### Find the volume of a pyramid formed by folding isometric pentagonal paper card

Question Sample Titled 'Find the volume of a pyramid formed by folding isometric pentagonal paper card'

In Figure (a), ${A}{B}{C}{D}{B}'$ is a pentagonal paper card. It is given that ${A}{B}={A}{B}'={38}$ $\text{cm}$ , ${B}{C}={B}'{D}={28}$ $\text{cm}$ and $\angle{A}{B}{C}={A}{B}'{D}={72}^{\circ}$ .

${A}$${B}$${C}$${D}$${B}'$Figure (a)Figure (b)${A}$${B}$${C}$${D}$

 (a) Suppose that ${95}^{\circ}\le\angle{B}{C}{D}\le{145}^{\circ}$ . (i) Find the distance between ${A}$ and ${C}$ . (ii) Find $\angle{A}{C}{B}$ . (iii) Describe how the area of the paper card varies when $\angle{B}{C}{D}$ increases from ${95}^{\circ}$ to ${145}^{\circ}$ .  (7 marks) (b) Suppose that $\angle{B}{C}{D}={132}^{\circ}$ .  The paper card in Figure (b) is folded along ${A}{C}$ and ${A}{D}$ such that ${A}{B}$ and ${A}{B}'$ join together to form a pyramid ${A}{B}{C}{D}$ as shown in Figure (b). Find the volume of the pyramid ${A}{B}{C}{D}$ .  (6 marks)

