Question Number 176942 by Ar Brandon last updated on 28/Sep/22 Answered by som(math1967) last updated on 28/Sep/22 $${HE}={GF}=\frac{\mathrm{1}}{\mathrm{2}}×{AC}=\frac{\mathrm{17}}{\mathrm{2}}{cm} \\ $$$${EF}={HG}=\frac{\mathrm{1}}{\mathrm{2}}×{DB}=\frac{\mathrm{13}}{\mathrm{2}}{cm} \\ $$$${perimeter}\:{EFGH}=\mathrm{2}\left(\frac{\mathrm{17}}{\mathrm{2}}\:+\frac{\mathrm{13}}{\mathrm{2}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{30}{cm} \\…
Question Number 176853 by Ar Brandon last updated on 27/Sep/22 Answered by som(math1967) last updated on 27/Sep/22 $$\:\frac{{PT}}{{TA}}=\frac{{PR}}{{RK}}=\frac{{PS}}{{SE}}=\frac{\mathrm{2}}{\mathrm{3}} \\ $$$${let}\:{PT}={PR}={PS}=\mathrm{2}{x}\: \\ $$$${TA}=\mathrm{3}{x} \\ $$$$\therefore\:{each}\:{side}\:{of}\:{cube}={PT}+{TA}=\mathrm{5}{x} \\…
Question Number 176848 by Ar Brandon last updated on 27/Sep/22 Answered by mr W last updated on 27/Sep/22 $$\beta=\frac{\pi}{\mathrm{2}}−\alpha \\ $$$$\mathrm{cos}\:\left(\frac{\pi}{\mathrm{2}}−\mathrm{2}\beta\right)=\mathrm{sin}\:\mathrm{2}\beta=\mathrm{sin}\:\left(\pi−\mathrm{2}\alpha\right) \\ $$$$=\mathrm{sin}\:\mathrm{2}\alpha=\mathrm{2}\:\mathrm{sin}\:\alpha\:\mathrm{cos}\:\alpha \\ $$$$=\mathrm{2}{x}\sqrt{\mathrm{1}−{x}^{\mathrm{2}}…
Question Number 176821 by Ar Brandon last updated on 27/Sep/22 Commented by som(math1967) last updated on 27/Sep/22 $$\:\frac{{AF}}{{AL}}=\frac{\frac{{BC}}{\mathrm{2}}+{CE}+\frac{{EK}}{\mathrm{2}}}{\frac{{BC}}{\mathrm{2}}+{CE}+{EK}+\frac{{KM}}{\mathrm{2}}} \\ $$$$\:\:=\frac{.\mathrm{5}+\mathrm{2}+\mathrm{2}}{.\mathrm{5}+\mathrm{2}+\mathrm{4}+\mathrm{8}}=\frac{\mathrm{4}.\mathrm{5}}{\mathrm{14}.\mathrm{5}}=\frac{\mathrm{9}}{\mathrm{29}} \\ $$ Commented by Ar…
Question Number 176820 by Ar Brandon last updated on 27/Sep/22 Commented by mr W last updated on 27/Sep/22 Commented by som(math1967) last updated on 27/Sep/22…
Question Number 176816 by Ar Brandon last updated on 27/Sep/22 Commented by MJS_new last updated on 27/Sep/22 $${y}={x}\wedge\mathrm{5}{x}+\mathrm{12}{y}=\mathrm{60}\:\Rightarrow\:{x}={y}=\frac{\mathrm{60}}{\mathrm{17}} \\ $$ Commented by Ar Brandon last…
Question Number 45313 by ajfour last updated on 11/Oct/18 Commented by ajfour last updated on 11/Oct/18 $${If}\:{all}\:{five}\:{areas}\:{are}\:{equal},\:{and} \\ $$$${circle}\:{has}\:{unit}\:{radius},\:{find} \\ $$$$\boldsymbol{{a}},{and}\:\boldsymbol{{b}}\:{of}\:{ellipse}. \\ $$ Answered by…
Question Number 45284 by ajfour last updated on 11/Oct/18 Commented by ajfour last updated on 11/Oct/18 $${DP}\:=\:{DT}\:=\:{b}−\mathrm{1}\:\:;\:\:\:\:{BT}\:=\:{a}−\mathrm{1} \\ $$$${BD}={DT}+{BT} \\ $$$$\Rightarrow\:\:\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }\:=\:{a}+{b}−\mathrm{2}\:\:\: \\ $$$$\Rightarrow\:\mathrm{2}{ab}−\mathrm{4}{a}−\mathrm{4}{b}+\mathrm{4}=\mathrm{0}…
Question Number 176354 by BaliramKumar last updated on 16/Sep/22 Answered by Frix last updated on 16/Sep/22 $$\mathrm{in}\:\bigtriangleup\:\mathrm{with}\:\mathrm{sides}\:{a},\:{b},\:{c} \\ $$$${r}=\frac{\sqrt{\left({a}+{b}+{c}\right)\left(−{a}+{b}+{c}\right)\left({a}−{b}+{c}\right)\left({a}+{b}−{c}\right)}}{\mathrm{2}\left({a}+{b}+{c}\right)} \\ $$$$\mathrm{in}\:\mathrm{this}\:\mathrm{case}\:{r}=\mathrm{15} \\ $$ Commented by…
Question Number 176281 by BaliramKumar last updated on 15/Sep/22 Commented by amin96 last updated on 15/Sep/22 Commented by BaliramKumar last updated on 16/Sep/22 $${thanks}\:{Sir} \\…