XYZ is an isosceles triangle such that XY = YZ = 15 cm and angle XYZ = 90°.  What is the length of the perpendicular drawn from Y on XZ?

\(\frac{15}{\sqrt {2}}\)

(XY)² + (YZ)² = (XZ)² → Pythagoras theorem

(15)² + (15)² = (XZ)²

225 + 225 = (XZ)²

450 = (XZ)²

Therefore,

XZ = 15\(\sqrt{2}\)

Area of Δ with base YZ = Area of Δ with base XZ

⇒\(\frac{1}{2}\) × 15 × 15 = \(\frac{1}{2}\) × YS × 15\(\sqrt{2}\)

⇒ YS =\(\frac{15}{\sqrt {2}}\)