# Finding the center of area (centroid) of a right triangle

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• annamal
In summary, to find the centroid of a right triangle, we can use the formula $$\frac{\int_{0}^{h} ydA}{\int dA} = \frac{\int_{0}^{h} yx(y)dy}{\int dA}$$ where ##x(y) = y\frac{b}{h}## if the triangle is sitting on its tip or ##x(y) = b - y\frac{b}{h}## if the triangle is sitting on its flat base. However, the formula $$\int_{0}^{h} ydA = \int_{0}^{b} y*ydx$$ can also be used if we present ##y## as
annamal
TL;DR Summary
Finding the centroid of a right triangle
To find the y value of the centroid of a right triangle we do
$$\frac{\int_{0}^{h} ydA}{\int dA} = \frac{\int_{0}^{h} yxdy}{\int dA}$$
What is wrong with using
$$\int_{0}^{h} ydA = \int_{0}^{b} y*ydx$$ as the numerator value instead especially since ydx and xdy are equal and where h is height of triangle, b is base of triangle

annamal said:
TL;DR Summary: Finding the centroid of a right triangle

To find the y value of the centroid of a right triangle we do
$$\frac{\int_{0}^{h} ydA}{\int dA} = \frac{\int_{0}^{h} yxdy}{\int dA}$$
So here the "##x##" is the width of the triangle at height ##y##. I would prefer to see it made more clear that ##x## is a function of ##y## rather than a simple constant.$$\frac{\int_{0}^{h} ydA}{\int dA} = \frac{\int_{0}^{h} yx(y)dy}{\int dA}$$
Beyond this, we actually know that ##x(y) = y\frac{b}{h}##. So we can proceed in the obvious fashion to get an integral of ##y^2## times a constant. [and possibly with a constant offset based on the Oops that I just editted in]

Edit: Oops. This formula for ##x## holds if the right triangle is sitting on its tip. If it is sitting on its flat base then ##x(y) = b - y\frac{b}{h}##

annamal said:
What is wrong with using
$$\int_{0}^{h} ydA = \int_{0}^{b} y*ydx$$ as the numerator value instead especially since ydx and xdy are equal and where h is height of triangle, b is base of triangle
Here again, one should present ##y## as an explicit function of ##x## and note that ##y(x) = x\frac{h}{b}##. [or ##y(x) = h - x\frac{h}{b}##]. When all is said and done, one will get an integral of ##x^2## times a [different] constant and possibly with a constant offset depending on the orientation of the triangle.

Yes, it looks to me as if both integrals will yield the same value when correctly evaluated.

Last edited:
topsquark
jbriggs444 said:
So here the "##x##" is the width of the triangle at height ##y##. I would prefer to see it made more clear that ##x## is a function of ##y## rather than a simple constant.$$\frac{\int_{0}^{h} ydA}{\int dA} = \frac{\int_{0}^{h} yx(y)dy}{\int dA}$$
Beyond this, we actually know that ##x(y) = y\frac{b}{h}##. So we can proceed in the obvious fashion to get an integral of ##y^2## times a constant. [and possibly with a constant offset based on the Oops that I just editted in]

Edit: Oops. This formula for ##x## holds if the right triangle is sitting on its tip. If it is sitting on a flat side then ##x(y) = b - y\frac{b}{h}##Here again, one should present ##y## as an explicit function of ##x## and note that ##y(x) = x\frac{h}{b}##. [or ##y(x) = h - x\frac{h}{b}##]. When all is said and done, one will get an integral of ##x^2## times a [different] constant and possibly with a constant offset depending on the orientation of the triangle.

