Question Number 160701 by tounghoungko last updated on 05/Dec/21
Answered by som(math1967) last updated on 05/Dec/21
![ABCD cyclic [∠ADC+∠ABD=180] ∴∠BDC=∠BAC=45 [subtend samesegment] ∠BCD=180−75=105 BD=radius of quater circle from △BCD ((BD)/(Sin105))=((BC)/(Sin45)) BD=((√2)/(1/( (√2))))×sin(180−75) ★ BD=2×sin75=2×(((√3)+1)/(2(√2)))=(((√3)+1)/( (√2))) area of quatercircle =(1/4)×π×((((√3)+1)/( (√2))))^2 =(1/4)×π×(2+(√3))sq units ★BC=(√(4/2))=(√2)](Q160704.png)
$$\boldsymbol{{ABCD}}\:\boldsymbol{{cyclic}}\:\:\left[\angle{ADC}+\angle{ABD}=\mathrm{180}\right] \\ $$$$\therefore\angle{BDC}=\angle{BAC}=\mathrm{45}\:\left[{subtend}\:{samesegment}\right] \\ $$$$\angle{BCD}=\mathrm{180}−\mathrm{75}=\mathrm{105} \\ $$$${BD}={radius}\:{of}\:{quater}\:{circle} \\ $$$${from}\:\bigtriangleup{BCD} \\ $$$$\:\:\frac{{BD}}{{Sin}\mathrm{105}}=\frac{{BC}}{{Sin}\mathrm{45}} \\ $$$$\:{BD}=\frac{\sqrt{\mathrm{2}}}{\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}}×{sin}\left(\mathrm{180}−\mathrm{75}\right)\:\bigstar \\ $$$${BD}=\mathrm{2}×{sin}\mathrm{75}=\mathrm{2}×\frac{\sqrt{\mathrm{3}}+\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}=\frac{\sqrt{\mathrm{3}}+\mathrm{1}}{\:\sqrt{\mathrm{2}}} \\ $$$${area}\:{of}\:{quatercircle} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}×\pi×\left(\frac{\sqrt{\mathrm{3}}+\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right)^{\mathrm{2}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}×\pi×\left(\mathrm{2}+\sqrt{\mathrm{3}}\right){sq}\:{units} \\ $$$$\bigstar{BC}=\sqrt{\frac{\mathrm{4}}{\mathrm{2}}}=\sqrt{\mathrm{2}} \\ $$
Commented by som(math1967) last updated on 05/Dec/21
Commented by mr W last updated on 05/Dec/21
$${or} \\ $$$${BD}×{AC}={AB}×{CD}+{BC}×{AD} \\ $$$${BD}×\mathrm{2}=\sqrt{\mathrm{2}}×\mathrm{1}+\sqrt{\mathrm{2}}×\sqrt{\mathrm{3}} \\ $$$${BD}=\frac{\mathrm{1}+\sqrt{\mathrm{3}}}{\:\sqrt{\mathrm{2}}} \\ $$
Commented by Tawa11 last updated on 05/Dec/21
$$\mathrm{Great}\:\mathrm{sirs} \\ $$