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Lösung 2.2:9a

Aus Online Mathematik Brückenkurs 1

(Unterschied zwischen Versionen)
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K (Lösning 2.2:9a moved to Solution 2.2:9a: Robot: moved page)
Zeile 1: Zeile 1:
 +
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]]
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<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.

Version vom 12:56, 18. Sep. 2008

We can start by drawing the points 14 , 33  and 10  in a coordinate system and draw lines between them, so that we get a picture of how thetriangle looks.

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

Area= 21 (base)∙(height),

it is clear that it is most appropriate to use the edge from 10  to 14  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 10  and 14 , i.e.

base =4−0=0. In addition, the triangle's height is the horizontal distance from the third corner point 33  to the base and we can read that off as the difference in the x -direction between 33  and the line x=1, i.e.

height =3−1=2.


Thus, the triangle's area is

Area= 21 (base)∙(height) \displaystyle =\frac{1}{2}\centerdot 4\centerdot 2=4 = area units.

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