Question Number 67849 by mr W last updated on 01/Sep/19 Commented by MJS last updated on 01/Sep/19 $$\mathrm{is}\:\mathrm{the}\:\mathrm{angle}\:\mathrm{in}\:\mathrm{the}\:\mathrm{center}\:\mathrm{90}°? \\ $$ Commented by mr W last…
Question Number 133371 by I want to learn more last updated on 21/Feb/21 Answered by mr W last updated on 22/Feb/21 Commented by mr W…
Question Number 67813 by mr W last updated on 31/Aug/19 Commented by mr W last updated on 31/Aug/19 $${shaded}\:{area}\:{A}=? \\ $$ Commented by Prithwish sen…
Question Number 67807 by TawaTawa last updated on 31/Aug/19 Commented by MJS last updated on 01/Sep/19 $$\mathrm{perimeter}\:\mathrm{or}\:\mathrm{area}? \\ $$ Commented by TawaTawa last updated on…
Question Number 67774 by TawaTawa last updated on 31/Aug/19 Answered by MJS last updated on 31/Aug/19 $${DE}={AB}=\mathrm{2}{r} \\ $$$$\mathrm{area}\:\mathrm{of}\:\mathrm{semicircle}\:=\frac{\pi}{\mathrm{2}}{r}^{\mathrm{2}} \\ $$$$\mathrm{area}\:\mathrm{of}\:\mathrm{triangle}\:=\frac{\mathrm{1}}{\mathrm{2}}\mid{AB}\mid{h}={rh} \\ $$$$\:\:\:\:\:\mathrm{with}\:{h}={r}\mathrm{tan}\:{x} \\ $$$$\:\:\:\:\:={r}^{\mathrm{2}}…
Question Number 67772 by TawaTawa last updated on 31/Aug/19 Commented by Prithwish sen last updated on 31/Aug/19 $$\boldsymbol{\mathrm{r}}=\boldsymbol{\mathrm{a}}\left(\frac{\sqrt{\mathrm{2}}−\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right) \\ $$ Commented by Prithwish sen last…
Question Number 67775 by TawaTawa last updated on 31/Aug/19 Answered by MJS last updated on 31/Aug/19 $${ABCD}\:\mathrm{is}\:\mathrm{a}\:\mathrm{square}\:\mathrm{with}\:\mathrm{side}\:{s}=\sqrt{\mathrm{196}}=\mathrm{14} \\ $$$$\Rightarrow\:\mathrm{the}\:\mathrm{yellow}\:\mathrm{sector}'\mathrm{s}\:\mathrm{area}\:=\frac{\pi}{\mathrm{8}}{s}^{\mathrm{2}} =\frac{\mathrm{49}\pi}{\mathrm{2}}\:\mathrm{minus} \\ $$$$\mathrm{the}\:\mathrm{white}\:\mathrm{segment}\:\mathrm{which}\:\mathrm{intersects}\:\mathrm{the} \\ $$$$\mathrm{diagonal}\:\mathrm{in}\:\mathrm{the}\:\mathrm{center}\:\mathrm{of}\:\mathrm{the}\:\mathrm{square} \\…
Question Number 67770 by TawaTawa last updated on 31/Aug/19 Answered by MJS last updated on 31/Aug/19 $${AD}+{DE}={AE}=\mathrm{6} \\ $$$$\Rightarrow\:{AD}={q};\:{DE}=\mathrm{6}−{q}\:\:\left[\Rightarrow\:{q}<\mathrm{6}\right] \\ $$$${AB}={DE}−{AD}=\mathrm{6}−\mathrm{2}{q}\:\:\left[\Rightarrow\:{q}<\mathrm{3}\right] \\ $$$${A}=\begin{pmatrix}{\mathrm{0}}\\{\mathrm{0}}\end{pmatrix}\:\:{D}=\begin{pmatrix}{{q}}\\{\mathrm{0}}\end{pmatrix}\:\:{E}=\begin{pmatrix}{\mathrm{6}}\\{\mathrm{0}}\end{pmatrix} \\ $$$$\mathrm{line}\:{AB}:\:{y}={x}\mathrm{tan}\:\mathrm{60}°\:\Leftrightarrow\:{y}=\sqrt{\mathrm{3}}{x}\:\:\left[\mathrm{1}\right]…
Question Number 67749 by TawaTawa last updated on 31/Aug/19 Answered by JDamian last updated on 31/Aug/19 $${Assuming}\:{A},\:{B}\:{and}\:{C}\:{points}\:{are}\:{the}\:{centers} \\ $$$${of}\:{their}\:{circles}: \\ $$$$ \\ $$$${AC}={R}_{{A}} −{R}_{{C}} =\mathrm{11}−\mathrm{5}=\mathrm{6}…
Question Number 2196 by Rasheed Soomro last updated on 08/Nov/15 $$\left({a}+{b}\right)^{\mathrm{2}} ={a}^{\mathrm{2}} +{b}^{\mathrm{2}} +\mathrm{2}{ab} \\ $$$$\left({a}+{b}+{c}\right)^{\mathrm{2}} ={a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +\mathrm{2}{ab}+\mathrm{2}{bc}+\mathrm{2}{ca} \\ $$$$\left({a}+{b}+{c}+{d}\right)^{\mathrm{2}} =? \\ $$$$\left({a}_{\mathrm{1}}…