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Question Number 208003 by efronzo1 last updated on 02/Jun/24
Answered by A5T last updated on 02/Jun/24
Commented by A5T last updated on 02/Jun/24
PM=((18)/s);Let line through Q parallel to KN meet KL at A. Then,QA=b_h =((50)/s) ((sin45)/(KQ))=((sinθ)/(NQ)); ((cosθ)/(NQ))=((sin45)/(PQ)) ⇒((1/(KQ))/(1/(PQ)))=((PQ)/(KQ))=tanθ=((s−((18)/s))/s)=((s^2 −18)/s^2 ) ⇒((KQ)/(KP))=(s^2 /(s^2 −18+s^2 ))=(b_h /s)⇒b_h =(s^3 /(2s^2 −18)) b_h =((50)/s)⇒(s^3 /(2s^2 −18))=((50)/s)⇒s=3(√(10))⇒s^2 =90
$${PM}=\frac{\mathrm{18}}{{s}};{Let}\:{line}\:{through}\:{Q}\:{parallel}\:{to}\:{KN}\:{meet} \\ $$$${KL}\:{at}\:{A}.\:{Then},{QA}={b}_{{h}} =\frac{\mathrm{50}}{{s}} \\ $$$$\frac{{sin}\mathrm{45}}{{KQ}}=\frac{{sin}\theta}{{NQ}};\:\frac{{cos}\theta}{{NQ}}=\frac{{sin}\mathrm{45}}{{PQ}} \\ $$$$\Rightarrow\frac{\frac{\mathrm{1}}{{KQ}}}{\frac{\mathrm{1}}{{PQ}}}=\frac{{PQ}}{{KQ}}={tan}\theta=\frac{{s}−\frac{\mathrm{18}}{{s}}}{{s}}=\frac{{s}^{\mathrm{2}} −\mathrm{18}}{{s}^{\mathrm{2}} } \\ $$$$\Rightarrow\frac{{KQ}}{{KP}}=\frac{{s}^{\mathrm{2}} }{{s}^{\mathrm{2}} −\mathrm{18}+{s}^{\mathrm{2}} }=\frac{{b}_{{h}} }{{s}}\Rightarrow{b}_{{h}} =\frac{{s}^{\mathrm{3}} }{\mathrm{2}{s}^{\mathrm{2}} −\mathrm{18}} \\ $$$${b}_{{h}} =\frac{\mathrm{50}}{{s}}\Rightarrow\frac{{s}^{\mathrm{3}} }{\mathrm{2}{s}^{\mathrm{2}} −\mathrm{18}}=\frac{\mathrm{50}}{{s}}\Rightarrow{s}=\mathrm{3}\sqrt{\mathrm{10}}\Rightarrow{s}^{\mathrm{2}} =\mathrm{90} \\ $$
Answered by mr W last updated on 02/Jun/24
Commented by mr W last updated on 02/Jun/24
say S=area of square 25+B=A+B+9 (=(S/2)) ⇒A=25−9=16 (A/(25))=(((NP)/(KL)))^2 =(((NM−PM)/(NM)))^2 =(1−((PM)/(NM)))^2 (A/(25))=(1−((2×9)/S))^2 ((16)/(25))=(1−((2×9)/S))^2 ⇒S=90 ✓
$${say}\:{S}={area}\:{of}\:{square} \\ $$$$\mathrm{25}+{B}={A}+{B}+\mathrm{9}\:\left(=\frac{{S}}{\mathrm{2}}\right) \\ $$$$\Rightarrow{A}=\mathrm{25}−\mathrm{9}=\mathrm{16} \\ $$$$\frac{{A}}{\mathrm{25}}=\left(\frac{{NP}}{{KL}}\right)^{\mathrm{2}} =\left(\frac{{NM}−{PM}}{{NM}}\right)^{\mathrm{2}} =\left(\mathrm{1}−\frac{{PM}}{{NM}}\right)^{\mathrm{2}} \\ $$$$\frac{{A}}{\mathrm{25}}=\left(\mathrm{1}−\frac{\mathrm{2}×\mathrm{9}}{{S}}\right)^{\mathrm{2}} \\ $$$$\frac{\mathrm{16}}{\mathrm{25}}=\left(\mathrm{1}−\frac{\mathrm{2}×\mathrm{9}}{{S}}\right)^{\mathrm{2}} \\ $$$$\Rightarrow{S}=\mathrm{90}\:\checkmark \\ $$
Commented by Tawa11 last updated on 21/Jun/24
Weldone sirs
$$\mathrm{Weldone}\:\mathrm{sirs} \\ $$
Answered by A5T last updated on 02/Jun/24
Commented by A5T last updated on 02/Jun/24
ML=s⇒PM=((18)/s)⇒PN=s−((18)/s) 25+B=A+B+9⇒A=16 ⇒(1/2)(s−((18)/s))(s−((50)/s))=16⇒s=3(√(10))⇒s^2 =90
$${ML}={s}\Rightarrow{PM}=\frac{\mathrm{18}}{{s}}\Rightarrow{PN}={s}−\frac{\mathrm{18}}{{s}} \\ $$$$\mathrm{25}+{B}={A}+{B}+\mathrm{9}\Rightarrow{A}=\mathrm{16} \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{2}}\left({s}−\frac{\mathrm{18}}{{s}}\right)\left({s}−\frac{\mathrm{50}}{{s}}\right)=\mathrm{16}\Rightarrow{s}=\mathrm{3}\sqrt{\mathrm{10}}\Rightarrow{s}^{\mathrm{2}} =\mathrm{90} \\ $$
Answered by efronzo1 last updated on 02/Jun/24

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