From Förberedande kurs i matematik 1

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-	{{NAVCONTENT_START}}	+	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.
-	[[Bild:2_2_9_a-1(2).gif|center]]	+
-	<center> [[Bild:2_2_9a-1(2).gif]] </center>	+
-	{{NAVCONTENT_STOP}}	+
-	{{NAVCONTENT_START}}	+
-	[[Bild:2_2_9_a-2(2).gif|center]]	+	[[Image:2_2_9_a-1(2).gif|center]]
-	<center> [[Bild:2_2_9a-2(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
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		+	{{Displayed math||<math>\text{Area} = \frac{1}{2}\cdot\text{(base)}\cdot\text{(height),}</math>}}
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		+	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.
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		+	{{Displayed math||<math>\text{base} = 4-0 = 4\,\textrm{.}</math>}}
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		+	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.
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		+	{{Displayed math||<math>\text{height} = 3-1 = 2\,\textrm{.}</math>}}
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		+
		+	[[Image:2_2_9_a-2(2).gif|center]]
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		+
		+	Thus, the triangle's area is
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		+	{{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>}}

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Current revision

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.

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

\displaystyle \text{Area} = \frac{1}{2}\cdot\text{(base)}\cdot\text{(height),}

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.

\displaystyle \text{base} = 4-0 = 4\,\textrm{.}

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 \displaystyle x=1, i.e.

\displaystyle \text{height} = 3-1 = 2\,\textrm{.}

Thus, the triangle's area is

\displaystyle \text{Area} = \tfrac{1}{2}\cdot\textrm{(base)}\cdot\textrm{(height)} = \tfrac{1}{2}\cdot 4\cdot 2 = 4\,\text{u.a.}