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base and height of triangle
(2,0) \rput(3,-0.3){base} \rput(0.75,0.9){$h_2$} \end{pspicture}](http://images.planetmath.org/cache/objects/11642/js/img1.png)
Considering the area of a triangle, one usually names a side of the triangle to be its base. For expressing the calculation way of the area of the triangle, one then uses the height, which means the perpendicular distance of the vertex, opposite to the base side, from the line determined by the base. In the above two triangles, the heights $h_1$ and $h_2$ correspond the horizontal bases.
The rule for the calculation reads $$\mbox{area \;=\; base times height divided by 2}$$
In the below figure, there is the illustration of the rule. The parallelogram $ABCD$ has been divided by the diagonal $BD$ into two triangles, which are congruent by the ASA criterion (see the alternate interior angles). Thus the both triangles have the areas half of the area of the parallelogram, which in turn has the common base $AB$ and the common height $h$ with the triangle $ABD$ .
(-0.5,0) \rput(-0.3,1.1){$h$} \end{pspicture}](http://images.planetmath.org/cache/objects/11642/js/img2.png)
Note. In an isosceles triangle, one sometimes calls the two equal sides the legs and the third side the base.
