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Question Number 96936 by joki last updated on 05/Jun/20

Commented by PRITHWISH SEN 2 last updated on 05/Jun/20

$$\mathrm{shorter}\mathrm{diagonal}=\mathrm{a}$$$$\mathrm{longer}\mathrm{diagonal}=\mathrm{a}\sqrt{3}$$$$\therefore \mathrm{Ratio}=\frac{1}{\sqrt{3}}$$

Answered by 1549442205 last updated on 05/Jun/20

$$\mathrm{Denote}\mathrm{by}\mathrm{a}\mathrm{the}\mathrm{length}\mathrm{of}\mathrm{the}\mathrm{side}\mathrm{of}\mathrm{parallelogram}.\mathrm{Then}$$$$\mathrm{the}\mathrm{length}\mathrm{of}\mathrm{shorter}\mathrm{diagonal}\mathrm{is}\mathrm{equal}\mathrm{to}\mathrm{a}$$$$(\mathrm{since}\mathrm{it}\mathrm{is}\mathrm{a}\mathrm{side}\mathrm{of}\mathrm{the}\mathrm{regular}\mathrm{triangle}.\mathrm{The}\mathrm{length}$$$$\mathrm{of}\mathrm{longer}\mathrm{diagonal}\mathrm{equal}\mathrm{to}\mathrm{two}\mathrm{times}\mathrm{the}\mathrm{length}\mathrm{of}$$$$\mathrm{altitude}\mathrm{of}\mathrm{regular}\mathrm{triangle}\mathrm{i}.\mathrm{e}\mathrm{it}\mathrm{is}\mathrm{equal}\mathrm{to}$$$$\mathrm{a}\sqrt{3}.\mathrm{Thus},\frac{{\mathrm{d}}_{\mathrm{sh}}}{{\mathrm{d}}_{\mathrm{l}}}=\frac{\mathrm{a}}{\mathrm{a}\sqrt{3}}=\frac{1}{\sqrt{3}}.\mathrm{Hence},\mathrm{choose}\mathrm{D}\mathrm{is}\mathrm{the}$$$$\mathrm{answer}$$

Answered by bobhans last updated on 06/Jun/20

$$\mathrm{length}\mathrm{of}\mathrm{shorter}\mathrm{diagonal}\mathrm{x}=\sqrt{{2\mathrm{s}}^2-{2\mathrm{s}}^2\times \cos {60}^{\mathrm{o}}}=\mathrm{s}$$$$\mathrm{length}\mathrm{of}\mathrm{longer}\mathrm{diagonal}\mathrm{y}=\sqrt{{2\mathrm{s}}^{\mathrm{s}}-{2\mathrm{s}}^2\times \cos {120}^{\mathrm{o}}}=\mathrm{s}\sqrt{3}$$$$\mathrm{then}\mathrm{ratio}=\frac{\mathrm{s}}{\mathrm{s}\sqrt{3}}=1:\sqrt{3}$$

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