Question 267116
Let us suppose that shortest side = x cm, middle side = y cm and longest side = z cm.

The shortest side of a triangle is one half the length of the middle side, plus 2 cm. This means {{{x = (1/2)*y + 2}}} _____ (1)

The longest side of the triangle is one and one-half times the length of the middle side, less 2 cm. This means, {{{z = (3/2)*y - 2}}} ______ (2)


Also given, perimeter = 60 cm i.e. {{{x + y + z = 60}}} _______ (3)


Substituting for x and z from (1) and (2) respectively into (3), we have
{{{(1/2)*y + 2 + y + (3/2)*y - 2 = 60}}}
{{{(1/2)*y + y + (3/2)*y = 60}}}
{{{3*y = 60}}}
{{{y= 60/3=20}}}


Hence, {{{x = (1/2)*20 + 2 = 12}}} and {{{z = (3/2)*20 - 2 = 28}}}


The sides of the triangle are: 12 cm, 20 cm and 28 cm.