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Question Number 156671 by cortano last updated on 14/Oct/21

Answered by som(math1967) last updated on 14/Oct/21

Commented by som(math1967) last updated on 14/Oct/21

$$cos\alpha =\frac{10}{12}=\frac{5}{6}$$$$sin\alpha =\frac{\sqrt{11}}{6}$$$$sin2\alpha =2\times \frac{\sqrt{11}}{6}\times \frac{5}{6}=\frac{5\sqrt{11}}{18}$$$$againsin2\alpha =\frac{LM}{AL}$$$$\frac{10}{AL}=\frac{5\sqrt{11}}{18}\Rightarrow AL=\frac{36}{\sqrt{11}}=\frac{36\sqrt{11}}{11}$$$${DL}^2={\left(\frac{36\sqrt{11}}{11}\right)}^2-{10}^2$$$$DL=\frac{14}{\sqrt{11}}=\frac{14\sqrt{11}}{11}$$$$LC=(10-\frac{14\sqrt{11}}{11})$$$$BN=\sqrt{{12}^2-{10}^2}=2\sqrt{11}$$$$CN=10-2\sqrt{11}$$$$perimeter=AL+LC+CN+AN$$$$=\frac{36\sqrt{11}}{11}+(10-\frac{14\sqrt{11}}{11})+10-2\sqrt{11}+12$$$$=32+\frac{36\sqrt{11}}{11}-\frac{36\sqrt{11}}{11}=32unit$$

Commented by cortano last updated on 14/Oct/21

$$\mathrm{oo}\mathrm{yes}\mathrm{gave}\mathrm{kudos}$$

Commented by otchereabdullai@gmail.com last updated on 14/Oct/21

$$\mathrm{wow}$$

Commented by Tawa11 last updated on 14/Oct/21

$$\mathrm{great}\mathrm{sir}$$

Answered by john_santu last updated on 15/Oct/21

Commented by cortano last updated on 15/Oct/21

$$\mathrm{nice}$$

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