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Category: Geometry

Question-212668

Question Number 212668 by Spillover last updated on 20/Oct/24 Answered by efronzo1 last updated on 21/Oct/24 $$\:\:\mathrm{BC}^{\mathrm{2}} =\:\mathrm{64}+\mathrm{25}−\mathrm{40}\:=\:\mathrm{49} \\ $$$$\:\:\mathrm{BC}^{\mathrm{2}} =\:\mathrm{2r}^{\mathrm{2}} +\mathrm{r}^{\mathrm{2}} \Rightarrow\mathrm{r}=\sqrt{\frac{\mathrm{BC}^{\mathrm{2}} }{\mathrm{3}}}=\frac{\mathrm{7}\sqrt{\mathrm{3}}}{\mathrm{3}} \\…

given-isoscele-triangle-with-sides-10-and-inradius-3-how-find-base-

Question Number 212533 by emilagazade last updated on 16/Oct/24 $${given}\:{isoscele}\:{triangle}\:{with}\:{sides}\:\mathrm{10}\:{and}\:{inradius}\:\mathrm{3}.\:{how}\:{find}\:{base}? \\ $$ Answered by A5T last updated on 16/Oct/24 $${Let}\:{base}={b}\:;{angle}\:{between}\:{non}-{equal}\:{sides}=\theta \\ $$$$\left[\bigtriangleup\right]=\frac{\mathrm{10}×\mathrm{10}{sin}\left(\mathrm{180}−\mathrm{2}\theta\right)}{\mathrm{2}}=\frac{{b}×\mathrm{10}×{sin}\theta}{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{50}×\mathrm{2}{cos}\theta=\mathrm{5}{b}\Rightarrow{cos}\theta=\frac{{b}}{\mathrm{20}}\Rightarrow{sin}\theta=\frac{\sqrt{\mathrm{400}−{b}^{\mathrm{2}} }}{\mathrm{20}}…