### Find the area of the rhombus in which each side is $25 \space cm$ long and one of whose diagonals is $14 \space cm$.

$336 \space cm^2$

Step by Step Explanation:
1. Let $ABCD$ be the given rhombus with $AB = 25 \space cm$ and $AC = 14 \space cm$.

Let the diagonals $AC$ and $BD$ bisect at a point $O$.
We know that the diagonals of a rhombus bisect each other at right angles.
So, $AO = \dfrac { 1 } { 2 } AC$ and $BO = \dfrac { 1 } { 2 } BD.$ $$\therefore AO = \dfrac { 1 } { 2 } \times 14 = 7 \space cm \text { and } \angle AOB = 90^\circ$$
2. Using Pythagous' theorem in right $\triangle AOB$, we have \begin{aligned} & AB^2 = AO^2 + BO^2 \\ \implies & (25)^2 = (7)^2 + BO^2 \\ \implies & 625 = 49 + BO^2 \\ \implies & 576 = BO^2 \\ \implies & 24 \space cm = BO \\ \end{aligned} As, $BO = \dfrac { 1 } { 2 } BD$, we have \begin{aligned} & BO = \dfrac { 1 } { 2 } BD \\ \implies & 24 = \dfrac { 1 } { 2 } BD \\ \implies & 2 \times 24 = BD \\ \implies & 48 \space cm = BD \\ \end{aligned}
3. We know, \begin{aligned} \text { Area of rhombus } & = \dfrac { 1 } { 2 } \times \text { Product of its diagonals } \\ & = \dfrac { 1 } { 2 } \times AC \times BD \\ & = \dfrac { 1 } { 2 } \times 14 \times 48 \space cm \\ & = 336 \space cm^2 \end{aligned} Thus, the area of the rhombus is $336 \space cm^2$.