In Triangle ARP, centroid E is on median AX¯¯¯¯¯¯¯¯ . AE = y+6 and EX = 3y-2. Find AX

Answer 1

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.