The second side of a triangle is 7 in. longer than the first. The length of the third side is 6 in. less than twice the length of the first. if the perimeter is 73., what is the length of each side?

Question

The second side of a triangle is 7 in. longer than the first. The length of the third side is 6 in. less than twice the length of the first. if the perimeter is 73., what is the length of each side?

Answer 1

Let's start by setting side 1 as X.

We know that side 2 is 7 inches longer than side 1 so we can write the equation for side 2 as (X + 7).

We are also told that side 3 is 6 inches less (-6) than twice the length of side 1 (2x) so the equation for side 3 would be (2x - 6).

We also know that the perimeter is 73 which is equal to all of the sides added together so we can combine our equations and set them equal to the perimeter. 73 = x + (x + 7) + (2x - 6)

From here, we can simplify the terms down to 73 = 4x + 1 and then we can solve for x.

73 = 4x + 1 subtract 1 from each side

72 = 4x then divide both sides by 4 to solve for x

18 = x

Now that we know the value of x, we can put it into our equations for each side of the triangle to determine their lengths.

Side 1: x

Side 2: x + 7

Side 3: 2x - 6

Answer 2

Think of a system of equations for legs 1,2, and 3.

Imagine that: leg 1 = x

leg 2 = x =7

leg 3 = y || or as the second condition describes: "six less than twice the length of the first"

which is ~ 2x-6

So we have equation 1: equation 3 for the perimeter of Triangle: x + (x+7) +y= 73

equation 2: y=2x-6

Substitute for 'y' in perimeter equation by plugging in equation 2: x + (x+7) + (2x-6) = 73

Solve for 'x', and we get x=18

Plug back in to different legs, and you've solved for legs 2 and 3 :)