Two sides of an isosceles triangle with perimeter 112 cm are in the ratio 3:2. Find ALL possible lengths for the base.

I guess the base is 3x and both sides are 2x

or
the base is 2x and both sides are 3x
so
3x + 4x = 112
or
2x + 6x = 112

x = 112/7 = 16 and 3x = 3*16
or
x = 112/8 = 14 and 2x = 2*14

An isosceles triangle has equal sides so they shouldn't be in the ratio of 3:2.

That is what the problem says...

To solve this problem, let's assume the two equal sides of the isosceles triangle have lengths 3x and 2x, where x is a positive number. The perimeter of the triangle is given as 112 cm, so we can write the equation:

3x + 3x + 2x = 112

Combining like terms, we get:

8x = 112

To isolate x, we divide both sides of the equation by 8:

x = 112 / 8
x = 14

Now that we know the value of x, we can find the lengths of the two equal sides:

Length of first side = 3x = 3 * 14 = 42 cm
Length of second side = 2x = 2 * 14 = 28 cm

Finally, we can calculate the possible lengths for the base of the isosceles triangle. Since the base must be smaller than the sum of the two equal sides and larger than their difference, the base must satisfy the following inequality:

|42 - 28| < base < 42 + 28

Which simplifies to:

14 < base < 70

Therefore, the possible lengths for the base are any value greater than 14 cm and less than 70 cm.