1. **State the problem:** We have a rhombus with diagonals intersecting perpendicularly. One diagonal is divided into segments of 6 cm and 4 cm (so its total length is $6 + 4 = 10$ cm). The other diagonal is divided into two equal segments of 3 cm each (so its total length is $3 + 3 = 6$ cm). 2. **Recall properties:** In a rhombus, the diagonals bisect each other at right angles. The diagonals are perpendicular, and their intersection divides each into two segments. 3. **Calculate the surface area (area of the rhombus):** The area of a rhombus is given by: $$\text{Area} = \frac{d_1 \times d_2}{2}$$ where $d_1$ and $d_2$ are the lengths of the diagonals. Here, $$d_1 = 10 \text{ cm}$$ $$d_2 = 6 \text{ cm}$$ So, $$\text{Area} = \frac{10 \times 6}{2} = \frac{60}{2} = 30 \text{ cm}^2$$ 4. **Calculate the perimeter (scope):** Each side of the rhombus can be found using the Pythagorean theorem because the diagonals intersect perpendicularly and form right triangles. Each side is the hypotenuse of a right triangle with legs equal to half the diagonals. Half of the longer diagonal: $$\frac{10}{2} = 5 \text{ cm}$$ Half of the shorter diagonal: $$\frac{6}{2} = 3 \text{ cm}$$ So the side length $s$ is: $$s = \sqrt{5^2 + 3^2} = \sqrt{25 + 9} = \sqrt{34} \approx 5.83 \text{ cm}$$ Perimeter (scope) is: $$P = 4 \times s = 4 \times 5.83 = 23.32 \text{ cm}$$ **Final answers:** - Area = $30$ cm$^2$ - Perimeter = $23.32$ cm