Question Number 223585 by hardmath last updated on 30/Jul/25
Commented by hardmath last updated on 30/Jul/25
$$\mathrm{Radius}\:\mathrm{of}\:\mathrm{the}\:\mathrm{circle}\:=\:\mathrm{R} \\ $$$$\mathrm{AB}\:\bot\:\mathrm{CD} \\ $$$$\mathrm{AC}\:=\:\mathrm{a} \\ $$$$\mathrm{BD}\:=\:\mathrm{b} \\ $$$$\mathrm{Prove}\:\mathrm{that}:\:\mathrm{R}\:=\:\frac{\sqrt{\mathrm{a}^{\mathrm{2}} \:+\:\mathrm{b}^{\mathrm{2}} }}{\mathrm{2}} \\ $$
Answered by mr W last updated on 31/Jul/25
Commented by mr W last updated on 31/Jul/25
$${set}\:{ED}//{AB} \\ $$$$\Rightarrow{ED}\bot{CD},\:{AE}={BD}={b},\: \\ $$$$\:\:\:\:\:{CE}={diameter}=\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} } \\ $$$$\Rightarrow{radius}=\frac{\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }}{\mathrm{2}} \\ $$
Commented by hardmath last updated on 31/Jul/25
$$\mathrm{thankyou}\:\mathrm{dear}\:\mathrm{professor}\:\mathrm{cool} \\ $$