Question Number 218736 by Spillover last updated on 14/Apr/25
Answered by mr W last updated on 15/Apr/25
Commented by mr W last updated on 15/Apr/25
![[4+(√(R^2 −(3+(9/2))^2 ))]^2 +((9/2))^2 =R^2 (√(R^2 −7.5^2 ))=2.5 R^2 =2.5^2 +7.5^2 =((125)/2) ⇒R=((5(√(10)))/2)](Q218751.png)
$$\left[\mathrm{4}+\sqrt{{R}^{\mathrm{2}} −\left(\mathrm{3}+\frac{\mathrm{9}}{\mathrm{2}}\right)^{\mathrm{2}} }\right]^{\mathrm{2}} +\left(\frac{\mathrm{9}}{\mathrm{2}}\right)^{\mathrm{2}} ={R}^{\mathrm{2}} \\ $$$$\sqrt{{R}^{\mathrm{2}} −\mathrm{7}.\mathrm{5}^{\mathrm{2}} }=\mathrm{2}.\mathrm{5} \\ $$$${R}^{\mathrm{2}} =\mathrm{2}.\mathrm{5}^{\mathrm{2}} +\mathrm{7}.\mathrm{5}^{\mathrm{2}} =\frac{\mathrm{125}}{\mathrm{2}} \\ $$$$\Rightarrow{R}=\frac{\mathrm{5}\sqrt{\mathrm{10}}}{\mathrm{2}} \\ $$
Commented by Spillover last updated on 15/Apr/25
$${correct}.{thanks} \\ $$
Answered by mr W last updated on 15/Apr/25
Commented by mr W last updated on 15/Apr/25
$${AC}=\sqrt{\left(\mathrm{9}+\mathrm{3}\right)^{\mathrm{2}} +\mathrm{4}^{\mathrm{2}} }=\mathrm{4}\sqrt{\mathrm{10}} \\ $$$$\mathrm{sin}\:\angle{ABC}=\mathrm{cos}\:\angle{DBC}=\frac{\mathrm{4}}{\mathrm{5}} \\ $$$${R}=\frac{{AC}}{\mathrm{2}\:\mathrm{sin}\:\angle{ABC}}=\frac{\mathrm{4}\sqrt{\mathrm{10}}}{\mathrm{2}×\frac{\mathrm{4}}{\mathrm{5}}}=\frac{\mathrm{5}\sqrt{\mathrm{10}}}{\mathrm{2}}\:\checkmark \\ $$
Answered by Spillover last updated on 15/Apr/25
Answered by Spillover last updated on 15/Apr/25
Answered by Spillover last updated on 15/Apr/25
