The length of each equal side of an isosceles triangle is 18 cm and angle between them is 90°. Find the length of the shortest altitude of the triangle.

9\(\sqrt {2}\) cm

Here, the triangle is right angle triangle and the smallest altitude will be on largest side of the triangle. So, in right angle triangle the largest side is hypotenuse. Therefore, BD is the smallest altitude.

Therefore,

AC = \(\sqrt {AB^2 + BC^2}\) = \(\sqrt {18^2 + 18^2}\) = 18\(\sqrt {2}\)

Area of triangle = \(\frac{1}{2}\) base × height

So,

⇒ \(\frac{1}{2}\) AB × BC = \(\frac{1}{2}\) AC × BD

⇒ 18 × 18 = 18\(\sqrt {2}\) × BD

⇒ BD = 9\(\sqrt {2}\)