The perimeter of a triangle is 40 inches. The second side exceeds twice the first day by 1 inch and the third side is 2 inches less than the second side. Find the length of each side of the triangle.

Answer 1

Let's assume that the first side of the triangle is x inches.

According to the problem, the second side exceeds twice the first side by 1 inch. So, the second side would be 2x+1 inches.

Similarly, the third side is 2 inches less than the second side, which means it would be 2x+1-2 = 2x-1 inches.

The perimeter of the triangle will be the sum of the lengths of all three sides:
x + (2x+1) + (2x-1) = 40 inches

Simplifying the equation:
5x = 40

Dividing both sides by 5:
x = 8

So, the first side of the triangle is 8 inches.
The second side would be 2x+1 = 2(8)+1 = 17 inches.
And the third side would be 2x-1 = 2(8)-1 = 15 inches.

Therefore, the length of each side of the triangle is 8 inches, 17 inches, and 15 inches.