Question Number 181070 by mr W last updated on 21/Nov/22 Commented by mr W last updated on 21/Nov/22 $$\left[{Q}\mathrm{181055}\:{modified}\right] \\ $$$${find}\:{the}\:{area}\:{of}\:{trapazoid}\:{ABCD}. \\ $$ Commented by…
Question Number 49987 by ajfour last updated on 12/Dec/18 Commented by ajfour last updated on 12/Dec/18 $${a}\:\neq\:{b}\:,\:{find}\:\boldsymbol{{c}}\:\:{in}\:{terms}\:{of}\:{a}\:{and}\:{b}. \\ $$ Commented by MJS last updated on…
Question Number 49984 by behi83417@gmail.com last updated on 12/Dec/18 Commented by behi83417@gmail.com last updated on 12/Dec/18 $${A}\overset{\bigtriangleup} {{B}C},{equilateral}.{AB}={a},{A}\overset{} {{C}E}=\alpha. \\ $$$${find}:{AD},{BD},{in}\:{terms}\:{of}\:{a}\:{and}\:\:\alpha. \\ $$ Answered by…
Question Number 181051 by mr W last updated on 20/Nov/22 Answered by a.lgnaoui last updated on 21/Nov/22 $${cote}\:{du}\:{carre}\:{bleu}\:\left[{CD}\right]={b}\:{comme}\:{il} \\ $$$${apparait}\:{dans}\:\left({image}\right){est} \\ $$$${CD}:{prolongement}\:{de}\left[\:{AC}\right]\left({AC}\right) \\ $$$$\left[{CD}\right]=\mathrm{2}×{AC} \\…
Question Number 181055 by Acem last updated on 21/Nov/22 Commented by Acem last updated on 21/Nov/22 $$\ast\:{DABC} \\ $$$${Please},\:{the}\:{comments}\:{section}\:{is}\:{only} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{for} \\ $$$$\:\:\:\:\:\boldsymbol{{inquiries}}\:\boldsymbol{{and}}\:\boldsymbol{{clarifications}} \\ $$$$…
Question Number 49970 by behi83417@gmail.com last updated on 12/Dec/18 Commented by behi83417@gmail.com last updated on 12/Dec/18 $${ABC},{equilateral}.{AD}={DC},\measuredangle{CDE}=\mathrm{30}^{\bullet} \\ $$$$\Rightarrow\:\:\:\:\:\:\:\frac{{area}\left[{ABED}\right]}{{area}\left[{EDC}\right]}=? \\ $$ Answered by mr W…
Question Number 181019 by cherokeesay last updated on 20/Nov/22 Answered by som(math1967) last updated on 20/Nov/22 $$\:\:{cos}\mathrm{120}=\frac{\mathrm{2}^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} −{AE}^{\mathrm{2}} }{\mathrm{2}.\mathrm{2}.\mathrm{2}} \\ $$$$−\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{8}=\mathrm{8}−{AE}^{\mathrm{2}} \\ $$$$\Rightarrow{AE}=\sqrt{\mathrm{12}}=\mathrm{2}\sqrt{\mathrm{3}} \\…
Question Number 180998 by HeferH last updated on 20/Nov/22 Answered by mr W last updated on 20/Nov/22 $$\left[\Delta{ADB}\right]=\left[\Delta{AEB}\right]+{green} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\left[\Delta{AEB}\right]+{magenta} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\left[\Delta{AFB}\right] \\ $$$$\Rightarrow{DF}//{AB} \\…
Question Number 180939 by mr W last updated on 19/Nov/22 Commented by mr W last updated on 19/Nov/22 $${next}\:{step}: \\ $$$${find}\:{the}\:{areas}\:{of}\:{the}\:{yellow}\:{squares} \\ $$$${in}\:{terms}\:{of}\:{triangle}'{s}\:{sides}\:{a},{b},{c}. \\ $$…
Question Number 180931 by ajfour last updated on 19/Nov/22 Commented by ajfour last updated on 19/Nov/22 $$\mathrm{3cos}\:\mathrm{15}°\:\:? \\ $$ Commented by mr W last updated…