Question-219337

Question Number 219337 by Spillover last updated on 23/Apr/25

Answered by mr W last updated on 23/Apr/25

Commented by mr W last updated on 23/Apr/25

a+s=4 tan 60°=4(√3) ((s/(2(√3))))^2 +(a+(s/2))^2 =R^2 =(4+(s/(2(√3))))^2 +((s/2))^2 ((s/(2(√3))))^2 +(4(√3)−(s/2))^2 =(4+(s/(2(√3))))^2 +((s/2))^2 ⇒s=2(√3) ⇒R^2 =(4+((2(√3))/(2(√2))))^2 +(((2(√3))/2))^2 =28 ⇒R=2(√7)

$${a}+{s}=\mathrm{4}\:\mathrm{tan}\:\mathrm{60}°=\mathrm{4}\sqrt{\mathrm{3}} \\ $$$$\left(\frac{{s}}{\mathrm{2}\sqrt{\mathrm{3}}}\right)^{\mathrm{2}} +\left({a}+\frac{{s}}{\mathrm{2}}\right)^{\mathrm{2}} ={R}^{\mathrm{2}} =\left(\mathrm{4}+\frac{{s}}{\mathrm{2}\sqrt{\mathrm{3}}}\right)^{\mathrm{2}} +\left(\frac{{s}}{\mathrm{2}}\right)^{\mathrm{2}} \\ $$$$\left(\frac{{s}}{\mathrm{2}\sqrt{\mathrm{3}}}\right)^{\mathrm{2}} +\left(\mathrm{4}\sqrt{\mathrm{3}}−\frac{{s}}{\mathrm{2}}\right)^{\mathrm{2}} =\left(\mathrm{4}+\frac{{s}}{\mathrm{2}\sqrt{\mathrm{3}}}\right)^{\mathrm{2}} +\left(\frac{{s}}{\mathrm{2}}\right)^{\mathrm{2}} \\ $$$$\Rightarrow{s}=\mathrm{2}\sqrt{\mathrm{3}} \\ $$$$\Rightarrow{R}^{\mathrm{2}} =\left(\mathrm{4}+\frac{\mathrm{2}\sqrt{\mathrm{3}}}{\mathrm{2}\sqrt{\mathrm{2}}}\right)^{\mathrm{2}} +\left(\frac{\mathrm{2}\sqrt{\mathrm{3}}}{\mathrm{2}}\right)^{\mathrm{2}} =\mathrm{28} \\ $$$$\Rightarrow{R}=\mathrm{2}\sqrt{\mathrm{7}} \\ $$

Answered by Spillover last updated on 24/Apr/25