### In a rhombus of side $52 \space cm$, one of the diagonals is $40 \space cm$ long. Find the length of the second diagonal.

$96 \space cm$

Step by Step Explanation:
1. Let $ABCD$ be the given rhombus whose diagonals intersect at $O$.

Then, $AB = 52\space cm$.

Let $AC = 40\space cm$ and $BD = 2x \space cm$.
2. We know that the diagonals of a rhombus bisect each other at right angles. \begin{aligned} \therefore \space OA = \dfrac { 1 } { 2 } AC = 20 \space cm, OB = \dfrac { 1 } { 2 } BD = x \space cm, and \space \angle AOB = 90^\circ \end{aligned}
3. From right $\Delta AOB$, we have \begin{aligned} AB^2 =& OA^2 + OB^2 \\ \implies OB^2 =& AB^2 - OA^2 \\ =& [(52)^2 - (20)^2] \space cm^2 = [ 2704 - 400 ] \space cm^2= 2304 \space cm^2 \\ \implies\space\space \space x^2 =& \space 2304 \implies x = \sqrt{ 2304 } = 48 \space cm. \\ \therefore \space\space \space OB =& \space 48 \space cm. \\ \therefore \space\space \space BD =& 2 \times OB = 2 \times 48 \space cm = 96 \space cm. \end{aligned} Hence, the length of the second diagonal is $96 \space cm.$