Question Number 103986 by ajfour last updated on 18/Jul/20 Commented by ajfour last updated on 18/Jul/20 $${If}\:{both}\:{coloured}\:{regions}\:{have}\:{equal} \\ $$$${area},\:{find}\:{x}_{{A}} ,\:{x}_{{B}} \:. \\ $$ Answered by…
Question Number 38289 by Tinkutara last updated on 23/Jun/18 Answered by tanmay.chaudhury50@gmail.com last updated on 23/Jun/18 $${centre}\left(−{g},−{f}\right)\:{lies}\:{on}\:{ax}+{by}+{d}=\mathrm{0} \\ $$$${a}\left(−{g}\right)+{b}\left(−{f}\right)+{d}=\mathrm{0}\: \\ $$$$\lambda\left(−{g}\right)−{a}\left(−{f}\right)+\mu{g}+{k}=\mathrm{0} \\ $$$${g}\left(−{a}\right)+{f}\left(−{b}\right)+{d}=\mathrm{0} \\ $$$${g}\left(−\lambda+\mu\right)+{f}\left({a}\right)+{k}=\mathrm{0}…
Question Number 38281 by Tinkutara last updated on 23/Jun/18 Answered by ajfour last updated on 23/Jun/18 $$\frac{{ap}_{{A}\rightarrow{BC}} }{\mathrm{2}}\:=\:\frac{{bp}_{{B}\rightarrow{CA}} }{\mathrm{2}}\:=\:\frac{{cp}_{{C}\rightarrow{AB}} }{\mathrm{2}}\:=\bigtriangleup \\ $$$$\frac{{ap}_{{P}\rightarrow{BC}} }{\mathrm{2}}\:+\:\frac{{bp}_{{P}\rightarrow{CA}} }{\mathrm{2}}\:+\:\frac{{cp}_{{P}\rightarrow{AB}} }{\mathrm{2}}\:=\:\Delta…
Question Number 103756 by ajfour last updated on 17/Jul/20 Commented by ajfour last updated on 17/Jul/20 $${Find}\:{radius}\:{r}\:{of}\:{shown}\:{circle}. \\ $$ Answered by mr W last updated…
Question Number 38006 by ajfour last updated on 20/Jun/18 Commented by ajfour last updated on 20/Jun/18 $${For}\:{each}\:{of}\:{the}\:{four}\:{single} \\ $$$${coloured}\:{areas}\:{to}\:{be}\:{equal},\:{locate} \\ $$$${P}\left({r},\theta\right)\:\:{and}\:{C}\left(\mathrm{0},{y}\right)\:{in}\:{terms}\:{of} \\ $$$${a}\:{and}\:{b}.\:\:\:\left(\:{r}=\:{OP}\:\:,\:\:{y}={OC}\:\right) \\ $$…
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Question Number 168980 by BHOOPENDRA last updated on 22/Apr/22 Commented by mr W last updated on 22/Apr/22 $${not}\:{clear}\:{what}\:{you}\:{want}\:{to}\:{show}… \\ $$ Commented by BHOOPENDRA last updated…
Question Number 168948 by cortano1 last updated on 22/Apr/22 Answered by mahdipoor last updated on 22/Apr/22 $${S}\left({ABCDEF}\right)=\mathrm{28}.\mathrm{75}=\frac{\mathrm{115}}{\mathrm{4}} \\ $$$$\left\{{consider}\:{a}\:{rectangle}\:{around}\:{it}\:{and}\:\right. \\ $$$$\left.{calculate}\:{S}\right\} \\ $$$${a}={coefficient}\:{of}\:{similar}=\frac{{AE}}{{AE}}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${S}\left({ABCDEF}\right)={a}^{\mathrm{2}}…
Question Number 37862 by ajfour last updated on 18/Jun/18 Commented by MrW3 last updated on 18/Jun/18 $${yes},\:\sqrt{{r}^{\mathrm{2}} +{q}^{\mathrm{2}} }<\mathrm{2}{a}. \\ $$ Commented by ajfour last…
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