# Finding centroid's coordinates using Pappus theorem

by Nicko   Last Updated September 11, 2019 09:20 AM - source

enter image description here

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 ($$\frac{2a}{3}$$, $$\frac{c+b}{3}$$).

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

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.)

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