Question Number 221592 by fantastic last updated on 08/Jun/25 Answered by mr W last updated on 08/Jun/25 Commented by mr W last updated on 08/Jun/25…
Question Number 221626 by fantastic last updated on 08/Jun/25 Answered by mr W last updated on 08/Jun/25 $${side}\:{length}\:{of}\:{square}\:=\mathrm{1} \\ $$$${shaded}\:{area}\:=\frac{\pi×\mathrm{1}^{\mathrm{2}} }{\mathrm{8}}+\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right)^{\mathrm{2}} −\frac{\pi}{\mathrm{4}}\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right)^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{4}} \\…
Question Number 221453 by gregori last updated on 06/Jun/25 Answered by Rasheed.Sindhi last updated on 06/Jun/25 $${x}^{\mathrm{3}} =−{x}^{\mathrm{2}} −{x}−\mathrm{1}\left({mod}\:{x}^{\mathrm{3}} +{x}^{\mathrm{2}} +{x}+\mathrm{1}\right) \\ $$$${x}^{\mathrm{4}} \equiv−{x}^{\mathrm{3}} −{x}^{\mathrm{2}}…
Question Number 221332 by fantastic last updated on 30/May/25 Answered by mr W last updated on 30/May/25 $${AB}={OB}=\sqrt{\mathrm{3}}\:{OD} \\ $$$$\frac{\Delta{DOE}}{\Delta{ABC}}=\left(\frac{{OD}}{{AB}}\right)^{\mathrm{2}} =\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\Rightarrow\Delta{DOE}=\frac{\mathrm{12}}{\mathrm{3}}=\mathrm{4}\:{sq}.\:{units} \\…
Question Number 221298 by ajfour last updated on 29/May/25 $${Find}\:{the}\:{area}\:{of}\:\bigtriangleup{ABC}. \\ $$$${sides}\:{are}\:\sqrt{\mathrm{20}},\:\sqrt{\mathrm{26}}.\:{and}\:\sqrt{\mathrm{34}}\:. \\ $$ Answered by Ghisom last updated on 29/May/25 $$\mathrm{area}\:\mathrm{of}\:\mathrm{triangle}:\:\mathrm{Heron}'\mathrm{s}\:\mathrm{Formula} \\ $$$$\mathrm{area}=\frac{\sqrt{\left({a}+{b}+{c}\right)\left({a}+{b}−{c}\right)\left({a}+{c}−{b}\right)\left({b}+{c}−{a}\right)}}{\mathrm{4}} \\…
Question Number 221132 by fantastic last updated on 25/May/25 Commented by fantastic last updated on 25/May/25 $${can}\:{someone}\:{give}\:{me}\:{a}\:{DETAILED}\:{and}\:{EASY}\:{SOLUTION}?\: \\ $$$${i}\:{solved}\:{it}\:{in}\:{very}\:{lengthy}\:{and}\:{hard}\:{method} \\ $$ Commented by mr W…
Question Number 221102 by fantastic last updated on 24/May/25 Answered by mehdee7396 last updated on 25/May/25 $${AB}=\mathrm{2}\sqrt{{ar}}\:\:\&\:\:{BC}=\mathrm{2}\sqrt{{br}}\:\:\:\&\:\:\:{AC}=\mathrm{2}\sqrt{{ab}} \\ $$$$\Rightarrow\sqrt{{ab}}=\left(\sqrt{{a}}+\sqrt{{b}}\right)\sqrt{{r}\:} \\ $$$$\Rightarrow{r}=\frac{{ab}}{{a}+{b}+\mathrm{2}\sqrt{{ab}}}\:\:{or}\:\:\frac{\mathrm{1}}{\:\sqrt{{r}}}=\frac{\mathrm{1}}{\:\sqrt{{a}}}+\frac{\mathrm{1}}{\:\sqrt{{b}}}\: \\ $$$$ \\ $$…
Question Number 221056 by fantastic last updated on 23/May/25 Answered by Frix last updated on 24/May/25 $$\mathrm{Area}\:\mathrm{of}\:\mathrm{trapezoid}\:=\frac{\mathrm{16}+\mathrm{22}}{\mathrm{2}}{h}=\mathrm{19}{h} \\ $$$${h}=\sqrt{\mathrm{4}^{\mathrm{2}} −\left(\frac{\mathrm{22}−\mathrm{16}}{\mathrm{2}}\right)^{\mathrm{2}} }=\sqrt{\mathrm{7}} \\ $$$${r}=\frac{\sqrt{\mathrm{7}}}{\mathrm{2}} \\ $$$$\mathrm{Shaded}\:\mathrm{area}\:=\:\mathrm{19}\sqrt{\mathrm{7}}−\frac{\mathrm{7}\pi}{\mathrm{4}}…
Question Number 220825 by fantastic last updated on 19/May/25 Answered by mr W last updated on 20/May/25 $$\frac{\mathrm{20}}{\mathrm{180}}×\mathrm{72}=\mathrm{8} \\ $$ Commented by mr W last…
Question Number 220602 by fantastic last updated on 16/May/25 Answered by A5T last updated on 16/May/25 $$\mathrm{Let}\:\mathrm{the}\:\mathrm{radius}\:\mathrm{and}\:\mathrm{circumference}\:\mathrm{of}\:\mathrm{semicircle}\:, \\ $$$$\mathrm{radius}\:\mathrm{and}\:\mathrm{circumference}\:\mathrm{of}\:\mathrm{circle}\:\mathrm{be}\:\mathrm{r}_{\mathrm{s}} \:\mathrm{and}\:\mathrm{c}_{\mathrm{s},} \: \\ $$$$\mathrm{r}\:\mathrm{and}\:\mathrm{c}\:\mathrm{respectively}. \\ $$$$\mathrm{c}_{\mathrm{s}}…