The lengths of the diagonals of a rhombus are 24 cm and 10 cm. Find each side of the rhombus.

Solution:

We know that a quadrilateral is said to a rhombus if all sides of the quadrilateral are equal. Diagonals of a rhombus bisect each other at right angles.

Quadrilateral ABCD is a rhombus and diagonals AC and BD intersect at point O.

As we defined above, we get $\mathrm{AB}=\mathrm{BC}=\mathrm{CD}=\mathrm{AD}, \mathrm{AO}=\mathrm{OC}, \mathrm{BO}=\mathrm{OD}$ and angle $\angle \mathrm{AOD}=\angle \mathrm{AOB}=\angle B O C=\angle C O D=90^{\circ}$.

We are given that AC = 10 cm and BD = 24 cm.

Therefore, we get, AO = OC = 5 cm and BO = OD = 12 cm.

Now we will use Pythagoras theorem in the right angled triangle AOD as below,

$\mathrm{AD}^{2}=\mathrm{AO}^{2}+\mathrm{OD}^{2}$....(1)

Now we will substitute the values of AO and OD in equation (1) we get,

$\mathrm{AD}^{2}=5^{2}+12^{2}$

$\mathrm{AD}^{2}=25+144$

 

$\mathrm{AD}^{2}=169$

Let us take the square root

AD = 13

Therefore, length of the side of the rhombus is $13 \mathrm{~cm}$.