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 3 sides be x, x+10,y
Atq
x+x+10=2y
2x+10=2y
x+5=y.....(1)
x+x+10+y=60(given)
2x+10+y=60
2x+10+x+5=60(from eq 1)
3x+15=60
3x=45
x=15
So other sides are 25 and 20

Answer 2

let sides of triangle be s , m and l
So perimeter is :
s + m + l = 60 (1)

Also the sum of the lengths of the longest and the shortest ones is twice the length of the other side, so
l + s = 2 * (m)
Substituting in (1)
2*(m) + m = 60
3m =60
m =20cm (2)

Since the longest side exceeds the shortest one by 10 cm, so
l = s + 10 (3)
Substituting in (1)
s + m + ( s+10) = 60
2s + m + 10 = 60
2s + m = 50
Substituting value from (2)
2s + 20 = 50
2s = 30
s = 15cm
Substituting value into (3)
l = 15 + 10
l = 25cm

So, the lengths of the sides are 15cm , 20cm and 25cm.