The diagonals of a rhombus are $10$ inches and $24$ inches. What is the perimeter of the rhombus, in inches?
The diagonals are perpendicular bisectors of each other, so therefore the side length of the rhombus can be calculated as $\sqrt{5^2+12^2} = 13$. Therefore, the perimeter of the rhombus is $4 \times 13 = \boxed{52}$ inches.
52
Level 4
Prealgebra
Problem: The diagonals of a rhombus are $10$ inches and $24$ inches. What is the perimeter of the rhombus, in inches?
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To find the perimeter of the rhombus, we first need to understand the properties of a rhombus. A rhombus has four equal sides, and its diagonals bisect each other at right angles. Given the lengths of the diagonals, we can determine the lengths of the sides using the Pythagorean theorem.

The diagonals of the rhombus are given as 10 inches and 24 inches. Each diagonal is divided into two equal segments at the intersection point. Therefore, one diagonal of 10 inches will create two segments of 5 inches each, and the diagonal of 24 inches will create two segments of 12 inches each.

Now, we can visualize one of the right triangles formed by the diagonals. One leg of the triangle is 5 inches, and the other leg is 12 inches. We can denote the length of each side of the rhombus as \( s \).

Using the Pythagorean theorem:
s^2 = (5 inches)^2 + (12 inches)^2
s^2 = 25 + 144
s^2 = 169
s = √169
s = 13 inches

Now that we have the length of one side of the rhombus, we can calculate the perimeter. The perimeter \( P \) of a rhombus is given by the formula:
P = 4s

Substituting the value of \( s \):
P = 4 * 13 inches
P = 52 inches

Thus, the final answer is 52.<|END_OF_LATENT|> The final answer is $52$.