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Question-211216




Question Number 211216 by Jubr last updated on 31/Aug/24
Answered by nikif99 last updated on 31/Aug/24
Commented by Jubr last updated on 31/Aug/24
Thank you sir
$${Thank}\:{you}\:{sir} \\ $$
Answered by mr W last updated on 01/Sep/24
Commented by mr W last updated on 01/Sep/24
BD=(√(a^2 +(2(√3)a)^2 ))=(√(13))a  (b/(3a))=(a/( (√(13))a)) ⇒b=((3a)/( (√(13))))  ((s+b)/(3a))=((3a)/( (√(13))a)) ⇒s=((9a)/( (√(13))))−((3a)/( (√(13))))=((6a)/( (√(13))))  (A_(shaded) =(((√3)s^2 )/4)=((√3)/4)×((36a^2 )/(13))=((9(√3)a^2 )/(13)))  fraction=(1/(18))((s/a))^2 =(1/(18))×((36)/(13))=(2/(13)) ✓
$${BD}=\sqrt{{a}^{\mathrm{2}} +\left(\mathrm{2}\sqrt{\mathrm{3}}{a}\right)^{\mathrm{2}} }=\sqrt{\mathrm{13}}{a} \\ $$$$\frac{{b}}{\mathrm{3}{a}}=\frac{{a}}{\:\sqrt{\mathrm{13}}{a}}\:\Rightarrow{b}=\frac{\mathrm{3}{a}}{\:\sqrt{\mathrm{13}}} \\ $$$$\frac{{s}+{b}}{\mathrm{3}{a}}=\frac{\mathrm{3}{a}}{\:\sqrt{\mathrm{13}}{a}}\:\Rightarrow{s}=\frac{\mathrm{9}{a}}{\:\sqrt{\mathrm{13}}}−\frac{\mathrm{3}{a}}{\:\sqrt{\mathrm{13}}}=\frac{\mathrm{6}{a}}{\:\sqrt{\mathrm{13}}} \\ $$$$\left({A}_{{shaded}} =\frac{\sqrt{\mathrm{3}}{s}^{\mathrm{2}} }{\mathrm{4}}=\frac{\sqrt{\mathrm{3}}}{\mathrm{4}}×\frac{\mathrm{36}{a}^{\mathrm{2}} }{\mathrm{13}}=\frac{\mathrm{9}\sqrt{\mathrm{3}}{a}^{\mathrm{2}} }{\mathrm{13}}\right) \\ $$$${fraction}=\frac{\mathrm{1}}{\mathrm{18}}\left(\frac{{s}}{{a}}\right)^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{18}}×\frac{\mathrm{36}}{\mathrm{13}}=\frac{\mathrm{2}}{\mathrm{13}}\:\checkmark \\ $$
Commented by Jubr last updated on 31/Aug/24
Thank you sir
$${Thank}\:{you}\:{sir} \\ $$

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