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Question Number 156509 by ARUNG_Brandon_MBU last updated on 12/Oct/21

Commented by mr W last updated on 12/Oct/21

$$bothsquaresareofequalsize?$$

Commented by ARUNG_Brandon_MBU last updated on 12/Oct/21

$$\mathrm{I}{\mathrm{don}}^{\prime }\mathrm{t}\mathrm{really}\mathrm{know}\mathrm{Sir}.\mathrm{I}\mathrm{think}\mathrm{so}\mathrm{too}$$

Commented by mr W last updated on 12/Oct/21

$$foruniquesolutionthesizeofboth$$$$squaresmustbegiven.$$

Commented by ARUNG_Brandon_MBU last updated on 12/Oct/21

$$\mathrm{OK}\mathrm{Sir}.\mathrm{But}\mathrm{the}\mathrm{question}\mathrm{is}\mathbf{How}\mathbf{big}\mathbf{is}\mathbf{y}?$$

Commented by mr W last updated on 12/Oct/21

$$thequestionishowbigcanybe.$$$$ifthesecondsquareisthesameas$$$$thefirstone,then1⩽y⩽1.6006.$$

Answered by mr W last updated on 12/Oct/21

Commented by mr W last updated on 12/Oct/21

$$CB=a\cos \alpha$$$$CD=a\sin \alpha$$$$x_F=1+a\cos \alpha +\sqrt{2}a\cos (45+90-\alpha )$$$$x_F=1+a\cos \alpha +\sqrt{2}a\sin (\alpha -45)$$$$x_F=1+a\cos \alpha +a(\sin \alpha -\cos \alpha )$$$$x_F=1+a\sin \alpha$$$$y_F=\sqrt{2}a\sin (45+90-\alpha )$$$$y_F=\sqrt{2}a\cos (\alpha -45)$$$$y_F=a(\cos \alpha +\sin \alpha )$$$$\frac{y-1}{1+a(\cos \alpha +\sin \alpha )}=\frac{a(\cos \alpha +\sin \beta )-1}{1+a\sin \alpha }$$$$fora=1:$$$$y=1+\frac{\sin 2\alpha }{1+\sin \alpha }$$$$y_{max}\approx 1.6006$$

Commented by ARUNG_Brandon_MBU last updated on 12/Oct/21

$$\mathrm{Thank}\mathrm{you}\mathrm{Sir}$$

Commented by Tawa11 last updated on 12/Oct/21

$$\mathrm{great}\mathrm{sir}.$$

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