1. **Problem 1: Area of quadrilateral EFGH** The quadrilateral has sides EH = 10 cm (vertical), HG = 11 cm (horizontal), and GF = 5 cm (right vertical segment tilted). Right angles are at H and F, and diagonal EG is drawn. 2. To find the area, split the quadrilateral into two right triangles by diagonal EG. 3. Use the given hint: $5 \times \tan 47^\circ = 5 \times 1.0724 = 5.362$ (approx). This suggests the horizontal projection of GF is about 5.362 cm. 4. The total horizontal length EH + HG = 10 + 11 = 21 cm, but since EH is vertical, the base is HG = 11 cm. 5. Calculate the area of triangle EHG (right angle at H): $$\text{Area}_{EHG} = \frac{1}{2} \times EH \times HG = \frac{1}{2} \times 10 \times 11 = 55 \text{ cm}^2$$ 6. Calculate the area of triangle GEF (right angle at F): Height = 5 cm (GF), base = horizontal projection $5.362$ cm $$\text{Area}_{GEF} = \frac{1}{2} \times 5 \times 5.362 = 13.405 \text{ cm}^2$$ 7. Total area of quadrilateral EFGH: $$55 + 13.405 = 68.405 \approx 68.4 \text{ cm}^2$$ --- 1. **Problem 2: Median of numbers 3.8, 7.5, 9.4, 5.3, 3.8, 8.9** 2. Sort the numbers: $$3.8, 3.8, 5.3, 7.5, 8.9, 9.4$$ 3. Median is the average of the middle two numbers (3rd and 4th): $$\frac{5.3 + 7.5}{2} = \frac{12.8}{2} = 6.4$$ --- 1. **Problem 3: Distance p between two parallel sides of a regular hexagon with side 6.3 cm** 2. The distance between two parallel sides (the height) of a regular hexagon is: $$p = \text{side} \times \sqrt{3} = 6.3 \times 1.732 = 10.9116$$ 3. Rounded to 2 d.p.: $$p = 10.91 \text{ cm}$$ --- 1. **Problem 4: Mean of other 5 temperatures given mean of 6 days is 2.1 and one day is 3.1** 2. Total sum of 6 days: $$6 \times 2.1 = 12.6$$ 3. Sum of other 5 days: $$12.6 - 3.1 = 9.5$$ 4. Mean of other 5 days: $$\frac{9.5}{5} = 1.9$$ --- 1. **Problem 5: Simplify $x$ in form $a + b\sqrt{c}$ given right triangle with hypotenuse $2x$, leg $x+10$, and angle opposite leg $30^\circ$** 2. Using sine definition: $$\sin 30^\circ = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{x+10}{2x}$$ 3. Since $\sin 30^\circ = \frac{1}{2}$, set equation: $$\frac{1}{2} = \frac{x+10}{2x}$$ 4. Cross multiply: $$2x \times \frac{1}{2} = x + 10 \Rightarrow x = x + 10$$ This is impossible, so check cosine instead (angle adjacent to leg $x+10$): 5. Using cosine: $$\cos 30^\circ = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{x+10}{2x}$$ 6. $\cos 30^\circ = \frac{\sqrt{3}}{2}$, so: $$\frac{\sqrt{3}}{2} = \frac{x+10}{2x}$$ 7. Cross multiply: $$2x \times \frac{\sqrt{3}}{2} = x + 10 \Rightarrow x \sqrt{3} = x + 10$$ 8. Rearrange: $$x \sqrt{3} - x = 10$$ $$x(\sqrt{3} - 1) = 10$$ 9. Solve for $x$: $$x = \frac{10}{\sqrt{3} - 1}$$ 10. Rationalize denominator: $$x = \frac{10}{\sqrt{3} - 1} \times \frac{\sqrt{3} + 1}{\sqrt{3} + 1} = \frac{10(\sqrt{3} + 1)}{(\sqrt{3})^2 - 1^2} = \frac{10(\sqrt{3} + 1)}{3 - 1} = \frac{10(\sqrt{3} + 1)}{2}$$ 11. Simplify: $$x = 5(\sqrt{3} + 1) = 5 + 5\sqrt{3}$$ 12. Therefore: $$a = 5, b = 5, c = 3$$