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Question-140139




Question Number 140139 by mathsuji last updated on 04/May/21
Answered by mr W last updated on 04/May/21
say radius of curcumcircle is r  Σsin^(−1) (a/(2r))=sin^(−1) (6/(2r))+sin (3/(2r))+sin^(−1) ((√(11))/(2r))+sin^(−1) (6/(2r))+sin^(−1) ((√2)/(2r))=π  ⇒2r=7.09124208  A=(1/4)Σa(√(4r^2 −a^2 ))      ≈23.81176
$${say}\:{radius}\:{of}\:{curcumcircle}\:{is}\:{r} \\ $$$$\Sigma\mathrm{sin}^{−\mathrm{1}} \frac{{a}}{\mathrm{2}{r}}=\mathrm{sin}^{−\mathrm{1}} \frac{\mathrm{6}}{\mathrm{2}{r}}+\mathrm{sin}\:\frac{\mathrm{3}}{\mathrm{2}{r}}+\mathrm{sin}^{−\mathrm{1}} \frac{\sqrt{\mathrm{11}}}{\mathrm{2}{r}}+\mathrm{sin}^{−\mathrm{1}} \frac{\mathrm{6}}{\mathrm{2}{r}}+\mathrm{sin}^{−\mathrm{1}} \frac{\sqrt{\mathrm{2}}}{\mathrm{2}{r}}=\pi \\ $$$$\Rightarrow\mathrm{2}{r}=\mathrm{7}.\mathrm{09124208} \\ $$$${A}=\frac{\mathrm{1}}{\mathrm{4}}\Sigma{a}\sqrt{\mathrm{4}{r}^{\mathrm{2}} −{a}^{\mathrm{2}} } \\ $$$$\:\:\:\:\approx\mathrm{23}.\mathrm{81176} \\ $$
Commented by mr W last updated on 04/May/21
Commented by mathsuji last updated on 05/May/21
thank you very much Sir
$${thank}\:{you}\:{very}\:{much}\:{Sir} \\ $$

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