We are given a rhombus $EFGH$ with side length $EF = 13$ cm and diagonal $EG = 10$ cm. We need to find the area of the rhombus.

1. Problem Description

We are given a rhombus $EFGH$ with side length $EF = 13$ cm and diagonal $EG = 10$ cm. We need to find the area of the rhombus.

2. Solution Steps

A rhombus has diagonals that bisect each other at right angles. Let $D$ be the intersection point of the diagonals $EG$ and $FH$.

Since $EG = 10$ cm, we have $ED = DG = \frac{1}{2} EG = \frac{1}{2}(10) = 5$ cm.

Consider the right triangle $EDF$. We know that $EF = 13$ cm and $ED = 5$ cm. Using the Pythagorean theorem, we can find $DF$.

$EF^2 = ED^2 + DF^2$

$13^2 = 5^2 + DF^2$

$169 = 25 + DF^2$

$DF^2 = 169 - 25 = 144$

$DF = \sqrt{144} = 12$ cm.

Since the diagonals bisect each other, $FH = 2 \times DF = 2 \times 12 = 24$ cm.

The area of a rhombus can be calculated using the lengths of its diagonals $d_1$ and $d_2$ as follows:

$Area = \frac{1}{2} \times d_1 \times d_2$

In our case, $d_1 = EG = 10$ cm and $d_2 = FH = 24$ cm.

$Area = \frac{1}{2} \times 10 \times 24 = \frac{1}{2} \times 240 = 120$ cm$^2$.

3. Final Answer

The area of the rhombus is 120 cm$^2$.