A triangle has corners at #(5 ,2 )#, #(9 ,6 )#, and #(8 ,5 )#. How far is the triangle's centroid from the origin?

Question

A triangle has corners at #(5  ,2  )#, #(9  ,6  )#, and #(8   ,5  )#.  How far is the triangle's centroid from the origin?

Jayden Jackson · Accepted Answer

The distance of triangle's centroid from the origin is #8.518# units

The centroid of a triangle whose vertices are #(x_1,y_1)#, (x_2,y_2)# and (x_3,y_3)# is given by #((x_1+x_2+x_3)/3,(y_1+y_2+y_3)/3)# Hence centrid of given triangle is #((5+9+8)/3,(2+6+5)/3)# or #(22/3,13/3)# and its distance from origin is #sqrt((22/3)^2+(13/3)^2)=sqrt(484/9+169/9)=sqrt(653/9)=1/3xx25.554=8.518#

Wyatt Anderson · Answer

undefined To find the centroid of the triangle with vertices at (5, 2), (9, 6), and (8, 5), you would first calculate the average of the x-coordinates and the average of the y-coordinates of the vertices. Then, these averages represent the coordinates of the centroid. Finally, you would find the distance from the centroid to the origin using the distance formula.

Let's denote the coordinates of the centroid as (Cx, Cy), then:

Cx = (5 + 9 + 8) / 3 = 22 / 3 ≈ 7.33
Cy = (2 + 6 + 5) / 3 = 13 / 3 ≈ 4.33

The coordinates of the centroid are approximately (7.33, 4.33).

Now, to find the distance from the centroid to the origin:

Distance = √((Cx - 0)^2 + (Cy - 0)^2)
         = √((7.33 - 0)^2 + (4.33 - 0)^2)
         = √(7.33^2 + 4.33^2)
         ≈ √(53.8289 + 18.7489)
         ≈ √72.5778
         ≈ 8.51 (approximately)

So, the distance from the triangle's centroid to the origin is approximately 8.51 units.