~~	+
-	in a coordinate system and draw lines between them, so that we get a picture of how ~~thetriangle~~ looks.	+

-	{{NAVCONTENT_START}}
	[[Image:2_2_9_a-1(2).gif\|center]]		[[Image:2_2_9_a-1(2).gif\|center]]

-	{{NAVCONTENT_STOP}}

	If we now think of how we should use the fact that the area of a triangle is given by the formula		If we now think of how we should use the fact that the area of a triangle is given by the formula

-	~~Area=~~	+	{{Displayed math\|\|<math>\text{Area} = \frac{1}{2}\cdot\text{(base)}\cdot\text{(height),}</math>}}
-	<math>\frac{1}{2}~~</math>~~	+
-	(base)∙(height),	+

-	it is clear that it is most appropriate to use the edge from	+	it is clear that it is most appropriate to use the edge from (1,0) to (1,4) as the base of the triangle. The base is then parallel with the ''y''-axis and we can read off its length as the difference in the ''y''-coordinate between the corner points (1,0) and (1,4), i.e.
-	~~<math>\left~~( 1 ~~\right.~~,~~\left.~~ 0 ~~\right~~)~~</math>~~	+
-	to	+
-	~~<math>\left~~( 1 ~~\right.~~,~~\left.~~ 4 ~~\right~~)~~</math>~~	+
-	as the base of the triangle.	+
-	The base is then parallel with the y-axis and we can read off its length as the difference in the	+

-	<math>y</math>	+	{{Displayed math\|\|<math>\text{base} = 4-0 = 4\,\textrm{.}</math>}}
-	-~~coordinate between the corner points~~	+
-	~~<math>\left( 1 \right.,\left.~~ 0 \~~right)</math>~~	+
-	~~and~~	+
-	~~<math>\left( 1 \right.~~,\~~left~~. ~~4 \right)~~</math>~~, i.e.~~	+

-	~~base~~	+	In addition, the triangle's height is the horizontal distance from the third corner point (3,3) to the base and we can read that off as the difference in the ''x''-direction between (3,3) and the line <math>x=1</math>, i.e.
-	~~<math>=4-0=0</math>.~~	+
-	In addition, the triangle's height is the horizontal distance from the third corner point	+
-	~~<math>\left~~( 3 ~~\right.~~,~~\left.~~ 3 ~~\right~~)~~</math>~~	+
-	to the base and we can read that off as the difference in the	+
-	~~<math>~~x~~</math>~~	+
-	-direction between	+
-	~~<math>\left~~( 3 ~~\right.~~,~~\left.~~ 3 ~~\right~~)~~</math>~~	+
-	and the line	+
-	<math>x=1</math>, i.e.	+

-	~~height~~	+	{{Displayed math\|\|<math>\text{height} = 3-1 = 2\,\textrm{.}</math>}}
-	<math>=3-1=2</math>.	+


-	{{NAVCONTENT_START}}
	[[Image:2_2_9_a-2(2).gif\|center]]		[[Image:2_2_9_a-2(2).gif\|center]]

-	{{NAVCONTENT_STOP}}

	Thus, the triangle's area is		Thus, the triangle's area is

-	~~Area=~~	+	{{Displayed math\|\|<math>\text{Area} = \tfrac{1}{2}\cdot\textrm{(base)}\cdot\textrm{(height)} = \tfrac{1}{2}\cdot 4\cdot 2 = 4\,\text{u.a.}</math>}}
-	<math>\~~frac~~{1}{2}~~</math>~~	+
-	(base)∙(height)	+
-	~~<math>~~=\~~frac~~{1}{2}\~~centerdot~~ 4\~~centerdot~~ 2=4</math>	+
-	~~= area units.~~	+

Ian at 12:56, 18 September 2008

Ian — Thu, 18 Sep 2008 12:56:09 GMT

←Older revision		Revision as of 12:56, 18 September 2008
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		+	We can start by drawing the points
		+	<math>\left( 1 \right.,\left. 4 \right)</math>,
		+	<math>\left( 3 \right.,\left. 3 \right)</math>
		+	and
		+	<math>\left( 1 \right.,\left. 0 \right)</math>
		+	in a coordinate system and draw lines between them, so that we get a picture of how thetriangle looks.
		+
	{{NAVCONTENT_START}}		{{NAVCONTENT_START}}
	[[Image:2_2_9_a-1(2).gif\|center]]		[[Image:2_2_9_a-1(2).gif\|center]]
-	~~<center> [[Image:2_2_9a-1(2).gif]] </center>~~	+
	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}
		+
		+	If we now think of how we should use the fact that the area of a triangle is given by the formula
		+
		+	Area=
		+	<math>\frac{1}{2}</math>
		+	(base)∙(height),
		+
		+	it is clear that it is most appropriate to use the edge from
		+	<math>\left( 1 \right.,\left. 0 \right)</math>
		+	to
		+	<math>\left( 1 \right.,\left. 4 \right)</math>
		+	as the base of the triangle.
		+	The base is then parallel with the y-axis and we can read off its length as the difference in the
		+
		+	<math>y</math>
		+	-coordinate between the corner points
		+	<math>\left( 1 \right.,\left. 0 \right)</math>
		+	and
		+	<math>\left( 1 \right.,\left. 4 \right)</math>, i.e.
		+
		+	base
		+	<math>=4-0=0</math>.
		+	In addition, the triangle's height is the horizontal distance from the third corner point
		+	<math>\left( 3 \right.,\left. 3 \right)</math>
		+	to the base and we can read that off as the difference in the
		+	<math>x</math>
		+	-direction between
		+	<math>\left( 3 \right.,\left. 3 \right)</math>
		+	and the line
		+	<math>x=1</math>, i.e.
		+
		+	height
		+	<math>=3-1=2</math>.
		+

	{{NAVCONTENT_START}}		{{NAVCONTENT_START}}
	[[Image:2_2_9_a-2(2).gif\|center]]		[[Image:2_2_9_a-2(2).gif\|center]]
-	~~<center> [[Image:2_2_9a-2(2).gif]] </center>~~	+
	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}
		+
		+	Thus, the triangle's area is
		+
		+	Area=
		+	<math>\frac{1}{2}</math>
		+	(base)∙(height)
		+	<math>=\frac{1}{2}\centerdot 4\centerdot 2=4</math>
		+	= area units.

Tekbot: Lösning 2.2:9a moved to Solution 2.2:9a: Robot: moved page

Tekbot — Tue, 09 Sep 2008 08:52:23 GMT

Lösning 2.2:9a moved to Solution 2.2:9a: Robot: moved page

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Revision as of 08:52, 9 September 2008

Tekbot: Robot: Automated text replacement (-[[Bild: +[[Image:)

Tekbot — Thu, 21 Aug 2008 06:43:19 GMT

Robot: Automated text replacement (-[[Bild: +[[Image:)

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Martin at 11:40, 15 August 2008

Martin — Fri, 15 Aug 2008 11:40:21 GMT

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Lina — Mon, 31 Mar 2008 15:09:47 GMT

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