length of AR is

Question

ABCD is a square. P, Q and O are the mid-points of AB, BC and AC respectively. AQ intersects PO at R. If AD = 9 cm, then the length of AR is

Answer 1

Since ABCD is a square.

∴ AB = AC = CD = AD = 9 cm.

Also,

P, Q and O are mid-points of AB,BC and AC of square ABCD.

∴ BQ = \(\frac{BC}{2}\) = \(\frac{9}{2}\) = 4.5 cm.

Now,

In triangles OQR and APR,

OQ || AB 

(∵ Q & Q are midpoints of diagonal AC and side BC)

∴ ∠OQR = ∠RAP (Alternate angles)

∠ORQ = ∠ARP (Vertically opposite angles)

∠QOR = ∠APR (Sum of angles in a triangle is 360°)

Also,

OQ = BP

(PBQO is a square)

∴ Δ OQR ≅ Δ APR

(By ASA congruence rule)

∴ AR = QR

Now,

In triangle ABQ,

∠B = 90°

∴ AQ² = AB² + BQ²

(By Pythagoras theorem)

= 9² + \((\frac{9}{2})^2\)

= 9² + (1 + \(\frac{1}{4}\))

= 9² x \(\frac{5}{4}\)

∴ AQ = 9 x \(\frac{\sqrt 5}{2}\) cm.

Since,

AR = QR 

= \(\frac{AQ}{2}\) 

= \(\frac{9\sqrt5}{4}\) cm