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Question Number 67422 by mr W last updated on 27/Aug/19

Commented by Prithwish sen last updated on 27/Aug/19

$$△\mathrm{AMN}\mathrm{is}\mathrm{an}\mathrm{isoceles}\mathrm{triangle}\mathrm{with}\mathrm{AM}=\mathrm{AN}$$$$\mathrm{AM}+\mathrm{BM}=\mathrm{AN}+{\mathrm{D}}^{{}^{\prime }}\mathrm{N}=16$$$$\mathrm{Area}{\mathrm{ABMND}}^{{}^{\prime }}=\mathrm{Area}\mathrm{of}\mathrm{ABCD}-\mathrm{Area}\mathrm{of}△\mathrm{AMN}.$$

Commented by mr W last updated on 27/Aug/19

$$apiecepapaerABCDisfoldedalong$$$$MNsuchthatpointCcoincideswith$$$$pointAandapentagon{ABMND}^{\prime }is$$$$formed.$$$$findtheareaofthepentagon.$$

Answered by mr W last updated on 27/Aug/19

$$letMC=MA=x$$$$BM=16-x$$$$x^2=4^2+{(16-x)}^2$$$$2x=17$$$$x=\frac{17}{2}$$$$AN=AM=\frac{17}{2}$$$${\mathrm{\Delta }}_{AMN}=\frac{1}{2}\times \frac{17}{2}\times 4=17$$$${ABMND}^{\prime }=16\times 4-17=47$$

Commented by Prithwish sen last updated on 28/Aug/19

$$\mathrm{thank}\mathrm{you}\mathrm{sir}.$$

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