# Triangle and sides.The first side of a triangle is 2 units less than twice the second side. The third side is 10 units longer than the second side. If the perimeter is 144 units, find the length of...

Triangle and sides.

The first side of a triangle is 2 units less than twice the second side. The third side is 10 units longer than the second side. If the perimeter is 144 units, find the length of each side.

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The three sides of the triangle are 66 for the first side, 34 for the second and 44 for the third. Here is how to find this:

Let A be side 1, B be side 2 and C be side 3.

We know that a + b + c = 144

We also know that

a = 2b - 2 and that c = b + 10

So we can substitute those values into the first equation.

2b - 2 + b + b + 10 = 144

4b + 8 = 144

4b = 136

b = 34

If b = 34, a = 66 and c = 44

The sum of those is 144.

Let the 2nd side be x units.

So By data 1st side =2x-2.

3rd side = 2nd side +10 = (2x-2)+10.

So the perimeter (algebraically) = (2x-2)+x+(x+10) which is given to be equal to the perimeter = 144. So the required equation is set up as below:

2x-2+x+ x+10 = 144. Or

4x+8 = 144. Or

4x = 144-8 = 136. Or

x = 136/4 = 34 is the second side.

So the 1st side = 2x-2 = 2*34-2 = 66, and the 3rd side = x+10 = 35+10 = 44.

We observe than the first and the third sides are longer than the second side.

We'll put the second side as x.

x=second side

2x-2=first side

x+10=third side.

The perimeter of a triangle is the sum of it's sides, so the formula is:

P=x+2x-2+x+10, where P=144

144=4x+8

4x=144-8

4x=136

**x=34**

**2x-2=2*34-2=2(34-1)=2*33=66**

**x+10=34+10=44**