Question and Answers Forum

All Questions      Topic List

Geometry Questions

Previous in All Question      Next in All Question      

Previous in Geometry      Next in Geometry      

Question Number 188336 by normans last updated on 28/Feb/23

Answered by mr W last updated on 16/Mar/23

Commented by mr W last updated on 16/Mar/23

$$AF=\sqrt{{\left(2k\right)}^2+{\left(3k\right)}^2}=\sqrt{13}k$$$$DE=EF=\frac{\sqrt{13}k}{\sqrt{2}}$$$$DB=3\sqrt{2}k$$$$\cos \mathrm{\angle }BDF=\frac{{\left(3\sqrt{2}k\right)}^2+{\left(\sqrt{13}k\right)}^2-k^2}{2\times 3\sqrt{2}k\times \sqrt{13}k}=\frac{5}{\sqrt{26}}$$$$EG=\frac{\sqrt{13}k}{\sqrt{2}}\sin (45°-\mathrm{\angle }BDF)$$$$=\frac{\sqrt{13}k}{\sqrt{2}}\times \frac{1}{\sqrt{2}}\times (\frac{5}{\sqrt{26}}-\frac{1}{\sqrt{26}})=\sqrt{2}k$$$$GF=\frac{\sqrt{13}k}{\sqrt{2}}-\sqrt{2}k$$$$\left[CEG\right]=\frac{\sqrt{2}k}{\frac{\sqrt{13}k}{\sqrt{2}}-\sqrt{2}k}\times 10=\frac{20}{\sqrt{13}-2}$$$$\sin \mathrm{\angle }CEG=\frac{2k}{\frac{\sqrt{13}k}{\sqrt{2}}}\times \sin 45°=\frac{2}{\sqrt{13}}$$$$HG=\sqrt{2}k\times \frac{2}{\sqrt{13}}=\frac{2\sqrt{2}k}{\sqrt{13}}$$$$GB=\frac{3k}{\sqrt{2}}-\frac{2\sqrt{2}k}{\sqrt{13}}$$$$\left[CHG\right]=\frac{\frac{2\sqrt{2}k}{\sqrt{13}}}{\frac{3k}{\sqrt{2}}-\frac{2\sqrt{2}k}{\sqrt{13}}}\times 15=\frac{60}{3\sqrt{13}-4}$$$$yellow=15+\frac{60}{3\sqrt{13}-4}-(\frac{20}{\sqrt{13}-2}-\frac{60}{3\sqrt{13}-4})$$$$=15+\frac{2\times 60}{3\sqrt{13}-4}-\frac{20}{\sqrt{13}-2}$$$$=\frac{1220\sqrt{13}+13915}{909}\approx 20.147$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com