Centroid of Isosceles Triangle

We will put the learnings of the previous post into practice by applying the principles to a triangle. We will consider an isosceles triangle, with its base aligned with the x-axis and symmetrical around the y-axis as below, and determine the centroid (x,y), with a base of b and height of h.

Since it is intuitively obvious that the x coordinate of the centroid is 0, we will focus our efforts on calculating the y-coordinate of the centroid.

Consider the right half of the triangle

In order to make the calculation a little easier, we will also only consider the right half of the triangle - hopefully it is clear that the y-coordinate of the centroid for half of the triangle is the same as that of the whole.

Before we get integrating..

\[A_{right-half}=\frac{1}{2}\times\frac{b}{2}\times h\]

(Everything from here onwards will be with respect to the right half of the triangle, unless otherwise specified).

The equation for the "edge" highlighted in red is:

\[y=\frac{-h}{b/2}x+h=\frac{-2h}{b}x+h\]

which means that:

\[x=\frac{b}{2h}(h-y)\]

The y-coordinate of the centroid is: \[\begin{equation} \begin{split} \bar{y}&=\frac{1}{A} \int y dA = \frac{4}{bh}\int _0^hy xdy\\ &=\frac{4}{bh}\int _0^h y \times\frac{b}{2h}(h-y)dy\\ &=\frac{2}{h^2}\int _0^h (hy - y^2)dy\\ &=\frac{2}{h^2}\int _0^h y(hy - y^2)dy\\ &=\frac{2}{h^2}\left[ \frac{h}{2}y^2 -\frac{1}{3}y^3\right]_0^h\\ &=\frac{2}{h^2}\frac{h^3}{6}\\ &=\frac{h}{3} \end{split} \end{equation} \]

In other words, the y-coordinate of the centroid of an isosceles triangle is a third way up from its base.