 (a)(i) By consine formula, ${A}{C}^{{2}}$ $={A}{B}^{{2}}+{B}{C}^{{2}}-{2}{\left({A}{B}\right)}{\left({B}{C}\right)}{\left({\cos}\angle{A}{B}{C}\right)}$ 1M ${A}{C}^{{2}}$ $={38}^{{2}}+{28}^{{2}}-{2}{\left({38}\right)}{\left({28}\right)}{\left({\cos}\angle{A}{B}{C}^{\circ}\right)}$ ${A}{C}$ $={39.62842207}$ $\text{cm}$ ${A}{C}$ $\approx{39.6}$ $\text{cm}$ 1A r.t. ${39.6}$ $\text{cm}$ Thus, the distance between ${A}$ and ${C}$ is ${39.6}$ $\text{cm}$ . (a)(ii) By sine formula, $\dfrac{{{\sin}\angle{A}{C}{B}}}{{{A}{B}}}$ $=\dfrac{{{\sin}\angle{A}{B}{C}}}{{{A}{C}}}$ 1M $\dfrac{{{\sin}\angle{A}{C}{B}}}{{{38}}}$ $=\dfrac{{{\sin{{72}}}^{\circ}}}{{{39.62842207}}}$ $\angle{A}{C}{B}$ $\approx{65.77978612}$ or $\angle{A}{C}{B}\approx{114.2202139}^{\circ}$ (rejected) $\angle{A}{C}{B}$ $\approx{65.8}^{\circ}$ 1A r.t. ${65.8}^{\circ}$  (a)(iii) $\angle{C}{A}{D}$ $={180}^{\circ}-{2}{\left(\angle{B}{C}{D}-\angle{A}{C}{B}\right)}$ base ∠s, isos. △ $={180}^{\circ}-{2}{\left(\angle{B}{C}{D}-{65.77978612}^{\circ}\right)}$ ∵  ${95}^{\circ}\le\angle{B}{C}{D}\le{145}^{\circ}$ , ∴  ${21.55957224}^{\circ}\le\angle{C}{A}{D}\le{121.5595722}^{\circ}$  The area of the paper card $={2}{\left(\dfrac{{1}}{{2}}{\left({38}\right)}{\left({28}\right)}{\sin{{72}}}^{\circ}\right)}+\dfrac{{1}}{{2}}{A}{C}^{{2}}{\sin}\angle{C}{A}{D}$ 1M $={1064}{{\sin{{72}}}^{\circ}+}\dfrac{{1}}{{2}}{A}{C}^{{2}}{\sin}\angle{C}{A}{D}$ Note that ${1064}{{\sin{{72}}}^{\circ}}$ and ${A}{C}$ are constants and hence the area of the paper card only varies as ${\sin}\angle{C}{A}{D}$ . 1M Also note that the area of the paper card is the greatest when $\angle{C}{A}{D}$ $={90}^{\circ}$ , as ${{\sin{{90}}}^{\circ}}$ is the greatest. When $\angle{C}{A}{D}={90}^{\circ}$ , $\angle{B}{C}{D}={65.77978612}^{\circ}+\dfrac{{{180}^{\circ}-{90}^{\circ}}}{{2}}={110.7797861}^{\circ}$ . r.t. ${111}^{\circ}$ Thus, when $\angle{B}{C}{D}$ increases from ${95}^{\circ}$ to ${110.7797861}^{\circ}$ , the area of the paper card increases. 1A f.t.; need not to specify the actual area When $\angle{B}{C}{D}$ increases from ${110.7797861}^{\circ}$ to ${145}^{\circ}$ , the area of the paper card decreases.   (b) Denote ${M}$ be the mid-point of ${C}{D}$ . $\angle{A}{C}{M}$ $={132}^{\circ}-{65.77978612}^{\circ}$ $={66.22021388}^{\circ}$ Consider $\triangle{A}{C}{M}$， ${\sin}\angle{A}{C}{M}$ $=\dfrac{{{A}{M}}}{{{A}{C}}}$ 1M ${{\sin{{66.22021388}}}^{\circ}}$ $=\dfrac{{{A}{M}}}{{39.62842207}}$ ${A}{M}$ $={36.26404755}$ $\text{cm}$ ${C}{M}$ $=\sqrt{{{A}{C}^{{2}}-{A}{M}^{{2}}}}$ ${C}{M}$ $=\sqrt{{{39.62842207}^{{2}}-{36.26404755}^{{2}}}}$  $={15.97907041}$ $\text{cm}$ Consider $\triangle{B}{C}{D}$ ， ${B}{M}$ $=\sqrt{{{B}{C}^{{2}}-{C}{M}^{{2}}}}$ ${B}{M}$ $=\sqrt{{{28}^{{2}}-{15.97907041}^{{2}}}}$ ${B}{M}$ $={22.99280994}$ $\text{cm}$ Consider $\triangle{A}{B}{M}$, by cosine formula, ${\cos}\angle{A}{M}{B}$ $=\dfrac{{{\left({A}{M}\right)}^{{2}}+{\left({B}{M}\right)}^{{2}}-{\left({A}{B}\right)}^{{2}}}}{{{2}{\left({A}{M}\right)}{\left({B}{M}\right)}}}$ 1M ${\cos}\angle{A}{M}{B}$ $\approx\dfrac{{{36.26404755}^{{2}}+{22.99280994}^{{2}}-{38}^{{2}}}}{{{2}{\left({36.26404755}\right)}{\left({22.99280994}\right)}}}$ $\angle{A}{M}{B}$ $\approx{76.13042869}^{\circ}$  The height of the pyramid ${A}{B}{C}{D}$ $={B}{M}{\sin}\angle{A}{M}{B}$ 1M accept ${B}{A}{\sin}\angle{B}{A}{M}$ $={22.99280994}{{\sin{{76.13042869}}}^{\circ}}$ $={22.32242985}$ $\text{cm}$  The area of $\triangle{A}{C}{D}$ $=\dfrac{{1}}{{2}}{\left({C}{D}\right)}{\left({A}{M}\right)}$ 1M accept $\dfrac{{1}}{{2}}{A}{C}^{{2}}{\sin}\angle{C}{A}{D}$ $=\dfrac{{1}}{{2}}{\left({2}{C}{M}\right)}{\left({36.26404755}\right)}$ $=\dfrac{{1}}{{2}}{\left({2}\sqrt{{{A}{C}^{{2}}-{A}{M}^{{2}}}}\right)}{\left({36.26404755}\right)}$ $=\dfrac{{1}}{{2}}{\left({2}\sqrt{{{39.62842207}^{{2}}-{36.26404755}^{{2}}}}\right)}{\left({36.26404755}\right)}$ $={579.4657692}$ $\text{cm}^{{2}}$  The volume of the pyramid ${A}{B}{C}{D}$ $=\dfrac{{1}}{{3}}{(}$the area of $\triangle{A}{C}{D}{)}{(}$the height of the pyramid${)}$ 1M $=\dfrac{{1}}{{3}}{\left({579.4657692}\right)}{\left({22.32242985}\right)}$ $={4311.694661}$ $\text{cm}^{{2}}$ $\approx{4310}$ $\text{cm}^{{3}}$ 1A r.t. ${4310}$ $\text{cm}^{{3}}$

${A}$${B}$${C}$${D}$${B}'$Figure (a)${38}$${28}$Figure (b)${A}$${B}$${C}$${D}$${M}$

 Caution: In (b), attempts to find the height of the pyramid by mistakenly considering the height as $\dfrac{{1}}{{2}}{B}{B}'$ in Figure (a) should receive zero marks.

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