Question

The perimeter of a triangle is 60 cm. The longest side exceeds the shortest one by 10 cm.and the sum of the lengths of the longest and the shortest ones is twice the length of the other side. Find the lengths of the sides.

Answer 1

Let the shortest side be x(cm).So the longest side equals to (x+10)cm and the other side equals to (x+x+10)/2

So perimeter of the triangle= x + (x+10)+[(x+x+10)/2] =60
2x+10+[(2x+10)/2] =60

Now let's multiply both sides by 2 to get,
2(2x+10) + (2x+10) = 60*2
4x+20 +2x+10 = 129
6x+30=120
6x=90
x= 15 cm

Answer 2

There's a typo in the in the fourth line from the bottom.It should be corrected as,
4x +20 +2x +10 = * 120

* is to indicate the place I've corrected