Step-by-step explanation:
Let ABCD be a rhombus. So, AC (AC = 14 cm) and BD (BD=48 cm) will be its diagonals. Let us assume that diagonals are intersecting at point O.
Since, diagonals of a rhombus are perpendicular bisector.
[tex] \because \: OA = \frac{1}{2} \times AC \\ \\
\therefore \: OA = \frac{1}{2} \times 14
\\ \\ \huge \red{ \boxed{\therefore \: OA = 7 \: cm}} \\ \\ \because \: OB = \frac{1}{2} \times BD \\ \\ \therefore \: OB = \frac{1}{2} \times 48 \\ \\ \huge \red{ \boxed{\therefore \: OB = 24 \: cm}} \\ \\ In \: \triangle OAB, \: \: \angle AOB=90° \\ \therefore \: by \: Pythagoras \: Theorem \\ AB= \sqrt{OA^2 +OB^2 } \\ = \sqrt{7^2 +24^2 } \\ = \sqrt{49+576 } \\ = \sqrt{625 } \\ \huge \orange{ \boxed{\ \therefore \:AB= 25 \: cm.}}[/tex]
Hence, length of a side (or all) is 25 cm.
