### 3D figure: Glasshouse

Question Sample Titled '3D figure: Glasshouse'

The figure shows a glasshouse built on a horizontal ground. It consists of two parts. The upper part is a right pyramid with a rectangular base of dimensions ${21}$ $\text{m}\times{43}$ $\text{m}$ and the slant height is ${32}$ $\text{m}$. The lower part is a cuboid with a base the same as the pyramid.

${B}$${O}$${A}$${C}$${D}$${21}$ $\text{m}$${43}$ $\text{m}$${32}$ $\text{m}$${8}$ $\text{m}$${V}$${T}$EN

 (a) Find the height of the glasshouse. (2 marks) (b) As shown in the figure, the edge ${O}{B}$ is along the north-south direction and the edge ${O}{A}$ is along the east-west direction. The overhead sun casts a shadow of the glasshouse on the ground. The angle of elevation from the tip ${T}$ of the shadow to the tip ${V}$ of the glasshouse is ${29}^{\circ}$ . The bearing of ${O}$ from ${T}$ is ${290}^{\circ}$ . Find the distance between ${O}$ and ${T}$ . (5 marks)

 (a) Let ${M}$ be the mid-point of ${C}{D}$, ${X}$ be the vertical projection of ${V}$ on the base of pyramid. In $\triangle{V}{C}{M}$ , ${V}{C}^{{2}}$ $={C}{M}^{{2}}+{V}{M}^{{2}}$ Pyth. theorem 1M ${V}{M}$ $=\sqrt{{{32}^{{2}}-{\left(\dfrac{{21}}{{2}}\right)}^{{2}}}}$  $=\dfrac{{\sqrt{{{3655}}}}}{{2}}$ $\text{m}$  In $\triangle{V}{M}{X}$ , ${V}{M}^{{2}}$ $={M}{X}^{{2}}+{V}{X}^{{2}}$ Pyth. theorem ${V}{X}$ $=\sqrt{{{\left(\dfrac{{\sqrt{{{3655}}}}}{{2}}\right)}^{{2}}-{\left(\dfrac{{43}}{{2}}\right)}^{{2}}}}$  $=\dfrac{{\sqrt{{{1806}}}}}{{2}}$ $\text{m}$  ∴  Required height $={\left(\dfrac{{\sqrt{{{1806}}}}}{{2}}+{8}\right)}$ $\text{m}$ 1A  (b) Let ${Z}$ be the vertical projection of ${V}$ on the ground. In $\triangle{V}{Z}{T}$ , ${\tan}\angle{V}{T}{Z}$ $=\dfrac{{{V}{Z}}}{{{Z}{T}}}$ ${Z}{T}$ $=\dfrac{{\dfrac{{\sqrt{{{1806}}}}}{{2}}+{8}}}{{{\tan{{29}}}^{\circ}}}$ $\text{m}$ 1A  Let ${N}$ be the mid-point of ${A}{O}$. In $\triangle{O}{N}{Z}$ , ${O}{Z}^{{2}}$ $={O}{N}^{{2}}+{N}{Z}^{{2}}$ Pyth. theorem 1M ${O}{Z}$ $=\sqrt{{{\left(\dfrac{{21}}{{2}}\right)}^{{2}}+{\left(\dfrac{{43}}{{2}}\right)}^{{2}}}}$  $=\dfrac{{\sqrt{{{2290}}}}}{{2}}$ $\text{m}$  $\angle{T}{O}{Z}$ $={180}^{\circ}-\angle{N}{O}{Z}+{\left({90}^{\circ}-{\left({360}^{\circ}-{290}^{\circ}\right)}\right)}$ 1M  $={200}^{\circ}-{{\tan}^{{-{1}}}{\left(\tfrac{{21.5}}{{10.5}}\right)}}$  Let ${x}$ be the distance between ${O}$ and ${T}$. By the cosine formula, ${Z}{T}^{{2}}$ $={O}{Z}^{{2}}+{x}^{{2}}-{2}{\left({O}{Z}\right)}{\left({x}\right)}{\cos}\angle{T}{O}{Z}$ 1M ${x}^{{2}}+{\left(-{2}{\left(\dfrac{{\sqrt{{{2290}}}}}{{2}}\right)}{\cos{{\left({200}^{\circ}-{{\tan}^{{-{1}}}{\left(\tfrac{{21.5}}{{10.5}}\right)}}\right)}}}\right)}{x}+{572.5}-{\left(\dfrac{{\dfrac{{\sqrt{{{1806}}}}}{{2}}+{8}}}{{{\tan{{29}}}^{\circ}}}\right)}^{{2}}={0}$ ${x}$ $={32.862}$ or ${x}=-{67.303}$ (rejected)  ∴  Required distance $={32.9}$ $\text{m}$ (cor. to 3 sig. fig.) 1A

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