Subjects geometry

Triangle Enlargement Prism

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Triangle Enlargement Prism


1. **Problem statement:** (a) Enlarge triangle A with scale factor $\frac{1}{3}$ and centre of enlargement $(2,6)$. (b) Prism X and prism Y are similar. (i) Complete the sentence: The height of prism Y is [ ] times the height of prism X. (ii) Given volume of prism Y is 95 cm³, calculate volume of prism X. --- 2. **Part (a) Enlargement of triangle A:** - The centre of enlargement is $(2,6)$. - Scale factor is $\frac{1}{3}$. - For each vertex $(x,y)$ of triangle A, the image $(x',y')$ is found by: $$x' = 2 + \frac{1}{3}(x - 2)$$ $$y' = 6 + \frac{1}{3}(y - 6)$$ - Vertices of triangle A are approximately: - $A_1 = (6,1)$ - $A_2 = (6,5)$ - $A_3 = (8,1)$ - Calculate new vertices: - For $A_1$: $$x' = 2 + \frac{1}{3}(6 - 2) = 2 + \frac{4}{3} = \frac{10}{3} \approx 3.33$$ $$y' = 6 + \frac{1}{3}(1 - 6) = 6 - \frac{5}{3} = \frac{13}{3} \approx 4.33$$ - For $A_2$: $$x' = 2 + \frac{1}{3}(6 - 2) = 3.33$$ $$y' = 6 + \frac{1}{3}(5 - 6) = 6 - \frac{1}{3} = \frac{17}{3} \approx 5.67$$ - For $A_3$: $$x' = 2 + \frac{1}{3}(8 - 2) = 2 + 2 = 4$$ $$y' = 6 + \frac{1}{3}(1 - 6) = 4.33$$ - So, the enlarged triangle vertices are approximately: $$\left(\frac{10}{3}, \frac{13}{3}\right), \left(\frac{10}{3}, \frac{17}{3}\right), (4, \frac{13}{3})$$ --- 3. **Part (b)(i) Surface area ratio and height ratio:** - Surface area ratio of prism X to prism Y is $1 : 5$. - For similar solids, surface area ratio = (linear scale factor)$^2$. - Let linear scale factor be $k$ from X to Y. Then: $$k^2 = 5 \implies k = \sqrt{5}$$ - Height of prism Y is $\sqrt{5}$ times the height of prism X. --- 4. **Part (b)(ii) Volume calculation:** - Volume ratio = (linear scale factor)$^3$. - Volume of prism Y = 95 cm³. - Volume of prism X = $V_X$. - Using ratio: $$\frac{V_X}{95} = \frac{1}{k^3} = \frac{1}{(\sqrt{5})^3} = \frac{1}{5\sqrt{5}}$$ - Calculate $V_X$: $$V_X = \frac{95}{5\sqrt{5}} = \frac{19}{\sqrt{5}}$$ - Rationalize denominator: $$V_X = \frac{19}{\sqrt{5}} \times \frac{\sqrt{5}}{\sqrt{5}} = \frac{19\sqrt{5}}{5} \approx 17.0 \text{ cm}^3$$ --- **Final answers:** - (a) Enlarged triangle vertices: $\left(\frac{10}{3}, \frac{13}{3}\right), \left(\frac{10}{3}, \frac{17}{3}\right), (4, \frac{13}{3})$ - (b)(i) Height of prism Y is $\sqrt{5}$ times height of prism X. - (b)(ii) Volume of prism X is approximately 17.0 cm³.