Question Number 220253 by fantastic last updated on 10/May/25 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 220231 by fantastic last updated on 09/May/25 Answered by mr W last updated on 09/May/25 $$\mathrm{tan}\:\alpha=\frac{\mathrm{1}.\mathrm{5}}{\mathrm{1}+\mathrm{1}.\mathrm{5}+\sqrt{\left(\mathrm{1}+\mathrm{1}.\mathrm{5}\right)^{\mathrm{2}} −\mathrm{1}.\mathrm{5}^{\mathrm{2}} }}=\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$?=\mathrm{2}{R}=\frac{\mathrm{2}×\mathrm{1}.\mathrm{5}}{\mathrm{sin}\:\mathrm{2}\alpha}=\frac{\mathrm{3}×\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }\right)}{\mathrm{2}×\frac{\mathrm{1}}{\mathrm{3}}}=\mathrm{5}\:\checkmark \\ $$…
Question Number 220208 by Tawa11 last updated on 08/May/25 Commented by Tawa11 last updated on 08/May/25 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{volume}\:\mathrm{and}\:\mathrm{area}\:\mathrm{of}\:\mathrm{the}\:\mathrm{shape}. \\ $$$$\mathrm{volume}\:\:=\:\:\frac{\mathrm{1}}{\mathrm{2}}\pi\mathrm{r}^{\mathrm{2}} \mathrm{h}\:\:\:\:\:\mathrm{but}\:\mathrm{I}\:\mathrm{am}\:\mathrm{not}\:\mathrm{getting}\:\mathrm{the}\:\mathrm{area}. \\ $$$$\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{area}\:\mathrm{please}? \\ $$ Commented…
Question Number 220016 by fantastic last updated on 04/May/25 Answered by efronzo1 last updated on 04/May/25 $$\:\frac{\mathrm{PO}}{\mathrm{sin}\:\mathrm{60}°}\:=\:\frac{\mathrm{8}}{\mathrm{sin}\:\mathrm{75}°}\:=\:\frac{\mathrm{QO}}{\mathrm{sin}\:\mathrm{45}°} \\ $$$$\:\mathrm{PO}\:=\:\frac{\mathrm{16}\sqrt{\mathrm{3}}}{\:\sqrt{\mathrm{6}}\:+\sqrt{\mathrm{2}}}\:=\mathrm{4}\left(\mathrm{3}\sqrt{\mathrm{2}}\:−\sqrt{\mathrm{6}}\:\right) \\ $$$$\:\mathrm{area}\:\Delta\mathrm{POQ}\:=\:\frac{\mathrm{1}}{\mathrm{2}}.\:\mathrm{8}.\:\mathrm{4}\sqrt{\mathrm{2}}\:\left(\mathrm{3}−\sqrt{\mathrm{3}}\right) \\ $$$$\:=\:\mathrm{16}\sqrt{\mathrm{2}}\:\left(\mathrm{3}−\sqrt{\mathrm{3}}\:\right) \\ $$…
Question Number 219998 by fantastic last updated on 04/May/25 Answered by mehdee7396 last updated on 04/May/25 $${AB}=\sqrt{\mathrm{61}}\:\:\&\:\:\:{AC}=\mathrm{6}\sqrt{\mathrm{10}} \\ $$$${S}_{{ABC}} =\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{13}×\mathrm{6}=\mathrm{39} \\ $$$${R}=\frac{{abc}}{\mathrm{4}{S}}=\frac{\mathrm{13}×\sqrt{\mathrm{61}}×\mathrm{6}\sqrt{\mathrm{10}}}{\mathrm{4}×\mathrm{39}}=\frac{\sqrt{\mathrm{610}}}{\mathrm{2}} \\ $$$$ \\…
Question Number 219879 by fantastic last updated on 03/May/25 Answered by A5T last updated on 03/May/25 $$\mathrm{Let}\:\mathrm{the}\:\mathrm{sides}\:\mathrm{of}\:\mathrm{the}\:\mathrm{rectngle}\:\mathrm{be}\:\mathrm{a}\:\mathrm{and}\:\mathrm{b}. \\ $$$$\sqrt{\left(\mathrm{1}.\mathrm{5}+\mathrm{2}\right)^{\mathrm{2}} −\left(\mathrm{a}−\mathrm{1}.\mathrm{5}−\mathrm{2}\right)^{\mathrm{2}} }+\sqrt{\left(\mathrm{3}+\mathrm{2}\right)^{\mathrm{2}} −\left(\mathrm{a}−\mathrm{3}−\mathrm{2}\right)^{\mathrm{2}} } \\ $$$$=\mathrm{b}−\mathrm{1}.\mathrm{5}−\mathrm{3}…
Question Number 219540 by BaliramKumar last updated on 28/Apr/25 Answered by som(math1967) last updated on 28/Apr/25 $${CD}^{\mathrm{2}} +{BC}^{\mathrm{2}} =\mathrm{2}×\left\{\left(\frac{\mathrm{25}}{\mathrm{2}}\right)^{\mathrm{2}} +\left(\frac{\mathrm{39}}{\mathrm{2}}\right)^{\mathrm{2}} \right\} \\ $$$$\:{CD}=\sqrt{\mathrm{289}}=\mathrm{17} \\ $$$${Ar}\:\bigtriangleup{BCD}=\sqrt{\mathrm{42}\left(\mathrm{42}−\mathrm{28}\right)\left(\mathrm{42}−\mathrm{17}\right)\left(\mathrm{42}−\mathrm{39}\right)}…
Question Number 219278 by Mingma last updated on 22/Apr/25 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 219088 by Spillover last updated on 19/Apr/25 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 219086 by Spillover last updated on 19/Apr/25 Answered by Spillover last updated on 19/Apr/25 v= [axbxsin60°]l V=[2x5xsin 60°]×10 cm^3 Answered by mr W…