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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

$$PM=\frac{18}{s};LetlinethroughQparalleltoKNmeet$$$$KLatA.Then,QA=b_h=\frac{50}{s}$$$$\frac{sin45}{KQ}=\frac{sin\theta }{NQ};\frac{cos\theta }{NQ}=\frac{sin45}{PQ}$$$$\Rightarrow \frac{\frac{1}{KQ}}{\frac{1}{PQ}}=\frac{PQ}{KQ}=tan\theta =\frac{s-\frac{18}{s}}{s}=\frac{s^2-18}{s^2}$$$$\Rightarrow \frac{KQ}{KP}=\frac{s^2}{s^2-18+s^2}=\frac{b_h}{s}\Rightarrow b_h=\frac{s^3}{2s^2-18}$$$$b_h=\frac{50}{s}\Rightarrow \frac{s^3}{2s^2-18}=\frac{50}{s}\Rightarrow s=3\sqrt{10}\Rightarrow s^2=90$$

Answered by mr W last updated on 02/Jun/24

Commented by mr W last updated on 02/Jun/24

$$sayS=areaofsquare$$$$25+B=A+B+9(=\frac{S}{2})$$$$\Rightarrow A=25-9=16$$$$\frac{A}{25}={\left(\frac{NP}{KL}\right)}^2={\left(\frac{NM-PM}{NM}\right)}^2={(1-\frac{PM}{NM})}^2$$$$\frac{A}{25}={(1-\frac{2\times 9}{S})}^2$$$$\frac{16}{25}={(1-\frac{2\times 9}{S})}^2$$$$\Rightarrow S=90✓$$

Commented by Tawa11 last updated on 21/Jun/24

$$\mathrm{Weldone}\mathrm{sirs}$$

Answered by A5T last updated on 02/Jun/24

Commented by A5T last updated on 02/Jun/24

$$ML=s\Rightarrow PM=\frac{18}{s}\Rightarrow PN=s-\frac{18}{s}$$$$25+B=A+B+9\Rightarrow A=16$$$$\Rightarrow \frac{1}{2}(s-\frac{18}{s})(s-\frac{50}{s})=16\Rightarrow s=3\sqrt{10}\Rightarrow s^2=90$$

Answered by efronzo1 last updated on 02/Jun/24

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