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Category: Geometry

Question-212668

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Question Number 212668 by Spillover last updated on 20/Oct/24 Answered by efronzo1 last updated on 21/Oct/24 BC2=64+25−40=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}} \…

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

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Question Number 212533 by emilagazade last updated on 16/Oct/24 givenisosceletrianglewithsides10andinradius3.howfindbase? Answered by A5T last updated on 16/Oct/24 Letbase=b;anglebetweennon−equalsides=θ[△]=10×10sin(180−2θ)2=b×10×sinθ2$$\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}}…

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