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Question Number 86379 by jagoll last updated on 28/Mar/20

Commented by jagoll last updated on 28/Mar/20

$$\mathrm{to}\mathrm{miss}\mathrm{tawa}$$

Commented by TawaTawa1 last updated on 28/Mar/20

$$\mathrm{Sir},\mathrm{am}\mathrm{here}$$

Answered by jagoll last updated on 28/Mar/20

$$\tan {30}^{\mathrm{o}}=\frac{9}{\mathrm{AS}}\Rightarrow \mathrm{AS}=9\sqrt{3}$$$$\tan {60}^{\mathrm{o}}=\frac{9}{\mathrm{DP}}\Rightarrow \mathrm{DP}=3\sqrt{3}$$$$\mathrm{perimeter}\mathrm{PQRS}=2\times (18\sqrt{3}+6\sqrt{3})$$$$=48\sqrt{3}\mathrm{cm}$$

Commented by TawaTawa1 last updated on 28/Mar/20

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}.\mathrm{I}\mathrm{appreciate}.$$

Commented by Prithwish Sen 1 last updated on 28/Mar/20

$${\mathrm{is}}^{\prime }\mathrm{nt}\mathrm{it}48\sqrt{3}\mathrm{cm}.$$

Commented by jagoll last updated on 28/Mar/20

$$\mathrm{oo}\mathrm{yes}.{\mathrm{i}}^{\prime }\mathrm{m}\mathrm{forgot}\mathrm{multiply}\mathrm{by}2$$

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