Question Number 198176 by Blackpanther last updated on 12/Oct/23
Answered by som(math1967) last updated on 13/Oct/23
![let side of ■PQRB=x unit side of ■ EFGH =y unit AB=AP+PB =xtan30 +x [ ∵ ((AP)/x)=tan30] again AB=AH+HB =ycosec 60+ysin30 [∵ ((AH)/y)=cosec 60,((HB)/y)=sin30] ∴x(tan30+1)=y(cosec 60+sin 30) ⇒(x/y)=(((4+(√3))/( 2(√3)))/((1+(√3))/( (√3)))) ⇒(x/y)=((4+(√3))/(2(1+(√3)))) ⇒(x^2 /y^2 )=((19+8(√3))/(4(4+2(√3))))=((19+8(√3))/(16+8(√3)))](Q198177.png)
$${let}\:{side}\:{of}\:\blacksquare{PQRB}={x}\:{unit} \\ $$$${side}\:{of}\:\blacksquare\:{EFGH}\:={y}\:{unit} \\ $$$$\:{AB}={AP}+{PB} \\ $$$$\:\:\:\:\:={xtan}\mathrm{30}\:+{x}\:\left[\:\because\:\frac{{AP}}{{x}}={tan}\mathrm{30}\right] \\ $$$$\:{again}\:{AB}={AH}+{HB} \\ $$$$\:\:\:\:\:\:\:={y}\mathrm{cosec}\:\mathrm{60}+{y}\mathrm{sin30} \\ $$$$\left[\because\:\frac{{AH}}{{y}}=\mathrm{cosec}\:\mathrm{60},\frac{{HB}}{{y}}={sin}\mathrm{30}\right] \\ $$$$\therefore{x}\left({tan}\mathrm{30}+\mathrm{1}\right)={y}\left(\mathrm{cosec}\:\mathrm{60}+\mathrm{sin}\:\mathrm{30}\right) \\ $$$$\Rightarrow\frac{{x}}{{y}}=\frac{\frac{\mathrm{4}+\sqrt{\mathrm{3}}}{\:\mathrm{2}\sqrt{\mathrm{3}}}}{\frac{\mathrm{1}+\sqrt{\mathrm{3}}}{\:\sqrt{\mathrm{3}}}} \\ $$$$\Rightarrow\frac{{x}}{{y}}=\frac{\mathrm{4}+\sqrt{\mathrm{3}}}{\mathrm{2}\left(\mathrm{1}+\sqrt{\mathrm{3}}\right)} \\ $$$$\Rightarrow\frac{{x}^{\mathrm{2}} }{{y}^{\mathrm{2}} }=\frac{\mathrm{19}+\mathrm{8}\sqrt{\mathrm{3}}}{\mathrm{4}\left(\mathrm{4}+\mathrm{2}\sqrt{\mathrm{3}}\right)}=\frac{\mathrm{19}+\mathrm{8}\sqrt{\mathrm{3}}}{\mathrm{16}+\mathrm{8}\sqrt{\mathrm{3}}} \\ $$
Commented by Blackpanther last updated on 13/Oct/23
$${Thanks}\:{but}\:{the}\:{given}\:{answer}\:{was}\:{choice} \\ $$$${C}. \\ $$
Commented by som(math1967) last updated on 13/Oct/23
$$\boldsymbol{{S}}{orry}\:{typo}\: \\ $$