In Triangle ARP, centroid E is on median AX¯¯¯¯¯¯¯¯ . AE = y+6 and EX = 3y-2. Find AX
In a triangle, the centroid is the point of intersection of the three medians of the triangle, and it divides each median into two segments whose lengths are in a 2:1 ratio, with the longer segment being closer to the vertex of the triangle.
Given that E is the centroid, and AE and EX are segments of the median AX from vertex A to side RP, we have AE:EX = 2:1. The problem states that AE = y+6 and EX = 3y-2.
Using the ratio, we can set up a proportion:
2/1 = (y+6)/(3y-2)
Cross multiply to solve for y:
2(3y - 2) = 1(y + 6)
6y - 4 = y + 6
Now, isolate y by subtracting y from both sides:
6y - y = 6 + 4
5y = 10
Divide both sides by 5:
y = 10/5
y = 2
Now that we have the value of y, we can calculate the lengths of AE and EX. Plugging y back into their expressions:
AE = y + 6 = 2 + 6 = 8
EX = 3y - 2 = 3(2) - 2 = 6 - 2 = 4
To find AX, we simply add AE and EX:
AX = AE + EX
AX = 8 + 4
AX = 12
Therefore, the length of AX is 12 units.