Lösung 2.2:9a

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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
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{{Displayed math||<math>\text{Area} = \frac{1}{2}\cdot\text{(base)}\cdot\text{(height),}</math>}}
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{{Abgesetzte Formel||<math>\text{Area} = \frac{1}{2}\cdot\text{(base)}\cdot\text{(height),}</math>}}
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.
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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{{Abgesetzte Formel||<math>\text{base} = 4-0 = 4\,\textrm{.}</math>}}
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.
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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{{Abgesetzte Formel||<math>\text{height} = 3-1 = 2\,\textrm{.}</math>}}
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Thus, the triangle's area is
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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{{Abgesetzte Formel||<math>\text{Area} = \tfrac{1}{2}\cdot\textrm{(base)}\cdot\textrm{(height)} = \tfrac{1}{2}\cdot 4\cdot 2 = 4\,\text{u.a.}</math>}}

Version vom 08:30, 22. Okt. 2008

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.}
Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-TU-Berlin/index.php/L%C3%B6sung_2.2:9a“