The area of an isosceles triangle is 12cm sq. If one side is 5cm, the length of the base is?

Having drawn your isosceles triangle with the length of the equal sides equal to 5 cm, the width of the base and the altitude equal to 2.x and h respectively, then we may write for the area A of the triangle

A = h.x = 12 cm²

so that h = 12/x

and from Pythagoras applied to either of the right-angled triangles

x² + h² = 5²

Substituting for h from above gives

x² + (144/x)² - 5² = 0

Multiplying through by x² and rearranging

x⁴ + 144 – 25x² = 0

which is a quadratic in z = x² with solutions

z² -25z + 144 = 0

z = [25 ±?(25² - 4*144)]/2

giving z = [25 ±7]/2 = 16 and 9

so that the solutions for x are 4 and 3, both of which are valid,

The base of the triangle may therefore be either 2.x = 8 cm or 6 cm in length.

As you might expect, the altitude h takes the values 8 cm and 6 cm respectively,
so that both triangles have areas of 12 cm².