Question Number 30719 by mrW2 last updated on 24/Feb/18 Commented by mrW2 last updated on 25/Feb/18 $${This}\:{question}\:{is}\:{not}\:{visible}\:{any}\:{more}. \\ $$$${The}\:{question}\:{is}: \\ $$$${Find}\:{the}\:{side}\:{lengths}\:{of}\:\Delta{PQR}\:{in} \\ $$$${terms}\:{of}\:{a},{b},{c}. \\ $$…
Question Number 161742 by mr W last updated on 21/Dec/21 Commented by mr W last updated on 21/Dec/21 $${find}\:{the}\:{radii}\:{of}\:{the}\:{circles}\:{inside} \\ $$$${the}\:{big}\:{quater}\:{circle}. \\ $$ Answered by…
Question Number 161482 by amin96 last updated on 18/Dec/21 Commented by amin96 last updated on 18/Dec/21 $${circles}\:{area}=? \\ $$ Answered by FongXD last updated on…
Question Number 161424 by cherokeesay last updated on 17/Dec/21 Answered by aleks041103 last updated on 17/Dec/21 $$\left({R}−\mathrm{2}\right)^{\mathrm{2}} +\left({R}−\mathrm{4}\right)^{\mathrm{2}} ={R}^{\mathrm{2}} \\ $$$${R}^{\mathrm{2}} −\mathrm{4}{R}+\mathrm{4}+{R}^{\mathrm{2}} −\mathrm{8}{R}+\mathrm{16}={R}^{\mathrm{2}} \\ $$$${R}^{\mathrm{2}}…
Question Number 161367 by amin96 last updated on 17/Dec/21 Commented by 1549442205PVT last updated on 17/Dec/21 $${I}\:{think}\:{that}\:{the}\:{condition}\:{of}\:{the}\:{probem}\:{isn}'{t}\:\:{clear} \\ $$$${the}\:{radius}\:{of}\:{the}\:{arc}\:{which}\:{limits}\:{small} \\ $$$${circle}\:{isn}'{t}\:{given}\:{yet} \\ $$ Answered by…
Question Number 95830 by I want to learn more last updated on 27/May/20 Answered by behi83417@gmail.com last updated on 28/May/20 $$\mathrm{4s}+\frac{\mathrm{16}^{\mathrm{2}} }{\mathrm{2}}=\mathrm{16}^{\mathrm{2}} \Rightarrow\mathrm{s}=\frac{\mathrm{16}^{\mathrm{2}} }{\mathrm{8}}=\mathrm{32} \\…
Question Number 95772 by I want to learn more last updated on 27/May/20 Commented by I want to learn more last updated on 27/May/20 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:\mathrm{shaded}…
Question Number 161296 by amin96 last updated on 15/Dec/21 Answered by mr W last updated on 16/Dec/21 $${A}\left({p}^{\mathrm{2}} ,{p}\right) \\ $$$${C}\left({q}^{\mathrm{2}} ,{q}\right) \\ $$$$\mathrm{tan}\:\theta=\frac{{dy}}{{dx}}=\frac{\mathrm{1}}{\mathrm{2}\sqrt{{x}}}=\frac{\mathrm{1}}{\mathrm{2}{p}}=\mathrm{1} \\…
Question Number 161257 by amin96 last updated on 15/Dec/21 Commented by amin96 last updated on 15/Dec/21 $${yes}\:.\:{solution}? \\ $$ Answered by som(math1967) last updated on…
Question Number 30090 by gopikrishnan005@gmail.com last updated on 16/Feb/18 $$\mathrm{if}\:\Delta\mathrm{ABC}\:\mathrm{similar}\:\Delta\mathrm{PQR}\:\mathrm{and}\:\mathrm{area}\:\mathrm{of}\:\Delta\mathrm{PQR}=\mathrm{4area}\left(\Delta\mathrm{ABC}\right)\:\mathrm{then}\:\mathrm{AB}:\mathrm{PQ}\:\mathrm{is} \\ $$ Answered by Rasheed.Sindhi last updated on 17/Feb/18 $$\mathrm{mAB}=\mathrm{c},\mathrm{mBC}=\mathrm{a},\mathrm{mAC}=\mathrm{b},\mathrm{s}=\frac{\mathrm{a}+\mathrm{b}+\mathrm{c}}{\mathrm{2}} \\ $$$$\mathrm{mPQ}=\mathrm{r},\mathrm{mQR}=\mathrm{p}\:,\mathrm{mPR}=\mathrm{q},\mathrm{S}=\frac{\mathrm{p}+\mathrm{q}+\mathrm{r}}{\mathrm{2}} \\ $$$$\blacktriangle\mathrm{ABC} \\…