The perimeter of a rhombus whose diagonals are $10\text{cm}$ and $24\text{cm}$ is

Let the lengths of the two diagonals of the rhombus be $d_1=10\text{cm}$ and $d_2=24\text{cm}$.A property of a rhombus states that its diagonals bisect each other at right angles ($90^{\circ }$). This splits the interior into four symmetrical right-angled triangles.The relationship connecting side length ($s$) and diagonals ($d_1,d_2$) using the Pythagorean theorem is:
$$s=\sqrt{{\left(\frac{d_1}{2}\right)}^2+{\left(\frac{d_2}{2}\right)}^2}$$
Substitute the diagonal values:
$$s=\sqrt{{\left(\frac{10}{2}\right)}^2+{\left(\frac{24}{2}\right)}^2}$$
$$s=\sqrt{(5{)}^2+(12{)}^2}$$
$$s=\sqrt{25+144}=\sqrt{169}=13\text{cm}$$
The perimeter of a rhombus with four equal sides is:
$$\text{Perimeter}=4\times s=4\times 13\text{cm}=52\text{cm}$$