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Category: Geometry

Question-185324

Question Number 185324 by cherokeesay last updated on 20/Jan/23 Answered by som(math1967) last updated on 20/Jan/23 $${AC}=\sqrt{\mathrm{2}^{\mathrm{2}} +\mathrm{1}^{\mathrm{2}} }=\sqrt{\mathrm{5}}={BC} \\ $$$$\:{r}×\frac{\mathrm{2}+\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{2}×\mathrm{1} \\ $$$$\Rightarrow{r}=\frac{\mathrm{2}}{\mathrm{3}+\sqrt{\mathrm{5}}} \\ $$$$\:{R}×\frac{\sqrt{\mathrm{5}}+\sqrt{\mathrm{5}}+\mathrm{2}}{\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{2}×\mathrm{2}…

Question-54206

Question Number 54206 by ajfour last updated on 31/Jan/19 Commented by ajfour last updated on 31/Jan/19 $${Find}\:{the}\:{area}\:{of}\:{trapezium}\:{ABCD} \\ $$$${in}\:{terms}\:{of}\:\boldsymbol{{a}}.\:\left({the}\:{red}\:{circles}\:{are}\:{each}\right. \\ $$$${within}\:{a}\:{half}\:{quarter}\:{circle}\:{and}\:{the} \\ $$$$\left.{outer}\:{boundary}\:{is}\:{a}\:{square}\:{of}\:{side}\:\boldsymbol{{a}}\right). \\ $$…