### Dimensions of symmetrical pyramid

Question Sample Titled 'Dimensions of symmetrical pyramid'

In figure (a), ${A}{T}{B}{C}{D}{T}'$ is a hexagonal paper card. It is given that ${A}{T}={D}{T}'={13}$ $\text{cm}$ and ${T}{B}={T}'{C}={9}$ $\text{cm}$ . ${A}{B}{C}{D}$ is a rectangle with ${A}{D}={4}$ $\text{cm}$ . Let $\angle{A}{T}{B}=\theta$ , where ${60}^{\circ}\le\theta\le{85}^{\circ}$ .

Figure (a)Figure (b)${A}$${B}$${C}$${D}$${T}$${T}'$${A}$${B}$${C}$${D}$${T}$

 (a) Suppose that $\theta={76}^{\circ}$ . (i) Find the length of ${A}{B}$ . (ii) Find the area of the paper card ${A}{T}{B}{C}{T}'{D}$ .  (4 marks) (b) Describe how the area of the paper card ${A}{T}{B}{C}{T}'{D}$ varies when $\theta$ increases from ${60}^{\circ}$ to ${85}^{\circ}$ . Explain your answer.  (3 marks) (c) Suppose that $\theta={75}^{\circ}$ . The paper card in figure (a) is folded along ${A}{B}$ and ${C}{D}$ such that ${T}$ and ${T}'$ join together to form a pyramid ${A}{B}{C}{D}{T}$ as shown in figure (b). Find the volume of the pyramid ${A}{B}{C}{D}{T}$ .  (6 marks)

 (ai) ${A}{B}^{{2}}$ $={A}{T}^{{2}}+{T}{B}^{{2}}-{2}{\left({A}{T}\right)}{\left({T}{B}\right)}{\cos}\angle{A}{T}{B}$  $={13}^{{2}}+{9}^{{2}}-{2}{\left({13}\right)}{\left({9}\right)}{{\cos{{76}}}^{\circ}}$ 1M  $={250}-{234}{{\cos{{76}}}^{\circ}}$ ${A}{B}$ $={13.9}$ $\text{cm}$ (cor. to 3 sig. fig.) 1A  (aii) Note that $\triangle{A}{T}{B}\stackrel{\sim}{=}\triangle{D}{T}'{C}$ .  SSS Area of ${A}{T}{B}{C}{T}'{D}$ $={2}\times{(}$Area of$\triangle{A}{T}{B}{)}+{(}$ Area of ${A}{B}{C}{D}{)}$ $={2}\times\dfrac{{1}}{{2}}{\left({13}\right)}{\left({9}\right)}{{\sin{{76}}}^{\circ}+}{\left({4}\right)}{\left({13.906483251695152}\right)}$ 1M $={169}$ $\text{cm}^{{2}}$ (cor. to 3 sig. fig.) 1A  (b) Area of ${A}{T}{B}{C}{T}'{D}$ $={2}\times\dfrac{{1}}{{2}}{\left({13}\right)}{\left({9}\right)}{\sin{\theta}}+{\left({4}\right)}\sqrt{{{250}-{234}{\cos{\theta}}}}$ $={\left({117}{\sin{\theta}}+{4}\sqrt{{{250}-{234}{\cos{\theta}}}}\right)}$ $\text{cm}^{{2}}$ 1M For ${60}^{\circ}\le\theta\le{85}^{\circ}$ , as $\theta$ increases, ${\sin{\theta}}$ increases while ${\cos{\theta}}$ decreases resulting that $\sqrt{{{250}-{234}{\cos{\theta}}}}$ increases.  1A Thus, when $\theta$ increases from ${60}^{\circ}$ to ${85}^{\circ}$ , the area of the paper card is increasing throughout the interval. 1A  (c) Let ${E}$ and ${F}$ be the points on ${A}{B}$ and ${C}{D}$ respectively such that ${T}{E}\bot{A}{B}$ and ${T}{F}\bot{C}{D}$ .  Let ${M}$ be the mid-point of ${E}{F}$ . Note that ${T}{E}$ $={T}{F}$ and $\angle{T}{M}{E}={90}^{\circ}$ . prop. of isos. △ The height of the pyramid is ${T}{M}$ . ${A}{B}$ $=\sqrt{{{250}-{234}{\cos{{75}}}^{\circ}}}$ 1A Consider the area of $\triangle{A}{T}{B}$ ,  $\dfrac{{1}}{{2}}{\left({A}{B}\right)}{\left({T}{E}\right)}$ $=\dfrac{{1}}{{2}}{\left({A}{T}\right)}{\left({T}{B}\right)}{{\sin{{75}}}^{\circ}}$ 1M ${T}{E}$ $=\dfrac{{{117}{\sin{{75}}}^{\circ}}}{{\sqrt{{{250}-{234}{\cos{{75}}}^{\circ}}}}}$  $\approx{8.211036614361895}$ $\text{cm}$ 1A ${E}{M}^{{2}}+{T}{M}^{{2}}$ $={T}{E}^{{2}}$ Pyth. theorem ${T}{M}$ $\approx\sqrt{{{8.211036614361895}^{{2}}-{2}^{{2}}}}$  $\approx{7.963737959174175}$ $\text{cm}$ 1A Thus, required volume $=\dfrac{{1}}{{3}}{\left({A}{D}\right)}{\left({A}{B}\right)}{\left({T}{M}\right)}$ 1M  $=\dfrac{{1}}{{3}}{\left({4}\right)}{\left({13.763587593574945}\right)}{\left({7.963737959174175}\right)}$  $={146}$ $\text{cm}^{{3}}$ (cor. to 3 sig. fig.) 1A

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