PQR is an isosceles triangle such that PQ = QR = 10 cm and ΔPQR = 90°. What is the length of the perpendicular drawn from Q on PR?

$5\sqrt{2}$ cm

We know that,

Area of triangle = (1/2) × Base × Height

(Hypotenuse)² = (Base)² + (Perpendicular)²

We have,

PQ = QR = 10 cm

∠PQR = 90°

PR² = 10² + 10²

= PR² = 100 + 100

= PR² = 200

= PR = 10√2

We know that,

= (1/2) × 10 × 10 = (1/2) × QM × 10√2 

QM = 5√2 cm.