1. **Problem statement:** We have a trapezoid with the following dimensions: top base $= 12.0$ cm, bottom base $= 6.0$ cm, left side $= 4.3$ cm, right side $= 6.0$ cm, and height $h = 4.0$ cm. We need to find: a) The perimeter of the trapezoid. b) The area of the trapezoid. c) The sum of the interior angles of the trapezoid. d) The number of sides a polygon must have for its interior angle sum to be $2520^\circ$. 2. **Formulas and rules:** - Perimeter of a trapezoid: sum of all sides. - Area of a trapezoid: $$\text{Area} = \frac{(\text{base}_1 + \text{base}_2)}{2} \times h$$ - Sum of interior angles of any quadrilateral (including trapezoid): $$360^\circ$$ - Sum of interior angles of an n-sided polygon: $$180^\circ \times (n-2)$$ 3. **Calculations:** a) Perimeter: $$\text{Perimeter} = 12.0 + 6.0 + 4.3 + 6.0 = 28.3\text{ cm}$$ b) Area: $$\text{Area} = \frac{(12.0 + 6.0)}{2} \times 4.0 = \frac{18.0}{2} \times 4.0 = 9.0 \times 4.0 = 36.0\text{ cm}^2$$ c) Vinkelsummen (sum of interior angles) of trapezoid: Since trapezoid is a quadrilateral, sum is always: $$360^\circ$$ d) Number of sides for polygon with interior angle sum $2520^\circ$: Using formula: $$180^\circ \times (n-2) = 2520^\circ$$ Divide both sides by 180: $$n - 2 = \frac{2520}{180} = 14$$ Add 2: $$n = 14 + 2 = 16$$ So, the polygon must have 16 sides. **Final answers:** a) Perimeter = $28.3$ cm b) Area = $36.0$ cm$^2$ c) Sum of interior angles = $360^\circ$ d) Number of sides for polygon with $2520^\circ$ interior angle sum = 16