Question

The perimeter of a triangle is 40 inches +2nd side exceeds twice the first side by 1 inch the third side is 2 inches less than the second side find the length of each side of the triangle

Answer 1

S2=2*S1+1

S3=3*S1-1

P=S1+S2+S3

42=S1+2*S1+1+3*S1-1

42=6*S1

S1=42/6

S1=7 ANS. FOR S1.

S2=2*7+1=14+1=15 ANS FOR S2.

S3=3*7-1=21-1=20 ANS. FOR S3.

PROOF:

42=7+15+20

42=42

Answer 2

Explanation:

The perimeter is the addition of all sides

x+y+z=40

where x=1st side, y=2nd side, z= 3rd side

2nd side exceeds twice the first side by 1 inch= y=2x+1

the third side is 2 inches less than the second side z=y-2=2x+1-2=2x-1

plugging in the perimeter equation

x+(2x+1)+(2x-1)=40

5x=40

x=8 inches (1st side)

y=2*8+1=17 inches (2nd side)

z=2*8-1=15 inches (3rd side)