Category: Geometry

Question Number 32659 by Joel578 last updated on 30/Mar/18 Commented by Joel578 last updated on 30/Mar/18 $$\mathrm{Red}\:\mathrm{circle}\:\mathrm{with}\:\mathrm{radius}\:\mathrm{2}\sqrt{\mathrm{2}}\:\mathrm{cm} \\ $$$$\mathrm{Blue}\:\mathrm{circle}\:\mathrm{with}\:\mathrm{radius}\:\mathrm{4}\:\mathrm{cm} \\ $$$${C}\:\mathrm{is}\:\mathrm{center}\:\mathrm{of}\:\mathrm{blue}\:\mathrm{circle} \\ $$$${AB}\:\mathrm{is}\:\mathrm{diameter}\:\mathrm{of}\:\mathrm{red}\:\mathrm{circle} \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:\mathrm{shaded}\:\mathrm{part}…

Question Number 163656 by mnjuly1970 last updated on 09/Jan/22 Answered by mr W last updated on 09/Jan/22 Commented by mr W last updated on 09/Jan/22…

Question Number 163556 by Ari last updated on 07/Jan/22 Answered by ajfour last updated on 07/Jan/22 $$\frac{{x}}{\mathrm{8}}+\frac{{y}}{\mathrm{15}}=\mathrm{1}\:\:\Rightarrow\:\:\mathrm{15}{x}+\mathrm{8}{y}=\mathrm{120} \\ $$$${r}=\frac{\mathrm{15}{r}+\mathrm{8}{r}−\mathrm{120}}{\:\sqrt{\mathrm{225}+\mathrm{64}}} \\ $$$$\Rightarrow\:\:{r}=\mathrm{20} \\ $$ Answered by…

Question Number 97929 by bobhans last updated on 10/Jun/20 $$\mathrm{Find}\:\mathrm{equation}\:\mathrm{of}\:\mathrm{line}\:\mathrm{through}\:\mathrm{A}\left(\mathrm{1},\mathrm{2},\mathrm{3}\right) \\ $$$$\mathrm{and}\:\mathrm{parallel}\:\mathrm{to}\:\mathrm{y}\:\mathrm{axis}\:?\: \\ $$ Commented by mr W last updated on 10/Jun/20 $$\frac{{x}−\mathrm{1}}{\mathrm{0}}=\frac{{y}−\mathrm{2}}{\mathrm{1}}=\frac{{z}−\mathrm{3}}{\mathrm{0}} \\ $$$${or}…

Question Number 163411 by amin96 last updated on 06/Jan/22 Terms of Service Privacy Policy Contact: info@tinkutara.com