BCDE is a square with side 90 cm and ∠F = 45°. The length of AF is :

1. $60\sqrt{2}$ cm

2. $90\sqrt{2}$ cm

3. $120\sqrt{2}$ cm

4. $180\sqrt{2}$ cm

Since, BCDE is a square.

∴ BE = CD = 90 cm.

We know that,

tan θ = $\dfrac{\text{Perpendicular}}{\text{Base}}$

From figure,

∠AEB = ∠EFD = 45° (Corresponding angles are equal)

In △ ABE,

⇒ tan 45° = $\dfrac{AB}{BE}$

⇒ $1 = \dfrac{AB}{90}$

⇒ AB = 90 cm.

From figure,

AC = AB + BC = 90 + 90 = 180 cm.

We know that,

sin θ = $\dfrac{\text{Perpendicular}}{\text{Hypotenuse}}$

In △ ACF,

⇒ sin 45° = $\dfrac{AC}{AF}$

⇒ $\dfrac{1}{\sqrt{2}} = \dfrac{180}{AF}$

⇒ AF = $180\sqrt{2}$ cm.

Hence, Option 4 is the correct option.