Question

Adjacent sides of a parallelogram are equal and one of diagonals is equal to any one of the sides of this parallelogram. Show that its diagonals are in the ratio $\sqrt{3}:1$.

Answer 1

Step 1: Proving that $\mathrm{ABCD}$ is a rhombus.

Here,

$\mathrm{ABCD}$ is a parallelogram.

We know that, the Opposite sides of the parallelogram are equal.

We are given that,

$\mathrm{AB}=\mathrm{DC}\mathrm{and}\mathrm{AB}=\mathrm{BC}$

Now, $\mathrm{AB}=\mathrm{BC}=\mathrm{CD}=\mathrm{DA}$

Also, $\mathrm{AB}=\mathrm{BC}=\mathrm{CD}=\mathrm{DA}=\mathrm{AC}(\mathrm{Diagonal}\mathrm{is}\mathrm{equal}\mathrm{to}\mathrm{the}\mathrm{side})$

Therefore, $\mathrm{ABCD}$ is a rhombus.

Step 2: Finding the ratio of length of diagonals

In a rhombus, the diagonals bisect each other at $90°$, so we can say that

$\mathrm{AB}=\mathrm{BC}=\mathrm{CD}=\mathrm{DA}=\mathrm{AC}=\mathrm{a}$

Consider $\mathrm{\Delta OAB}$,

Applying Pythagoras theorem, we get

$$\begin{gathered} \mathrm{AB}²=\mathrm{OA}²+\mathrm{OB}² \\ \Rightarrow \mathrm{a}²=\mathrm{OB}²+{\left(\frac{\mathrm{a}}{2}\right)}^2 \\ \Rightarrow \mathrm{OB}²=\mathrm{a}²–\frac{{\mathrm{a}}^2}{4} \\ \Rightarrow \mathrm{OB}=\frac{\mathrm{a}\sqrt{3}}{2} \\ \therefore \mathrm{BD}=2.\mathrm{OB} \\ \Rightarrow \mathrm{BD}=\mathrm{a}\surd 3 \end{gathered}$$

The length of diagonals are $\mathrm{a}$ and $\mathrm{a}\surd 3$.

The ratio of the Diagonals are

$$\begin{gathered} \frac{\mathrm{BD}}{\mathrm{AC}}=\frac{\mathrm{a}\sqrt{3}}{\mathrm{a}} \\ \frac{\mathrm{BD}}{\mathrm{AC}}=\frac{\sqrt{3}}{1} \\ \therefore \mathrm{BD}:\mathrm{AC}=\surd 3:1 \end{gathered}$$

Hence Proved