CBSE Class 9 Answered

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.