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Question Number 177353 by HeferH last updated on 04/Oct/22

Commented by HeferH last updated on 04/Oct/22

$$:)$$

Answered by BaliramKumar last updated on 04/Oct/22

$$\frac{3}{13}???$$

Commented by mr W last updated on 04/Oct/22

$$yes!buthowdidyougetthat?$$

Commented by BaliramKumar last updated on 04/Oct/22

$$similartriangle\mathrm{\&}cosinerule$$$$seemysolution\downdownarrows$$

Answered by mr W last updated on 04/Oct/22

Commented by mr W last updated on 04/Oct/22

$$R=radius$$$$s=\sqrt{3}R$$$$\sin \alpha =\frac{x}{2R}$$$$\sin \beta =\frac{3x}{2R}$$$$\alpha +\beta =\frac{\pi }{3}$$$$\cos (\alpha +\beta )=\cos \frac{\pi }{3}=\frac{1}{2}$$$$\sqrt{(1-\frac{x^2}{4R^2})(1-\frac{9x^2}{4R^2})}-\frac{x}{2R}\times \frac{3x}{2R}=\frac{1}{2}$$$$\sqrt{(4R^2-x^2)(4R^2-9x^2)}=2R^2+3x^2$$$$16R^4-40R^2{x}^2+9x^4=4R^4+12R^2{x}^2+9x^4$$$$3R^2=13x^2$$$$\Rightarrow x=\sqrt{\frac{3}{13}}R$$$$\sin \beta =\frac{3x}{2R}=3\sqrt{\frac{3}{52}}\Rightarrow \cos \beta =\frac{5}{\sqrt{52}}$$$$\frac{b}{s}=\frac{\sin \frac{\pi }{3}}{\sin (\frac{\pi }{3}+\beta )}$$$$\frac{a+b}{s}=\frac{\sin (\frac{\pi }{3}+\beta )}{\sin \frac{\pi }{3}}$$$$\frac{a+b}{b}={\left[\frac{\sin (\frac{\pi }{3}+\beta )}{\sin \frac{\pi }{3}}\right]}^2={[\cos \beta +\frac{\sin \beta }{\tan \frac{\pi }{3}}]}^2$$$$\frac{a}{b}={[\cos \beta +\frac{\sin \beta }{\tan \frac{\pi }{3}}]}^2-1$$$$={[\frac{5}{\sqrt{52}}+\frac{3}{\sqrt{3}}\sqrt{\frac{3}{52}}]}^2-1$$$$={\left(\frac{8}{\sqrt{52}}\right)}^2-1=\frac{3}{13}✓$$

Commented by Tawa11 last updated on 04/Oct/22

$$\mathrm{Great}\mathrm{sir}$$

Answered by BaliramKumar last updated on 04/Oct/22

Commented by BaliramKumar last updated on 04/Oct/22

Commented by mr W last updated on 04/Oct/22

$$niceapproach!$$

Commented by BaliramKumar last updated on 04/Oct/22

$$thankssir$$

Commented by HeferH last updated on 04/Oct/22

$$Looksnice,IdiditbypureSimilar$$$$triangles$$

Answered by HeferH last updated on 04/Oct/22

Commented by HeferH last updated on 04/Oct/22

$$\bullet ab=c(d-c)$$$$\bullet \frac{3x}{d}=\frac{c}{b}\Rightarrow 3xb=dc(\mathbf{÷})$$$$\bullet \frac{x}{d}=\frac{a}{c}\Rightarrow xc=ad$$$$ab=\frac{c^2}{3}$$$$\frac{c^2}{3}=c(d-c)$$$$d=\frac{4c}{3}$$$$x=\frac{ad}{c}\Rightarrow x=\frac{4a}{3}$$$$\bullet \frac{a+b}{x}=\frac{4a}{a}$$$$(a+b)\cdot \frac{1}{a}=\frac{16a}{3}\cdot \frac{1}{a}$$$$1+\frac{b}{a}=\frac{16}{3}$$$$\frac{b}{a}=\frac{16-3}{3}=\frac{13}{3}\Rightarrow \frac{a}{b}=\frac{3}{13}$$$$$$

Commented by Tawa11 last updated on 04/Oct/22

$$\mathrm{Great}\mathrm{sir}$$

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