Aus Online Mathematik Brückenkurs 1

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

• 2.2.9a - Solution - A triangle with vertical base 4 and height 2
• 2.2.9a - Solution - A triangle with vertices in (1,0), (1,4) and (3,3)
• Vorlage:Abgesetzte Formel

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