math question!!! HELP!!!!!!!?

Question

one of the equal sides of an isosceles triangle is 3m less than twice its base. the perimater is 44m. find the lengths of the sides.

how do you solve the question and whats the answer

Answer 1

This one isn't too hard.

Let b be the length of the base.

The way you compute the perimeter of a triangle is by adding the lengths of all 3 sides.

We're calling the length of the base b, and the problem says that one of the EQUAL sides is twice the length of the base (2b) less 3 (2b - 3). Since it's an isoceles triangle, the other one must also be 2b - 3.

So the sum of the sides is the perimeter, so it must equal 44:

b + (2b - 3) + (2b - 3) = 44
5b - 6 = 44
5b - 6 + 6 = 44 + 6
5b = 50
5b/5 = 50/5
b = 10

So the other sides are 2b - 3 = 2(10) - 3 = 20 - 3 = 17.

The sides are 10, 17, and 17.

Answer 2

Let x be the length of the base. We can write an equation:

2(2x-3) + x = 44

x = 10

2x - 3 = 17

Therefore, the three sides are 10 cm, 17 cm and 17 cm.

Answer 3

let x equal one of the equal sides
let y equal the base

Perimeter of the triangle: 2x+y=44

x=2y-3

2(2y-3)+y=44

4y-6+y=44

5y-6=44

5y=50

y=10m

then just subtitute 10 for "y" in the original equation

2x+10=44

2x=34

x=17

therefore, the base is 10m and the other two sides are 17m long