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Question Number 17729 by Tinkutara last updated on 09/Jul/17

$$\mathrm{A}\mathrm{lotus}\mathrm{plant}\mathrm{in}\mathrm{a}\mathrm{pool}\mathrm{of}\mathrm{water}\mathrm{is}\frac{1}{2}$$$$\mathrm{cubit}\mathrm{above}\mathrm{water}\mathrm{level}.\mathrm{When}$$$$\mathrm{propelled}\mathrm{by}\mathrm{air},\mathrm{the}\mathrm{lotus}\mathrm{sinks}\mathrm{in}\mathrm{the}$$$$\mathrm{pool}2\mathrm{cubits}\mathrm{away}\mathrm{from}\mathrm{its}\mathrm{position}.$$$$\mathrm{Find}\mathrm{the}\mathrm{depth}\mathrm{of}\mathrm{water}\mathrm{in}\mathrm{the}\mathrm{pool}.$$

Commented by alex041103 last updated on 10/Jul/17

$$\mathrm{What}\mathrm{does}\mathrm{a}{}^{\prime }{\mathrm{cubit}}^{\prime }\mathrm{mean}?{\mathrm{I}}^{\prime }\mathrm{m}\mathrm{sorry}\mathrm{for}\mathrm{my}\mathrm{english}.$$

Commented by mrW1 last updated on 10/Jul/17

$$1\mathrm{cubit}=45.72\mathrm{cm}$$

Commented by mrW1 last updated on 10/Jul/17

Answered by mrW1 last updated on 10/Jul/17

$$\mathrm{D}=\mathrm{water}\mathrm{depth}$$$${\mathrm{D}}^2+2^2={(\mathrm{D}+\frac{1}{2})}^2$$$$\Rightarrow \frac{1}{2}(2\mathrm{D}+\frac{1}{2})=4$$$$\Rightarrow \mathrm{D}=\frac{15}{4}=3.75\mathrm{cubit}$$

Commented by Tinkutara last updated on 11/Jul/17

$$\mathrm{Thanks}\mathrm{Sir}!$$

Commented by ajfour last updated on 11/Jul/17

$${\mathrm{couldn}}^{\prime }\mathrm{t}\mathrm{follow}\mathrm{the}\mathrm{geometry}..$$

Commented by Tinkutara last updated on 11/Jul/17

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