Yes, it looks to me as if both integrals will yield the same value when correctly evaluated.
$$x(y) = b - y\frac{b}{h}$$
$$y(x) = \frac{-h}{b}x +h$$
No they would be different because $$\int_{0}^{h} y*x(y)dy = \int_{0}^{h} y*(b - y\frac{b}{h})dy$$
this does not equal $$\int_{0}^{b} y(x)*y(x)dx = \int_{0}^{b} (\frac{-h}{b}x +h)^2 dx$$

annamal said:
$$x(y) = b - y\frac{b}{h}$$
$$y(x) = \frac{-h}{b}x +h$$
No they would be different because $$\int_{0}^{h} y*x(y)dy = \int_{0}^{h} y*(b - y\frac{b}{h})dy$$
this does not equal $$\int_{0}^{b} y(x)*y(x)dx = \int_{0}^{b} (\frac{-h}{b}x +h)^2 dx$$
Out by a factor of two? Because the latter imputes the height of a vertical strip as the height of the top of that strip instead of the height of the strip's centroid halfway up.

topsquark
jbriggs444 said:
Out by a factor of two? Because the latter imputes the height of a vertical strip as the height of the top of that strip instead of the height of the strip's centroid halfway up.
Yes off by a factor of two. Can you explain that more clearly? What do you mean by "instead of the height of the strip's centroid halfway up?"

annamal said:
Yes off by a factor of two. Can you explain that more clearly? What do you mean by "instead of the height of the strip's centroid halfway up?"
A centroid is the average position of the area in a shape.

If you are averaging vertical position over a bunch of horizontal strips, there is no ambiguity about the vertical position of each strip. The vertical position of every point on a horizontal strip is the same. So that is the vertical position of the strip.

It is a weighted average, of course. So you weight each strip by the incremental area of that strip.

If you are averaging the vertical position over a bunch of vertical strips there is room for ambiguity. What is the vertical position of a vertical strip? Is it at the top of the strip? Or at the bottom of the strip. The correct choice is to use the middle of the strip (the centroid of the strip) as its vertical position. So you are effectively averaging a set of average positions. Which is fine as long as you are weighting both the average and the average of the averages by the same measure (area in this case).

topsquark, annamal and Lnewqban
I guess you want to calculate the centroid of the triangle given by
$$0<x<b, \quad 0<y<h(1-x/b).$$
Then you get
$$\vec{X}=\int_0^ b \mathrm{d} x \int_0^{h(1-x/b)} \mathrm{d} y \begin{pmatrix} x\\y \end{pmatrix} = \frac{h b}{6} \begin{pmatrix}b \\ h \end{pmatrix}.$$
The centroid thus is
$$\vec{x}_c=\frac{1}{A} \vec{X}=\frac{2}{bh} \vec{X} = \frac{1}{3} \begin{pmatrix}b \\ h \end{pmatrix}.$$

dextercioby
You could use the simpler pappus' theorem which goes like this : when you rotate a 2d shape to generate a solid figure, distance travelled by centroid multiplied by area of the 2d shape is equal to volume of 3d shape generated, in case of triangle :
Let us say centroid is x distance in x-axis from orgin( at the vertex opposite of hyp), we rotate this triangle to generate a right circular cone, then the distance travelled by centroid is 2 pi x, area of triangle = 1/2ab and volume of cone = 1/3 pi a^2 b. Equating we get : 2 pi x ×1/2 ab = 1/3 pi a^2b; simplifying we get : x = 1/3 a, similarly we can get y = 1/3b

## 1. What is the formula for finding the centroid of a right triangle?

The formula for finding the centroid of a right triangle is (1/3)(base) x (height), where the base is the length of the triangle's base and the height is the length of the triangle's height.

## 2. Why is finding the centroid important in geometry?

Finding the centroid is important in geometry because it is the point where the triangle's three medians intersect. This point is useful in determining the triangle's properties, such as its center of mass and its balance point.

## 3. How do you find the centroid of a right triangle with non-integer coordinates?

To find the centroid of a right triangle with non-integer coordinates, you can use the formula (1/3)(x1 + x2 + x3) x (y1 + y2 + y3), where x1, x2, x3 are the x-coordinates of the triangle's vertices and y1, y2, y3 are the y-coordinates of the vertices.

## 4. Can the centroid of a right triangle be outside of the triangle?

No, the centroid of a right triangle will always be inside the triangle. This is because the centroid is the intersection of the triangle's medians, which are always contained within the triangle.

## 5. How is the centroid of a right triangle related to its area?

The centroid of a right triangle is located at (1/3) of the distance from each vertex to the midpoint of the opposite side. This means that the centroid divides the triangle's area into three equal parts, making it a useful point in finding the triangle's area.

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