Question

In the given figure, AB = BD, BC = 20 cm and ∠D = 45°, the length of AC is ______.

Options

• 54.64 cm

• 48.28 cm

• 40 cm

• 14.64 cm

Answer 1

In the given figure, AB = BD, BC = 20 cm and ∠D = 45°, the length of AC is 48.28 cm. 

Explanation:

We know that,

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

From figure,

In \[\triangle BCD,\]

\[\Rightarrow \sin 45^\circ = \frac{BC}{BD}\]

\[\Rightarrow \frac{1}{\sqrt{2}} = \frac{20}{BD}\]

\[\Rightarrow BD = 20\sqrt{2}\ \text{m}.\]

From figure,

\[\Rightarrow AB = BD = 20\sqrt{2}\ \text{m}\]

⇒ AC = AB + BC = \[20\sqrt{2}\] + 20 = 28.28 + 20 = 48.28 cm.