Answer to Question #223145 in Differential Equations for evatanshinda

Question

Consider ABC an equilateral triangle of sides 4cm.

(a) determine the position of the center of gravity of the equilateral triangle

(b) determine and construct the set of points C such that IMA + MB + MCII = 12/√3

(c) name the set of points C

Answer 1

a.

The position of the center of mass is $(\frac{l}{2},x).$ Wher l is the length of the sides of the triangle and x is gotten as follows.

$\frac{2}{2\sqrt3}=\frac{x}{2}\\ x=\frac{2}{3}\sqrt3$

Therefore, the position of the center of mass is $(2,\frac{2}{3}\sqrt3)$ .

b.

$|MA|=2\sqrt3-\frac{2}{3}\sqrt3=\frac{4}{3}\sqrt3\\ |MB|=2\sqrt3-\frac{2}{3}\sqrt3=\frac{4}{3}\sqrt3\\ |MA+MB+MC|=\frac{12}{\sqrt3}\\ \frac{4}{3}\sqrt3+\frac{4}{3}\sqrt3+|MC|=4\sqrt3\\ |MC|=\frac{4}{3}\sqrt3$

Therefore, the set of points of C is $(4,0)$ . Since

$MA=\sqrt{(2-4)^2+(\frac{2}{3}\sqrt3-0)^2}=\frac{4}{3}\sqrt3$

c.

the set of point of C is $(4,0)$