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} \\ $$$$\left({a}+{b}+{c}\right)^{\mathrm{2}} ={a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +\mathrm{2}{ab}+\mathrm{2}{bc}+\mathrm{2}{ca} \\ $$$$\left({a}+{b}+{c}+{d}\right)^{\mathrm{2}} =? \\ $$$$\left({a}_{\mathrm{1}}…
Question Number 67728 by TawaTawa last updated on 30/Aug/19 Commented by mr W last updated on 30/Aug/19 $${question}\:{is}\:{not}\:{clear}. \\ $$$${is}\:\angle{BAD}=\mathrm{90}°\:{as}\:{marked}\:{or}\:{is} \\ $$$$\angle{BAC}=\mathrm{90}°\:{as}\:{shown}? \\ $$ Commented…
Question Number 67716 by TawaTawa last updated on 30/Aug/19 Answered by mind is power last updated on 30/Aug/19 $${let}\:{c}_{\mathrm{1}} ={demi}\:{circle}\:{of}\:{r}=\mathrm{2} \\ $$$${c}\mathrm{2}={circle}\:{r}=\mathrm{1} \\ $$$${c}_{\mathrm{1}} ..{x}^{\mathrm{2}}…
Question Number 67692 by mr W last updated on 30/Aug/19 Answered by mr W last updated on 30/Aug/19 Commented by mr W last updated on…
Question Number 2138 by Filup last updated on 06/Nov/15 $$\mathrm{Is}\:\mathrm{the}\:\mathrm{following}\:\mathrm{proof}\:\mathrm{correct}? \\ $$$$ \\ $$$$\Delta=\underset{{i}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{i}} \mathrm{2}^{{i}} =\mathrm{1}−\mathrm{2}+\mathrm{4}−\mathrm{8}+\mathrm{16}−\mathrm{32}+… \\ $$$$ \\ $$$$\mathrm{Let}: \\ $$$$\Delta_{\mathrm{1}} =\mathrm{1}−\mathrm{2}+\mathrm{4}−\mathrm{8}+\mathrm{16}−\mathrm{32}+……
Question Number 67623 by mr W last updated on 29/Aug/19 Commented by mr W last updated on 29/Aug/19 $${find}\:{the}\:{shaded}\:{area}\:{A}=? \\ $$ Commented by Prithwish sen…
Question Number 67615 by TawaTawa last updated on 29/Aug/19 Commented by TawaTawa last updated on 29/Aug/19 $$\mathrm{i}\:\mathrm{did}.\:\:\:\:\:\:\mathrm{A}\:\mathrm{of}\:\mathrm{Big}\:\mathrm{semi}\:\mathrm{circle}\:=\:\frac{\pi\mathrm{r}^{\mathrm{2}} }{\mathrm{2}}\:\:=\:\:\frac{\pi.\mathrm{2}^{\mathrm{2}} }{\mathrm{2}}\:\:=\:\:\mathrm{2}\pi \\ $$$$\mathrm{A}\:\mathrm{of}\:\mathrm{second}\:\mathrm{big}\:\mathrm{semicircle}\:\:=\:\:\frac{\pi\mathrm{r}^{\mathrm{2}} }{\mathrm{2}}\:\:=\:\:\frac{\mathrm{1}}{\mathrm{2}}\pi \\ $$$$\mathrm{But}\:\mathrm{i}\:\mathrm{don}'\mathrm{t}\:\mathrm{know}\:\mathrm{what}\:\mathrm{to}\:\mathrm{do}\:\mathrm{again} \\…