Category: Mensuration

Question Number 125593 by shaker last updated on 12/Dec/20 Answered by MJS_new last updated on 12/Dec/20 $$\int\frac{{dx}}{\left({x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{2}\right)^{\mathrm{2}} }= \\ $$$$\:\:\:\:\:\left[\mathrm{Ostrogradski}'\mathrm{s}\:\mathrm{Method}\:\left({search}\:{the}\:{www}\:{for}\:{it}\right)\right] \\ $$$$=\frac{{x}+\mathrm{1}}{\mathrm{2}\left({x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{2}\right)}+\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{dx}}{{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{2}}=…

Question Number 191076 by Mingma last updated on 17/Apr/23 Answered by mr W last updated on 18/Apr/23 $${touching}\:{point}\:{P}\left({p},\frac{\mathrm{1}}{{p}^{\mathrm{2}} }\right) \\ $$$$\mathrm{tan}\:\theta=−\frac{{dy}}{{dx}}=\frac{\mathrm{2}}{{p}^{\mathrm{3}} } \\ $$$${r}={p}−{r}\:\mathrm{sin}\:\theta={p}−\frac{\mathrm{2}{r}}{\:\sqrt{\mathrm{4}+{p}^{\mathrm{6}} }}…

Question Number 191062 by Mingma last updated on 17/Apr/23 Answered by mr W last updated on 17/Apr/23 Commented by mr W last updated on 17/Apr/23…

Question Number 125444 by ajfour last updated on 11/Dec/20 Commented by ajfour last updated on 11/Dec/20 $${Find}\:{radius}\:{of}\:{the}\:{quarter}\:{circle}. \\ $$ Answered by ajfour last updated on…

Question Number 190845 by Rupesh123 last updated on 13/Apr/23 Answered by som(math1967) last updated on 13/Apr/23 Commented by som(math1967) last updated on 13/Apr/23 $${ar}\bigtriangleup{ABC}=\frac{\mathrm{1}}{\mathrm{2}}{ADEB}=\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{10}=\mathrm{5}{squ} \\…