one-vertex-of-a-equilateral-triangle-lies-on-one-vertex-of-a-square-and-two-anothers-lie-on-opposite-sides-of-square-such-that-triangle-have-the-maximum-area-find-1-ratio-of-square-

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Question Number 49731 by behi83417@gmail.com last updated on 09/Dec/18

$$\mathbf{one}\mathbf{vertex}\mathbf{of}\mathbf{a}\mathbf{equilateral}\mathbf{triangle}\mathbf{lies}$$$$\mathbf{on}\mathbf{one}\mathbf{vertex}\mathbf{of}\mathbf{a}\mathbf{square}\mathbf{and}\mathbf{two}$$$$\mathbf{anothers}\mathbf{lie}\mathbf{on}\mathbf{opposite}\mathbf{sides}\mathbf{of}\mathbf{square}$$$$\mathbf{such}\mathbf{that}\mathbf{triangle}\mathbf{have}\mathbf{the}\mathbf{maximum}$$$$\mathbf{area}.$$$$\mathbf{find}:$$$$1.\mathbf{ratio}\mathbf{of}:\frac{\mathbf{square}\mathbf{side}}{\mathbf{triangle}\mathbf{side}}$$$$2.\mathbf{angle}\mathbf{between}\mathbf{square}\mathbf{side}\mathbf{and}\mathbf{triangle}$$$$\mathbf{side}.[\mathbf{need}\mathbf{additional}\mathbf{data}?]$$

Answered by mr W last updated on 10/Dec/18

$$thereisonlyonesuchequilateral$$$$triangle.$$$$$$$$2)let\theta =anglebetweensquareside$$$$andtriangleside,$$$$2\theta +60°=90°$$$$\Rightarrow \theta =15°$$$$$$$$1)$$$$\frac{squareside}{triangleside}=\cos \theta =\cos 15°=\sqrt{\frac{1+\cos 30°}{2}}=\frac{\sqrt{2}+\sqrt{6}}{4}\approx 0.966$$

Commented by behi83417@gmail.com last updated on 10/Dec/18

$$thanksinadvancedearmaster.$$$$whatisyourideaiftrianglebe$$$$isoscale?$$

Commented by mr W last updated on 10/Dec/18

$$ifthetriangleshouldonlybeisosceles,the$$$$largesttriangleisobviouslyhalfof$$$$thesquare.$$

Commented by behi83417@gmail.com last updated on 10/Dec/18

$$thanksdearmaster.$$

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