Question Number 197670 by mr W last updated on 26/Sep/23 Commented by mr W last updated on 26/Sep/23 $${unsolved}\:{old}\:{question}\:{Q}#\mathrm{197017} \\ $$ Commented by ajfour last…
Question Number 197644 by gungun last updated on 25/Sep/23 Commented by gungun last updated on 25/Sep/23 $${find}\:{the}\:{pink}\:{area} \\ $$ Answered by Rasheed.Sindhi last updated on…
Question Number 197636 by universe last updated on 25/Sep/23 Answered by mr W last updated on 25/Sep/23 $${s}={x}+{y} \\ $$$$\left({x}+\mathrm{3}\right)^{\mathrm{2}} +{y}^{\mathrm{2}} =\left(\mathrm{8}\sqrt{\mathrm{2}}\right)^{\mathrm{2}} \\ $$$$\left({x}+\mathrm{3}\right)^{\mathrm{2}} +\left({s}−{x}\right)^{\mathrm{2}}…
Question Number 197595 by universe last updated on 23/Sep/23 Commented by universe last updated on 23/Sep/23 $$\mathrm{find}\:\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:\mathrm{equilateral}\:\mathrm{triangle}\:? \\ $$ Answered by MM42 last updated on…
Question Number 197582 by cherokeesay last updated on 22/Sep/23 Answered by som(math1967) last updated on 23/Sep/23 Commented by som(math1967) last updated on 23/Sep/23 $${let}\:{rad}\:{of}\:{big}\:{semicircle}={R} \\…
Question Number 197541 by a.lgnaoui last updated on 20/Sep/23 $$\boldsymbol{\mathrm{Montrer}}\:\boldsymbol{\mathrm{que}} \\ $$$$\:\:\:\:\:\boldsymbol{\mathrm{x}}=\frac{\boldsymbol{\mathrm{an}}+\boldsymbol{\mathrm{bm}}}{\boldsymbol{\mathrm{m}}+\boldsymbol{\mathrm{n}}} \\ $$ Commented by a.lgnaoui last updated on 20/Sep/23 Commented by a.lgnaoui last…
Question Number 197499 by cherokeesay last updated on 19/Sep/23 Answered by HeferH last updated on 19/Sep/23 $$\: \\ $$$$\:\frac{{x}}{{r}}\:=\:\frac{\mathrm{4}{r}}{\mathrm{5}{r}}\:\:\Rightarrow\:{x}\:=\:\frac{\mathrm{4}{r}}{\mathrm{5}}\: \\ $$$$\:{Green}\:=\:{Sqr}/\mathrm{2}\:−\:\frac{\mathrm{4}{r}}{\mathrm{5}}\:\centerdot\:\mathrm{4}{r}\:\centerdot\frac{\mathrm{1}}{\mathrm{2}}\: \\ $$$$\:{Green}\:=\:\frac{\mathrm{16}{r}^{\mathrm{2}} }{\mathrm{2}}\:−\frac{\mathrm{16}{r}^{\mathrm{2}} }{\mathrm{10}}\:=\:\frac{\mathrm{4}\centerdot\mathrm{16}{r}^{\mathrm{2}}…
Question Number 197287 by MM42 last updated on 13/Sep/23 $${answer}\:{to}\:{the}\:{question}\:{number} \\ $$$$\mathrm{197017} \\ $$$${AF}={FI}\:\&\:\:{AG}={GJ}\Rightarrow{FG}=\frac{\mathrm{1}}{\mathrm{2}}{IJ}=\frac{\mathrm{1}}{\mathrm{6}}{BC} \\ $$$$\bigtriangleup{FGH}\:\:{is}\:\:{squilatral}\:\Rightarrow\:\bigtriangleup{FGH}\approx\bigtriangleup{ABC} \\ $$$$\Rightarrow\frac{{S}_{{FGH}} }{{S}_{{SBC}} }\:=\frac{\mathrm{1}}{\mathrm{36}\:}\:\checkmark \\ $$$$ \\ $$ Commented…
Question Number 197157 by uchihayahia last updated on 09/Sep/23 $$ \\ $$$$\:{what}\:{is}\:{the}\:{area}\:{of}\:{triangle}\:{A}\:{and}\:{B}? \\ $$$$\:{i}\:{already}\:{found}\:{area}\:{of}\:{triangle}\:{A}\:{is}\:\mathrm{208} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$ Commented…
Question Number 197095 by Mingma last updated on 07/Sep/23 Answered by mahdipoor last updated on 07/Sep/23 $$\frac{{AB}}{{sin}\mathrm{4}{a}}=\frac{{BM}}{{sina}}\:\:\: \\ $$$$\:\frac{{CD}}{{sin}\mathrm{4}{a}}=\frac{{MC}}{{sin}\mathrm{2}{a}}\Rightarrow\frac{\mathrm{2}.{CD}.{cosa}}{{sin}\mathrm{4}{a}}=\frac{{MC}}{{sina}}\Rightarrow \\ $$$$\frac{{MC}}{{sina}}+\frac{{BM}}{{sina}}=\frac{{CB}}{{sina}}=\frac{\mathrm{2}.{CD}.{cosa}}{{sin}\mathrm{4}{a}}+\frac{{AB}}{{sin}\mathrm{4}{a}} \\ $$$$\Rightarrow\frac{{sin}\mathrm{4}{a}}{{sina}}=\mathrm{4}{cos}\mathrm{2}{a}.{cosa}=\mathrm{2}{cosa}+\mathrm{1}\Rightarrow \\ $$$$\mathrm{8}{cos}^{\mathrm{3}}…