1. The lengths of the sides of isosceles triangle $\triangle ABC$ are $3x+62$, $7x+30$, and $5x+50$ feet. What is the least possible number of feet in the perimeter of $\triangle ABC$?
2. A $100$-foot rope is cut into three pieces. The first piece is three times as long as the second, and the third piece is $20$ feet shorter than the second. How many feet long is the shortest piece?
+150 For number 2) We use algebra. Pretend the second piece is X. The first piece would be 3x, and the third would be x-20. Add these all up, you get 5x-20. 5x-20 is equal to 100 feet. Therefore, 5x=120 feet. So, x=24 feet. The second piece is 24 feet. We minus 20 from 24 and the shortest piece is 4 feet
1) Alright we have 3 possibilities. 3x+62=7x+30, 7x+30=5x+50, or 5x+50 = 3x+62. Lets see the first one.
First one: If 3x+62=7x+30, 32=4x. x=8, and the perimeter would be: 24+62+56+30+40+50, or 272.
Second one: 7x+30=5x+50. x=10. 100+100+92= 292.
Third one: 5x+50 = 3x+62. x=6. 80+80+72 = 232. So the least possible feet is 232.
+150 For number 2) We use algebra. Pretend the second piece is X. The first piece would be 3x, and the third would be x-20. Add these all up, you get 5x-20. 5x-20 is equal to 100 feet. Therefore, 5x=120 feet. So, x=24 feet. The second piece is 24 feet. We minus 20 from 24 and the shortest piece is 4 feet
1) Alright we have 3 possibilities. 3x+62=7x+30, 7x+30=5x+50, or 5x+50 = 3x+62. Lets see the first one.
First one: If 3x+62=7x+30, 32=4x. x=8, and the perimeter would be: 24+62+56+30+40+50, or 272.
Second one: 7x+30=5x+50. x=10. 100+100+92= 292.
Third one: 5x+50 = 3x+62. x=6. 80+80+72 = 232. So the least possible feet is 232.
