# Finding centroid's coordinates using Pappus theorem

The task is to find the centroid of the given triangle (see the image above). We also should use the fact that the volume of a cone of radius $$r$$ and height $$h$$ is $$V = \frac{1}{3}\Pi r^2h$$. My solution:

$$1)~$$ Denote the dots as following: $$~O(0,0),~~ R(a, 0),~~ Q(a, b),~~ P(a, c)~$$;

$$2)~$$ Find the volume of the cone with radius OR and height PR:

$$V = \frac{1}{3}\Pi r^2h = \frac{1}{3}\Pi a^2c$$

$$3)~$$ Find the volume of the cone with radius OR and height RQ:

$$V = \frac{1}{3}\Pi r^2h = \frac{1}{3}\Pi a^2b$$

$$4)~$$ If we rotate triangle OQP about the line $$~x = a~$$, then according to Pappus theorem it's volume is equal to:

$$V = 2\Pi pA$$, where

$$p -~$$ distance from centroid of OQP to axis of revolution (x = a);
$$A -~$$ area of triangle OQP;

$$5)~$$ The next step is to find area of OQP. We will do this with formula for right triangle $$\frac{1}{2}ab$$ $$(a$$ and $$b$$ are edges$$)$$. The area of triangle ORP is equal to $$\frac{1}{2} ac$$, and the area of OQR is equal to $$\frac{1}{2} ab$$. Thus, the area of OQP is equal:

$$A = \frac{1}{2} ac - \frac{1}{2} ab = \frac{1}{2} a(c-b)$$

$$6)~$$ We can now find the volume of solid, which will be created by revolving triangle OQP about line $$x = a$$ using the following approaches:

$$V = \frac{1}{3}\Pi a^2c - \frac{1}{3}\Pi a^2b$$ (difference of cone volumes)

$$V = 2\Pi pA = 2\Pi p\frac{1}{2} a(c-b)$$ (Pappus Theorem)

This to expressions are equal. We simplify the equality and receive that $$p = \frac{a}{3}$$ (the distance from centroid to axis of revolution $$(x = a)$$.

$$7)~$$ As this distance is equal to $$p = \frac{a}{3}$$, then $$x$$ coordinate of centroid is equal to $$p = \frac{2a}{3}$$. Also, we know that centroid of triangle is on median. So we draw the line between dots $$(0, 0)$$ and $$(a, \frac{c+b}{2})$$. This line has an equation $$y = \frac{c+b}{2a}x$$. If we replace $$x$$ with $$\frac{2a}{3}$$, the $$y$$ is equal $$\frac{c+b}{3}$$. So, the coordinates of centroid is $$\left(\frac{2a}{3} , \frac{c+b}{3}\right)~$$.

However, the textbook I use give the following answer $$\left(\frac{2a(a-b}{3(c-b)} , \frac{c+b}{3} \right)~$$.

I will be very grateful if anybody can explain why the answers is different. Thank you in advance! (Please, note that y coordinate is correct.)

• The centroid of a triangle has barycentric coordinates $1:1:1$, which here corresponds to $\frac13(O+P+Q)$. Your solution is correct. – amd Sep 11 at 9:59