Question

If one angle of a rhombus is ${60}^o$, then the ratio of the lengths of its diagonals is

• A. $\sqrt{6}:1$
• B. $2:1$
• C. $2:3$
• D. $\sqrt{3}:1$

Answer 1

The correct option is D $\sqrt{3}:1$

Let $AB=BC=CD=DA=x$

sides of rhombus are equal

$\angle BAD={60}^{\circ }$

In $△ABD,AD=AB$

$\therefore \angle ABD=\angle ADB\to \left(1\right)$ [Angles opposite to equal sides]

Now, $\angle ABD+\angle ADB+\angle BAD={180}^{\circ }$

$2\angle ABD+{60}^{\circ }={180}^{\circ }\Rightarrow \angle ABD=\frac{{170}^{\circ }}{2}={60}^{\circ }$

$\angle ABD=\angle ADB={60}^{\circ }$

$\therefore △ABD$ is an equilateral triangle.

$\therefore BD=AD=x$

Diagonals of rhombus bisect each other, therefore

$OD=\frac{BD}{2}=\frac{x}{2}$ , diagonals are perpendicular to each other.

Therefore, $\angle AOD={90}^{\circ }$

In $△AOD$, using Pythagoras theorem

$O{A}^2+O{D}^2=A{D}^2\Rightarrow \left({\frac{AC}{2}}^2\right)+\left({\frac{x}{2}}^2\right)=x^2$

$\left({\frac{AC}{2}}^2\right)=x^2-{\frac{x}{4}}^2=\frac{3x^2}{4}\Rightarrow \frac{AC}{2}=\frac{\sqrt{3}x}{2}\Rightarrow AC=\sqrt{3}x$

$\frac{AC}{BD}=\frac{\sqrt{3}x}{x}=\frac{\sqrt{3}}{1}=\sqrt{3}:1$