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Unnecessarily Complicated

How can we make things unnecessarily complicated? Here we use integration to calculate the area of a triangle isosceles.

The area \(A_{abc}\) of the triangle in figure 1 is

$$A_{abc} = \frac{1}2 bh \tag{1}$$

We can see one side of the triangle as a linear equation of the form

$$y = mx +c \tag{2}$$

Moving the point A to the origin and letting \(c=0\) we can rewrite equation (2) as

$$f(x)= mx + 0$$

$$f(x)= mx$$

We find the slope \(m\) substituting the coordinates of the point B

$$m = \frac{h}{1b/2} = \frac{2h}b$$

Hence, we rewrite (2) as

$$f(x) = \frac{2h}bx \tag{3}$$

Now, to compute the area under this function we can integrate

$$\int_0^{\frac12b} \frac{2h}bx \,dx$$

we take out the constant term

$$\frac{2h}b\int_0^{\frac12b} x \,dx= \frac{2h}b\left[ \frac{1}{2}x^2 \right]_0^\frac{b}2$$

which simplifies to

$$\frac{2h}b \frac12(\frac12b)^2 - \frac{2h}b \frac120^2 = \frac14bh $$

At this point we have the area under the function (3) which is half the area \(A_{abc}\) of the triangle isosceles. To find the area of the whole triangle we multiply by 2 to obtain

$$A_{abc} = 2\frac14bh = \frac12bh$$

which is the formula for the area of the triangle \(A_{abc}\) as reported in (1).