The first side of a triangle is 3 inches shorter than the second side, and 2 inches longer than the third side. How long is each side, if the triangle has a perimeter of 28 inches?
f=s-3
f=t+2
f+t+s=28
putting in matrix form
f -s +0t=-3
f+0s-t=2
f+s+t=28
add 1 and 3
2f+t=25
now add that to 2)
3f=27
f=27/3
then f=9
then put that in 1 and solve for s, and finally solve for t in equation 2
2nd side = X in.
1st. side = x - 3 in.
3rd side = (x-3) - 2 = x - 5 in.
x + (x-3) + (x-5) = 28.
X = 12 in.
x-3 = 12 - 3 = 9 in.
x - 5 = 12 - 5 = 7 in.
To solve this problem, let's assign variables to the lengths of the sides of the triangle. Let's say the first side is 'x' inches long, the second side is 'x+3' inches long, and the third side is 'x-2' inches long.
The perimeter of a triangle is the sum of the lengths of its sides. In this case, the perimeter is given as 28 inches. So we can write the equation:
x + (x+3) + (x-2) = 28
Now, let's simplify the equation:
3x + 1 = 28
Subtracting 1 from both sides of the equation:
3x = 27
Dividing both sides of the equation by 3:
x = 9
Now that we've found the value of 'x', we can find the lengths of the other sides:
First side: x = 9 inches
Second side: x+3 = 9 + 3 = 12 inches
Third side: x-2 = 9 - 2 = 7 inches
So, the lengths of the sides of the triangle are 9 inches, 12 inches, and 7 inches.