If point R is the centroid of triangle ABC what is the perimeter of triangle ABC given that segments CF DB and AE are equal to 2 3 and 3 respectively?

≈ 9.27

Step-by-step explanation:

suppose BC:a  AC:b AB: c     AE: ma   BD: mb    CF: mc

ma = mb = 3             mc = 2

by the Apollonius' Theorem:

b² + c² = 2(ma² + (a/2)² = 1/2*a² + 2ma² = 1/2*a² + 18   ...(1)

a² + c² = 2(mb² + (b/2)² = 1/2*b² + 2mb² = 1/2*b² + 18  ...(2)

a² + b² = 2(mc² + (c/2)² = 1/2*c² + 2mc² = 1/2*c² + 8    ...(3)

2(a² + b² + c²) = 1/2 (a² + b² + c²) + 44   ...(1)+(2)+(3)

a² + b² + c² = 88/3    ...(4)

(4)-(3): c² = 88/3 - 1/2*c² - 8 = - 1/2*c² + 64/3

3/2*c² = 64/3              c² = 128/9           c = 8√2 / 3 ≈ 3.77

Use the same calculation: a = b = 2√17/3 ≈ 2.75

perimeter ≈ 2.75+2.75+3.77 ≈ 9.27

I think you have to check the accuracy of the calculation. If there is calculation error, I apologize. But.. the process should be right.

3 1/3 x 2 3/5 = 8 2/3

step-by-step explanation:

just multiply the first number by the second number to get your answer.