If in the following figure (not to the scale), ∠ACB = 135° and the radius of the circle is $2\sqrt{2}$ cm, then the length of the chord AB is:

4 cm

We have,

Radius of the circle = 2√2 cm

∠ACB = 135

We know that,

The sum of the opposite angle of the cycle quadrilateral is 180.

The angle at the center of a circle = Twice the angle at the circumference

∠ACB + ∠ADB = 180^o

= 135 + ∠ADB = 180^o

= ∠ADB = 180 – 135 = 45^o

We know that,

O is the center of the circle.

∠BOA = 2 ∠BDA

= ∠BOA = 2 × 45

= ∠BOA = 90^o

We know that,

OA = OB = 2√2

According to the Pythagoras theorem,

AB² = OB² + OA²

= AB² = (2√2)² + (2√2)²

= AB² = 8 + 8

= AB = √16

 AB = 4