centroid

[logistic_guy]

here is the question

How to find the centroid of a \(\displaystyle 60-60-60\) triangle?


my attemb
i've to solve some questions involving the centroid of an equilateral triangle
but i can't derive it
do i have to use calculus or there's alternative like geometry?
i'm bad in calculus and very bad in geometry

 

[R.M.]

There's a big clue in its definition. Perhaps you should start there.

 

[logistic_guy]

thank

There's a big clue in its definition. Perhaps you should start there.

i look up the formula for centroid, it give me this

the cetroid is the point \(\displaystyle (\bar{x}, \bar{y})\)

\(\displaystyle \bar{x} = \frac{\iint\limits_R xdA}{\iint\limits_R dA}\)

\(\displaystyle \bar{y} = \frac{\iint\limits_R ydA}{\iint\limits_R dA}\)

if i've equilateral triangle, how to find
\(\displaystyle \iint\limits_R \) and \(\displaystyle x\) and \(\displaystyle y\) and \(\displaystyle dA\)

 

[Dr.Peterson]

i've to solve some questions involving the centroid of an equilateral triangle
but i can't derive it
do i have to use calculus or there's alternative like geometry?
i'm bad in calculus and very bad in geometry

As always, first look up the definition. But also consider the context, because there can be different (usually related) definitions.

In particular, if this problem comes from a geometrical context, then you will likely want to use the geometrical definition, which bypasses the more general calculus definition and is much easier to apply to triangles. For example,

Triangle centroid definition - Math Open Reference

This can be derived from the calculus definition, or from the basic idea of balancing (center of gravity), but is suitable for students who have not seen calculus yet, or for applications that don't require that much generality. And even if the context were general, since the problem is about a specific triangle, I would absolutely use the geometrical definition for the triangle!

(On the other hand, if your goal is to really learn everything, then do it both ways -- and don't ask others to do it for you. Give it a serious try!)

 

[logistic_guy]

As always, first look up the definition. But also consider the context, because there can be different (usually related) definitions.

In particular, if this problem comes from a geometrical context, then you will likely want to use the geometrical definition, which bypasses the more general calculus definition and is much easier to apply to triangles. For example,

Triangle centroid definition - Math Open Reference

This can be derived from the calculus definition, or from the basic idea of balancing (center of gravity), but is suitable for students who have not seen calculus yet, or for applications that don't require that much generality. And even if the context were general, since the problem is about a specific triangle, I would absolutely use the geometrical definition for the triangle!

(On the other hand, if your goal is to really learn everything, then do it both ways -- and don't ask others to do it for you. Give it a serious try!)

i'll see the websites
then i'll try to solve the integration \(\displaystyle \iint\limits_R xdA\) on paper
then i'll show you my thinking in latex

thank Dr.

 

[logistic_guy]

i'll see the websites
then i'll try to solve the integration \(\displaystyle \iint\limits_R xdA\) on paper
then i'll show you my thinking in latex

thank Dr.

i'm back

i'll start by the calculus approach, then i'll try to understand the geometrical one

\(\displaystyle \iint\limits_R xdA\)

this integration look fishy, so i've to break it down further. before i break it down, i'll just say something logical about centroid

if the equilateral triangle with leg \(\displaystyle a\) is placed in such a way the base is one of its legs, logic say \(\displaystyle \bar{x} = \frac{a}{2}\), it's only the location of \(\displaystyle \bar{y}\) is difficult to find
even i already know what \(\displaystyle \bar{x}\) by logic, i'll find it again anyway

\(\displaystyle \iint\limits_R xdA = \int_{0}^{h}\int_{0}^{a} x dx dy = \int_{0}^{a}\frac{a^2}{2} dy = \frac{a^2}{2}h\)

\(\displaystyle \iint\limits_R dA = \frac{1}{2}ah\)

\(\displaystyle \bar{x} = \frac{\iint\limits_R xdA}{\iint\limits_R dA} = \frac{\frac{a^2}{2}h}{\frac{1}{2}ah} = a\)

that's my first attemp on paper. logic tell me my integration have something wrong
i've to think more and i've remember i must treat the triangle like a function
so my second attemb is invent equation of one of the triangle sides

i'll place my triangle this time in the cartesian coordinate so the \(\displaystyle y\)-axis is dividing it by half
so the vertices of the triangle is \(\displaystyle (-a/2, 0)\) and \(\displaystyle (a/2,0)\) and \(\displaystyle (0,h)\)

the equation of a line is
\(\displaystyle y - y_0 = m(x - x_0)\)

equation of slope is
\(\displaystyle m = \frac{y_1 - y_0}{x_1 - x_0} = \frac{h - 0}{0 - -a/2} = 2\frac{h}{a}\)

\(\displaystyle y - h = 2\frac{h}{a}(x - 0)\)

\(\displaystyle y = 2\frac{h}{a}x + h\)

this equation describe the leg of the triangle that go from
the point \(\displaystyle (-a/2,0)\) to \(\displaystyle (0,h)\) which is sifficient to help find the centroid

\(\displaystyle \iint\limits_R xdA = 2\int_{-a/2}^{0}\int_{0}^{2\frac{h}{a}x + h} x dy dx = 2\int_{-a/2}^{0} 2\frac{h}{a}x^2 + hx dx = 2(\frac{h}{a}\frac{x^3}{3} + h\frac{x^2}{2})\bigg|_{-a/2}^{0}\)

\(\displaystyle = 2(\frac{h}{a}\frac{a^3}{3(8)} - h\frac{a^2}{2(4)}) = \frac{h}{a}\frac{a^3}{3(4)} - h\frac{a^2}{(4)} = \frac{ha^2}{12} - \frac{ha^2}{(4)}\)

\(\displaystyle = \frac{ha^2}{12} - \frac{3ha^2}{(12)} = -\frac{ha^2}{6}\)

\(\displaystyle \bar{x} = \frac{\iint\limits_R xdA}{\iint\limits_R dA} = \frac{-\frac{ha^2}{6}}{\frac{1}{2}ah} = -\frac{a}{3}\)

where's my mistake?