Question

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

Options

• \(60 \sqrt{2} \mathrm{~cm}\)

• \(90 \sqrt{2} \mathrm{~cm}\)

• \(120 \sqrt{2} \mathrm{~cm}\)

• \(180 \sqrt{2} \mathrm{~cm}\)

Answer 1

BCDE is a square with side 90 cm and ∠F = 45°. The length of AF is \(\underline{\boldsymbol{180 \sqrt{2} \mathrm{~cm}}}\).

Explanation:

Since, BCDE is a square.

\[\therefore BE = CD = 90\ \text{cm}.\]

We know that,

\[\tan \theta = \frac{\text{Perpendicular}}{\text{Base}}\]

From figure,

\[\angle AEB = \angle EFD = 45^\circ\] (Corresponding angles are equal)

In \[\triangle ABE,\]

\[\Rightarrow \tan 45^\circ = \frac{AB}{BE}\]

\[\Rightarrow 1 = \frac{AB}{90}\]

\[\Rightarrow AB = 90\ \text{cm}.\]

From figure,

\[AC = AB + BC = 90 + 90 = 180\ \text{cm}.\]

We know that,

\[\sin \theta = \frac{\text{Perpendicular}}{\text{Hypotenuse}}\]

In \[\triangle ACF,\]

\[\Rightarrow \sin 45^\circ = \frac{AC}{AF}\]

\[\Rightarrow \frac{1}{\sqrt{2}} = \frac{180}{AF}\]

\[\Rightarrow AF = 180\sqrt{2}\ \text{cm}.\]