Question-219232

Question Number 219232 by Spillover last updated on 20/Apr/25

Answered by A5T last updated on 21/Apr/25

Commented by A5T last updated on 21/Apr/25

AD=bsinθ ; AE=bcosθ ; AC=asinθ ; AB=acosθ In △BDC,[BCD]= ((BD×BC×CD)/(4R))=((BD×BCsinθ)/2) ⇒CD=2Rsinθ ⇒(√((acosθ−bsinθ)^2 +a^2 −2a(acosθ−bsinθ)cosθ))=2Rsinθ ⇒(√((a^2 +b^2 )sin^2 θ))=2Rsinθ ⇒a^2 +b^2 =4R^2

$$\mathrm{AD}=\mathrm{bsin}\theta\:;\:\mathrm{AE}=\mathrm{bcos}\theta\:;\:\mathrm{AC}=\mathrm{asin}\theta\:;\:\mathrm{AB}=\mathrm{acos}\theta \\ $$$$\mathrm{In}\:\bigtriangleup\mathrm{BDC},\left[\mathrm{BCD}\right]=\:\frac{\mathrm{BD}×\mathrm{BC}×\mathrm{CD}}{\mathrm{4R}}=\frac{\mathrm{BD}×\mathrm{BCsin}\theta}{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{CD}=\mathrm{2Rsin}\theta \\ $$$$\Rightarrow\sqrt{\left(\mathrm{acos}\theta−\mathrm{bsin}\theta\right)^{\mathrm{2}} +\mathrm{a}^{\mathrm{2}} −\mathrm{2a}\left(\mathrm{acos}\theta−\mathrm{bsin}\theta\right)\mathrm{cos}\theta}=\mathrm{2Rsin}\theta \\ $$$$\Rightarrow\sqrt{\left(\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} \right)\mathrm{sin}^{\mathrm{2}} \theta}=\mathrm{2Rsin}\theta \\ $$$$\Rightarrow\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} =\mathrm{4R}^{\mathrm{2}} \\ $$

Commented by Spillover last updated on 21/Apr/25

$${thank}\:{you}\:{AST} \\ $$

Answered by mr W last updated on 21/Apr/25

Commented by mr W last updated on 21/Apr/25

make FD//BA FB=DE=b ∠FBC=∠FDC=∠BAC=90° ⇒FC=diameter=2R ⇒a^2 +b^2 =(2R)^2 =4R^2 ✓

$${make}\:{FD}//{BA} \\ $$$${FB}={DE}={b} \\ $$$$\angle{FBC}=\angle{FDC}=\angle{BAC}=\mathrm{90}° \\ $$$$\Rightarrow{FC}={diameter}=\mathrm{2}{R} \\ $$$$\Rightarrow{a}^{\mathrm{2}} +{b}^{\mathrm{2}} =\left(\mathrm{2}{R}\right)^{\mathrm{2}} =\mathrm{4}{R}^{\mathrm{2}} \:\checkmark \\ $$

Commented by Spillover last updated on 21/Apr/25