Category: Geometry

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 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 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 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} \\…

Question Number 161033 by cherokeesay last updated on 11/Dec/21 Answered by FongXD last updated on 11/Dec/21 Commented by Tawa11 last updated on 11/Dec/21 $$\mathrm{Great}\:\mathrm{sir} \\…