1. **Problem 1: Find the area of the trapezium-shaped window.** The trapezium consists of a rectangle and two identical right-angled triangles. 2. **Formula for area of trapezium:** $$\text{Area} = \frac{1}{2} (a + b) h$$ where $a$ and $b$ are the lengths of the parallel sides and $h$ is the height. 3. **Identify dimensions:** - Top base $a = 110$ cm - Bottom base $b = 40$ cm (width of rectangle) - Height $h = 70$ cm 4. **Calculate area:** $$\text{Area} = \frac{1}{2} (110 + 40) \times 70 = \frac{1}{2} \times 150 \times 70 = 75 \times 70 = 5250 \text{ cm}^2$$ --- 1. **Problem 2: Find the surface area of the cuboid.** 2. **Formula for surface area of cuboid:** $$\text{Surface Area} = 2(lw + lh + wh)$$ where $l$, $w$, and $h$ are length, width, and height. 3. **Given dimensions:** - Length $l = 20$ cm - Width $w = 5$ cm - Height $h = 30$ cm 4. **Calculate surface area:** $$2(20 \times 5 + 20 \times 30 + 5 \times 30) = 2(100 + 600 + 150) = 2(850) = 1700 \text{ cm}^2$$ --- 1. **Problem 3: Analyze running times before and after training.** 2. **Given times before training:** 15, 13, 28, 25, 24, 22, 24, 17 minutes. 3. **(a) Calculate mean before training:** $$\text{Mean} = \frac{15 + 13 + 28 + 25 + 24 + 22 + 24 + 17}{8} = \frac{168}{8} = 21 \text{ minutes}$$ 4. **(b) Calculate range before training:** $$\text{Range} = \text{max} - \text{min} = 28 - 13 = 15 \text{ minutes}$$ 5. **(c) Comment comparing before and after training:** - Mean time decreased from 21 to 18 minutes, showing improvement in average speed. - Range increased from 15 to 19 minutes, indicating more variation in times after training. --- **Final answers:** - Area of window: $5250$ cm$^2$ - Surface area of cuboid: $1700$ cm$^2$ - Mean time before training: $21$ minutes - Range before training: $15$ minutes - Comment: Mean time improved, but range increased after training.