SOLUTION: The perimeter of a triangle is 51 cm. The lengths of its sides are consecutive odd integers. Find the Lengths of all three sides.

Question 157002: The perimeter of a triangle is 51 cm. The lengths of its sides are consecutive odd integers. Find the Lengths of all three sides.
Found 3 solutions by orca, CHICKEN-MATH-PRO, ikleyn:
Answer by orca(409)   (Show Source): You can put this solution on YOUR website!
As the lengths of its sides are consecutive odd integers, we suppose that the lengths of the 3 sides are n-2, n, n+2.
As their sum is 51, we have
n-2+n+n+2=51
Solving for n, we obtain
3n = 51
n = 17
So the lengths of the three sides are 15, 17, 19.

Answer by CHICKEN-MATH-PRO(1)   (Show Source): You can put this solution on YOUR website!
IDk I can try,try searching it up on a different website even though this website is the best or seek another tutor. BTW i did the problem and got the lengths of the three sides are 15, 17, 19. Srry for not showing the work but could help in other problems. TYSM for understanding (idk if you do or not)
-CHICKEN-MATH-PRO
Answer by ikleyn(52781)   (Show Source): You can put this solution on YOUR website!
.

The side lengths are three consecutive odd integer summed up to 51: (x-2), x and (x+2), (x-2) + x + (x+2) = 51 3x = 51 ====> x = = 17 ===> the numbers are 17-2 = 15, 17 and 17+2 = 19.

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1.   Solved.

2.   There is NOTHING more easy than that.

3.   The lessons to learn are THESE:

If three consecutive integer sum up to the integer N, then a) N is a multiple of 3; b) the medium number is ; c) the three numbers are , and . If three consecutive ODD integers sum up to the integer N, then a) N is a multiple of 3; b) the medium number is ; c) the three numbers are , and . If three consecutive EVEN integers sum up to the integer N, then a) N is a multiple of 3; b) the medium number is ; c) the three numbers are , and .

MEMORIZE it for all your life.

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