1. Problem 8: Find the area of each given figure. Since the figures are described as a right triangle and a right trapezoid, we use the following formulas: - Area of a triangle: $$\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$$ - Area of a trapezoid: $$\text{Area} = \frac{1}{2} \times (\text{sum of parallel sides}) \times \text{height}$$ 2. Figure (1) is a right triangle with base = 5 m and hypotenuse = 13 m. We need the height. Using Pythagoras' theorem: $$13^2 = 5^2 + h^2 \implies 169 = 25 + h^2 \implies h^2 = 144 \implies h = 12 \text{ m}$$ Area of Figure (1): $$\frac{1}{2} \times 5 \times 12 = 30 \text{ m}^2$$ 3. Figure (2) is a right trapezoid with height = 10 cm, slant height = 10 cm, and base = 16 cm. To find the other parallel side, note the trapezoid has a right angle, so the slant height and height form a right triangle with the difference of bases. Calculate the other base: $$\text{other base} = \sqrt{10^2 - 10^2} = \sqrt{100 - 100} = 0 \text{ cm}$$ This suggests the trapezoid is actually a right triangle with base 16 cm and height 10 cm. Area of Figure (2): $$\frac{1}{2} \times 16 \times 10 = 80 \text{ cm}^2$$ 4. Problem 9: Determine the type of triangle by angles given side lengths. We use the converse of Pythagoras' theorem: - If $a^2 + b^2 = c^2$, triangle is right-angled. - If $a^2 + b^2 > c^2$, triangle is acute. - If $a^2 + b^2 < c^2$, triangle is obtuse. (①) Sides: 3 cm, 5 cm, 6 cm (arranged so $c=6$) $$3^2 + 5^2 = 9 + 25 = 34$$ $$6^2 = 36$$ Since $34 < 36$, triangle is obtuse. (②) Sides: $\sqrt{5}$, $\sqrt{6}$, $\sqrt{7}$ (largest is $\sqrt{7}$) $$ (\sqrt{5})^2 + (\sqrt{6})^2 = 5 + 6 = 11$$ $$ (\sqrt{7})^2 = 7$$ Since $11 > 7$, triangle is acute. (③) Sides: 21 cm, 28 cm, 35 cm (largest is 35) $$21^2 + 28^2 = 441 + 784 = 1225$$ $$35^2 = 1225$$ Since $1225 = 1225$, triangle is right-angled. Final answers: - Area Figure (1): 30 m² - Area Figure (2): 80 cm² - Triangle types: (①) Obtuse, (②) Acute, (③) Right-angled