1. The problem states that the two shapes have equal areas. We are asked to find the ratio $k : m$. 2. The left shape is a right-angled triangle with base $k$ cm and height 12 cm. Its area is $$\text{Area}_{\text{triangle}} = \frac{1}{2} \times k \times 12 = 6k.$$ 3. The right shape is an L-shape composed of two rectangles stacked vertically. The width of both rectangles is $m$. The heights are 2 cm and 3 cm respectively. Hence, the total height is $2 + 3 = 5$ cm. One side length is labeled $k$, which matches the width of the taller part. 4. The total area of the L-shape is the sum of the areas of the two rectangles: $$\text{Area}_{\text{L-shape}} = m \times 2 + m \times 3 = 5m.$$ 5. Since the areas are equal: $$6k = 5m.$$ 6. To find the ratio $k : m$, isolate $k/m$: $$\frac{k}{m} = \frac{5}{6}.$$ 7. Therefore, $$k : m = 5 : 6.$$ --- 8. Next, for question 9, the graph of the quadratic function is given: $$y = 0.5x^2 - 6x + 12.$$ 9. The problem involves analyzing this parabola. To understand its key features: 10. Find the vertex using the formula: $$x = -\frac{b}{2a} = -\frac{-6}{2 \times 0.5} = \frac{6}{1} = 6.$$ 11. Calculate $y$-value at $x=6$: $$y = 0.5(6)^2 - 6(6) + 12 = 0.5 \times 36 - 36 + 12 = 18 - 36 + 12 = -6.$$ 12. The vertex is at $(6, -6)$. Since $a=0.5 > 0$, the parabola opens upwards. 13. Find the $y$-intercept by evaluating $y$ at $x=0$: $$y = 0.5 \times 0 - 6 \times 0 + 12 = 12.$$ So the $y$-intercept is $(0,12)$. 14. Find the $x$-intercepts by solving $0 = 0.5x^2 - 6x + 12$: Multiply all terms by 2: $$0 = x^2 - 12x + 24.$$ 15. Use the quadratic formula: $$x = \frac{12 \pm \sqrt{(-12)^2 - 4 \times 1 \times 24}}{2} = \frac{12 \pm \sqrt{144 - 96}}{2} = \frac{12 \pm \sqrt{48}}{2} = \frac{12 \pm 4\sqrt{3}}{2} = 6 \pm 2\sqrt{3}.$$ These are the two $x$-intercepts. 16. Summary: - Vertex: $(6, -6)$ (minimum point) - $y$-intercept: $(0, 12)$ - $x$-intercepts: $(6 - 2\sqrt{3}, 0)$ and $(6 + 2\sqrt{3}, 0)$ This information matches the graph described.