Question

Prove that the sum of 3 sides of a triangle is greater than the sum of its medians.
( I will mark the best answer as brainliest )

Answer 1

Step-by-step explanation:

let's assume it is an equilateral traingle for solving ..

1. here the length of the sides let be x
2. a median divides the opposite side into two parts of equal length.
3. a median is also a perpendicular bisector in the equilateral traingle.
4. the angle is 60 let's take tan 60° = P/B
5. tan 60° = length of median / one part of side which a median divides.
6. length of median =
7. $\sqrt{3}x \div 2$
8. sum of median =
9. $3 \sqrt{3}x \div 2$
10. sum of length of sides of traingle is 3x
11. therefore can see that the sum of length of side of traingle is greater then sum of the length of medians..