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Question Number 157379 by quvonch3737 last updated on 22/Oct/21

Commented by quvonch3737 last updated on 22/Oct/21

$$provehelp$$

Answered by mr W last updated on 22/Oct/21

$$R=\frac{3+4}{\sqrt{2}}=\frac{7}{\sqrt{2}}$$$$\tan \alpha =\frac{4-3.5}{3.5}=\frac{1}{7}\Rightarrow \cos \alpha =\frac{7}{\sqrt{50}}$$$$A_{yellow}=\frac{R^2}{2}[\sin (45+\alpha )+\sin (45-\alpha )]-\frac{R^2}{2}$$$$A_{yellow}=\frac{R^2}{2}(\sqrt{2}\cos \alpha -1)$$$$A_{yellow}=\frac{R^2}{2}(\sqrt{2}\times \frac{7}{\sqrt{50}}-1)$$$$A_{yellow}=\frac{R^2}{5}$$$$i.e.shadedareais20\mathrm{\%}ofthesquare.$$$$A_{yellow}=\frac{1}{5}\times {\left(\frac{7}{\sqrt{2}}\right)}^2=\frac{49}{10}$$

Commented by mr W last updated on 22/Oct/21

Commented by Tawa11 last updated on 23/Oct/21

$$\mathrm{Great}\mathrm{sirs}$$

Answered by ajfour last updated on 22/Oct/21

Commented by ajfour last updated on 22/Oct/21

$$letscallsquaresides.$$$$diagonals\sqrt{2}=3+4$$$$\Rightarrow s=\frac{7}{\sqrt{2}}$$$$D[\frac{3s}{7},\frac{4s}{7}]$$$$slopeofAD$$$$\tan \theta =\frac{s-\frac{4s}{7}}{s-\frac{3s}{7}}=\frac{3}{4}$$$$A[s(1-\cos \theta ),s(1-\sin \theta )]$$$$A[\frac{s}{5},\frac{2s}{5}]$$$$Area△ABC$$$$=\frac{s^2}{2}-\frac{s}{2}(x_A+y_A)$$$$=\frac{s^2}{5}★$$$$$$

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