Question

An isosceles triangle has \( AB = AC = 13 \) cm. If the length of the perpendicular from vertex \( A \) to \( BC \) is 5 cm, find the length of \( BC \).

Hint:

In an isosceles triangle, the altitude bisects the base.

Answer 1

In an isosceles triangle, the perpendicular from the vertex bisects the base.
Using Pythagoras' theorem:
\[ AB^2 = AD^2 + BD^2. \] \[ 13^2 = 5^2 + BD^2. \] \[ 169 = 25 + BD^2. \] \[ BD^2 = 144. \] \[ BD = 12. \] Since \( BC = 2BD \):
\[ BC = 24 \text{ cm}. \]