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Question Number 74747 by mr W last updated on 30/Nov/19

Answered by mind is power last updated on 30/Nov/19

$$\mathrm{let}\mathrm{p},\mathrm{q},\mathrm{r}\mathrm{side}\mathrm{of}\mathrm{red}\mathrm{squar},\mathrm{a},\mathrm{b},\mathrm{c}\mathrm{blue}$$$$\alpha ,\beta ,\varsigma \mathrm{angle}\mathrm{thetriangl}\mathrm{inside}\mathrm{red}\mathrm{squar}$$$${\mathrm{a}}^2={\mathrm{p}}^2+{\mathrm{q}}^2+2\mathrm{pqcos}\left(\alpha \right)$$$${\mathrm{b}}^2={\mathrm{p}}^2+{\mathrm{r}}^2+2\mathrm{prcos}\left(\beta \right)$$$${\mathrm{c}}^2={\mathrm{r}}^2+{\mathrm{q}}^2+2\mathrm{rqcos}\left(\varsigma \right)$$$$\cos \left(\alpha \right)=\frac{{\mathrm{p}}^2+{\mathrm{q}}^2-{\mathrm{r}}^2}{2\mathrm{pq}},\cos \left(\beta \right)=\frac{{\mathrm{p}}^2+{\mathrm{r}}^2-{\mathrm{q}}^2}{2\mathrm{pr}}$$$$\cos \left(\varsigma \right)=\frac{{\mathrm{q}}^2+{\mathrm{r}}^2-{\mathrm{p}}^2}{2\mathrm{qr}}$$$$\Rightarrow {\mathrm{a}}^2={2\mathrm{p}}^2+{2\mathrm{q}}^2-{\mathrm{r}}^2,{\mathrm{b}}^2={2\mathrm{p}}^2+{2\mathrm{r}}^2-{\mathrm{q}}^2,{\mathrm{c}}^2={2\mathrm{r}}^2+{2\mathrm{q}}^2-{\mathrm{p}}^2$$$$\Rightarrow {\mathrm{a}}^2+{\mathrm{b}}^2+{\mathrm{c}}^2=3({\mathrm{p}}^2+{\mathrm{q}}^2+{\mathrm{r}}^2)$$$$\Rightarrow \frac{{\mathrm{a}}^2+{\mathrm{b}}^2+{\mathrm{c}}^2}{{\mathrm{p}}^2+{\mathrm{q}}^2+{\mathrm{r}}^2}=3$$$$$$

Commented by mr W last updated on 30/Nov/19

$$thankssir!$$$$igotthesameresultinQ74726.$$

Commented by mind is power last updated on 30/Nov/19

$${\mathrm{y}}^{\prime }\mathrm{re}\mathrm{welcom}$$

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