Question: 13 Work out the area of each compound shape. Give your answer to part b correct to two decimal places (2 d.p.). a 10 cm 8 cm 16 cm b 12 cm 8 cm 14 Here is a function: 6x + 5y = 60. a Show that the point (5, 6) is on a graph of 6x + 5y = 60. b Work out where a graph of 6x + 5y = 60 crosses the coordinate axes. c Draw a graph of 6x + 5y = 60. 15 ABC is an equilateral triangle. Each side is 20 cm. Work out a the height AD b the area of the triangle. 16 The diagram shows a shape formed by a semicircle with diameter AB joined to a right-angled triangle ABC. AB = 12 cm and BC = 15 cm. Calculate the total area of the shape. 17 Without using a calculator, decide if these fractions are terminating or recurring decimals. a 8/3 b 21/5 c 28/9 d 39/12

1. Problem 13: Find the area of each compound shape. **Part a:** The shape can be divided into a rectangle and a triangle. - Rectangle area = length \times width = $16 \times 8 = 128$ cm$^2$. - Triangle area = $\frac{1}{2} \times$ base $\times$ height = $\frac{1}{2} \times 10 \times 8 = 40$ cm$^2$. - Total area = $128 + 40 = 168$ cm$^2$. **Part b:** The shape can be divided into a rectangle and a semicircle. - Rectangle area = $12 \times 8 = 96$ cm$^2$. - Semicircle radius = $\frac{12}{2} = 6$ cm. - Semicircle area = $\frac{1}{2} \pi r^2 = \frac{1}{2} \pi \times 6^2 = 18\pi$ cm$^2$. - Total area = $96 + 18\pi \approx 96 + 56.55 = 152.55$ cm$^2$ (to 2 d.p.). 2. Problem 14: Function $6x + 5y = 60$. **a.** Check if point $(5,6)$ lies on the graph. - Substitute $x=5$, $y=6$ into the equation: $$6(5) + 5(6) = 30 + 30 = 60$$ - Since LHS = RHS, point $(5,6)$ is on the graph. **b.** Find intercepts. - $x$-intercept: set $y=0$: $$6x + 5(0) = 60 \Rightarrow 6x = 60 \Rightarrow x = 10$$ - $y$-intercept: set $x=0$: $$6(0) + 5y = 60 \Rightarrow 5y = 60 \Rightarrow y = 12$$ **c.** Graph passes through $(10,0)$ and $(0,12)$. 3. Problem 15: Equilateral triangle ABC with side $20$ cm. **a.** Find height AD. - Height splits triangle into two 30-60-90 right triangles. - Height $AD = 20 \times \sin 60^\circ = 20 \times \frac{\sqrt{3}}{2} = 10\sqrt{3}$ cm. **b.** Area of triangle: - $$\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 20 \times 10\sqrt{3} = 100\sqrt{3} \approx 173.21 \text{ cm}^2$$ 4. Problem 16: Shape formed by semicircle with diameter AB and right triangle ABC. - AB = 12 cm, BC = 15 cm. - Find AC using Pythagoras: $$AC = \sqrt{15^2 + 12^2} = \sqrt{225 + 144} = \sqrt{369} = 3\sqrt{41}$$ cm. - Area of triangle ABC: $$\frac{1}{2} \times AB \times BC = \frac{1}{2} \times 12 \times 15 = 90 \text{ cm}^2$$ - Area of semicircle: $$\frac{1}{2} \pi r^2 = \frac{1}{2} \pi \times 6^2 = 18\pi \approx 56.55 \text{ cm}^2$$ - Total area = $90 + 56.55 = 146.55$ cm$^2$. 5. Problem 17: Determine if fractions are terminating or recurring decimals. - A fraction in lowest terms has a terminating decimal if its denominator's prime factors are only 2 and/or 5. **a.** $\frac{8}{3}$ denominator 3 (prime factor 3) → recurring decimal. **b.** $\frac{21}{5}$ denominator 5 → terminating decimal. **c.** $\frac{28}{9}$ denominator 9 = $3^2$ → recurring decimal. **d.** $\frac{39}{12}$ denominator 12 = $2^2 \times 3$ → recurring decimal because of factor 3. Final answers: - 13a: $168$ cm$^2$ - 13b: $152.55$ cm$^2$ - 14a: Point $(5,6)$ lies on graph. - 14b: $x$-intercept $10$, $y$-intercept $12$. - 15a: Height $10\sqrt{3}$ cm. - 15b: Area $100\sqrt{3} \approx 173.21$ cm$^2$. - 16: Total area $146.55$ cm$^2$. - 17a: Recurring - 17b: Terminating - 17c: Recurring - 17d: Recurring