Question

1) The diagonals of a chombus are 4 cm and 6 cm, then, length of its side is (A) 13 (B) 10 (C) 2 (D) 2root13 ​

Answer 1

Given: The diagonals of a rhombus are $4$cm and $6$cm.

We have to find the length of its side.

We are solving in the following way:

Let ABCD be the rhombus

where AC=$4$cm and BD=$6$cm

Let AC and BD intersect each other at O.Now, diagonals of a rhombus bisect each other at right angles.

Thus, we have

$\mathrm{AO}=\frac{1}{2} \times \mathrm{AC}=\frac{1}{2} \times 4=2\mathrm{~cm} \text { and }\\\mathrm{BO}=\frac{1}{2} \times \mathrm{BD}=\frac{1}{2} \times 6=3\mathrm{~cm}$

$\text { In right angled } \triangle \mathrm{AOB} \text {, }$

$\begin{array}{l}\Rightarrow(\mathrm{AB})^{2}=(\mathrm{AO})^{2}+(\mathrm{BO})^{2} \\\Rightarrow(\mathrm{AB})^{2}=(2)^{2}+(3)^{2} \\\Rightarrow(\mathrm{AB})^{2}=4+9\\\Rightarrow(\mathrm{AB})^{2}=13\end{array}$

$\therefore AB=\sqrt{13}$cm

∴  The length of each side of the rhombus is $\sqrt{13}$cm.

Hence, the length of the rhombus will be $\sqrt{13}$cm.

The following fig. will help you to understand: