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 178947 by HeferH last updated on 23/Oct/22

Answered by mr W last updated on 23/Oct/22

Commented by mr W last updated on 23/Oct/22

((sin β)/(4k))=((sin ((π/4)+β))/(GB)) ((sin α)/(3k))=((sin ((π/4)+α))/(GB)) ((3 sin β)/(4 sin α))=((sin ((π/4)+β))/(sin ((π/4)+α)))=((cos β+sin β)/(cos α+sin α)) α+β=(π/2) ((3 cos α)/(4 sin α))=1 ⇒tan α=(3/4) ((4k)/(sin β))=((GB)/(sin ((π/4)+β))) ⇒GB=((4k (cos β+sin β))/( (√2) sin β))=((7(√2)k)/( 2)) GH=GB−HB=((7(√2)k)/2)−((4(√2)k)/2)=((3(√2)k)/2) HE=GH tan α=((3(√2)k)/2)×(3/4)=((9(√2)k)/8) magenta area=((GB×HE)/2) =(1/2)×((7(√2)k)/( 2))×((9(√2)k)/8)=((63k^2 )/(16)) area of square =(4k)^2 =16k^2 yellow area=16k^2 −((63k^2 )/(16))=((193k^2 )/(16)) ((magenta)/(yellow))=((63)/(196)) ✓

$$\frac{\mathrm{sin}\:\beta}{\mathrm{4}{k}}=\frac{\mathrm{sin}\:\left(\frac{\pi}{\mathrm{4}}+\beta\right)}{{GB}} \\ $$$$\frac{\mathrm{sin}\:\alpha}{\mathrm{3}{k}}=\frac{\mathrm{sin}\:\left(\frac{\pi}{\mathrm{4}}+\alpha\right)}{{GB}} \\ $$$$\frac{\mathrm{3}\:\mathrm{sin}\:\beta}{\mathrm{4}\:\mathrm{sin}\:\alpha}=\frac{\mathrm{sin}\:\left(\frac{\pi}{\mathrm{4}}+\beta\right)}{\mathrm{sin}\:\left(\frac{\pi}{\mathrm{4}}+\alpha\right)}=\frac{\mathrm{cos}\:\beta+\mathrm{sin}\:\beta}{\mathrm{cos}\:\alpha+\mathrm{sin}\:\alpha} \\ $$$$\alpha+\beta=\frac{\pi}{\mathrm{2}} \\ $$$$\frac{\mathrm{3}\:\mathrm{cos}\:\alpha}{\mathrm{4}\:\mathrm{sin}\:\alpha}=\mathrm{1} \\ $$$$\Rightarrow\mathrm{tan}\:\alpha=\frac{\mathrm{3}}{\mathrm{4}} \\ $$$$\frac{\mathrm{4}{k}}{\mathrm{sin}\:\beta}=\frac{{GB}}{\mathrm{sin}\:\left(\frac{\pi}{\mathrm{4}}+\beta\right)} \\ $$$$\Rightarrow{GB}=\frac{\mathrm{4}{k}\:\left(\mathrm{cos}\:\beta+\mathrm{sin}\:\beta\right)}{\:\sqrt{\mathrm{2}}\:\mathrm{sin}\:\beta}=\frac{\mathrm{7}\sqrt{\mathrm{2}}{k}}{\:\mathrm{2}} \\ $$$${GH}={GB}−{HB}=\frac{\mathrm{7}\sqrt{\mathrm{2}}{k}}{\mathrm{2}}−\frac{\mathrm{4}\sqrt{\mathrm{2}}{k}}{\mathrm{2}}=\frac{\mathrm{3}\sqrt{\mathrm{2}}{k}}{\mathrm{2}} \\ $$$${HE}={GH}\:\mathrm{tan}\:\alpha=\frac{\mathrm{3}\sqrt{\mathrm{2}}{k}}{\mathrm{2}}×\frac{\mathrm{3}}{\mathrm{4}}=\frac{\mathrm{9}\sqrt{\mathrm{2}}{k}}{\mathrm{8}} \\ $$$${magenta}\:{area}=\frac{{GB}×{HE}}{\mathrm{2}} \\ $$$$\:\:\:\:=\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{7}\sqrt{\mathrm{2}}{k}}{\:\mathrm{2}}×\frac{\mathrm{9}\sqrt{\mathrm{2}}{k}}{\mathrm{8}}=\frac{\mathrm{63}{k}^{\mathrm{2}} }{\mathrm{16}} \\ $$$${area}\:{of}\:{square}\:=\left(\mathrm{4}{k}\right)^{\mathrm{2}} =\mathrm{16}{k}^{\mathrm{2}} \\ $$$${yellow}\:{area}=\mathrm{16}{k}^{\mathrm{2}} −\frac{\mathrm{63}{k}^{\mathrm{2}} }{\mathrm{16}}=\frac{\mathrm{193}{k}^{\mathrm{2}} }{\mathrm{16}} \\ $$$$\frac{{magenta}}{{yellow}}=\frac{\mathrm{63}}{\mathrm{196}}\:\checkmark \\ $$

Commented by Tawa11 last updated on 23/Oct/22

Great sir.

$$\mathrm{Great}\:\mathrm{sir}. \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com