Question-219921

Question Number 219921 by Spillover last updated on 03/May/25

Answered by mr W last updated on 04/May/25

Commented by mr W last updated on 04/May/25

cos α=((a^2 +((a/( (√2))))^2 −((a/2))^2 )/(2a×(a/( (√2)))))=((5(√2))/8) CL=a cos α=((5(√2)a)/8) OL=((5(√2)a)/8)−(a/( (√2)))=(((√2)a)/8) OH=(√2) OL=(a/4) HK=(a/2)−(a/4)=(a/4)=KG x=(a/2)−(a/4)=(a/4) ✓

$$\mathrm{cos}\:\alpha=\frac{{a}^{\mathrm{2}} +\left(\frac{{a}}{\:\sqrt{\mathrm{2}}}\right)^{\mathrm{2}} −\left(\frac{{a}}{\mathrm{2}}\right)^{\mathrm{2}} }{\mathrm{2}{a}×\frac{{a}}{\:\sqrt{\mathrm{2}}}}=\frac{\mathrm{5}\sqrt{\mathrm{2}}}{\mathrm{8}} \\ $$$${CL}={a}\:\mathrm{cos}\:\alpha=\frac{\mathrm{5}\sqrt{\mathrm{2}}{a}}{\mathrm{8}} \\ $$$${OL}=\frac{\mathrm{5}\sqrt{\mathrm{2}}{a}}{\mathrm{8}}−\frac{{a}}{\:\sqrt{\mathrm{2}}}=\frac{\sqrt{\mathrm{2}}{a}}{\mathrm{8}} \\ $$$${OH}=\sqrt{\mathrm{2}}\:{OL}=\frac{{a}}{\mathrm{4}} \\ $$$${HK}=\frac{{a}}{\mathrm{2}}−\frac{{a}}{\mathrm{4}}=\frac{{a}}{\mathrm{4}}={KG} \\ $$$${x}=\frac{{a}}{\mathrm{2}}−\frac{{a}}{\mathrm{4}}=\frac{{a}}{\mathrm{4}}\:\checkmark \\ $$

Answered by Spillover last updated on 04/May/25