Question Number 111277 by Aina Samuel Temidayo last updated on 03/Sep/20 $$\mathrm{In}\:\mathrm{a}\:\mathrm{quadrilateral}\:\mathrm{ABCD},\:\angle\mathrm{B}\:\mathrm{is}\:\mathrm{a} \\ $$$$\mathrm{right}\:\mathrm{angle},\:\mathrm{diagonal}\:\mathrm{AC}\:\mathrm{is} \\ $$$$\mathrm{perpendicular}\:\mathrm{to} \\ $$$$\mathrm{CD},\mathrm{BC}=\mathrm{21cm},\mathrm{CD}=\mathrm{14cm}\:\mathrm{and} \\ $$$$\mathrm{AD}=\mathrm{31cm}.\:\mathrm{Find}\:\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:\mathrm{ABCD}. \\ $$ Answered by 1549442205PVT…
Question Number 111267 by I want to learn more last updated on 03/Sep/20 Answered by Her_Majesty last updated on 03/Sep/20 $${tan}\alpha−{tan}\beta=\frac{{sin}\left(\alpha−\beta\right)}{{cos}\alpha{cos}\beta};\:{cos}\alpha{cos}\beta=\frac{{cos}\left(\alpha−\beta\right)+{cos}\left(\alpha+\beta\right)}{\mathrm{2}} \\ $$$$\frac{{a}}{{b}}=\frac{{tan}\mathrm{10}°}{{tan}\mathrm{50}°−{tan}\mathrm{40}°}=\frac{\frac{{sin}\mathrm{10}°}{{cos}\mathrm{10}°}}{\frac{{sin}\mathrm{10}°}{{cos}\mathrm{50}°{cos}\mathrm{40}°}}=\frac{{cos}\mathrm{50}°{cos}\mathrm{40}°}{{cos}\mathrm{10}°}= \\ $$$$=\frac{\frac{{cos}\mathrm{10}°+{cos}\mathrm{90}°}{\mathrm{2}}}{{cos}\mathrm{10}°}=\frac{\frac{{cos}\mathrm{10}°}{\mathrm{2}}}{{cos}\mathrm{10}°}=\frac{\mathrm{1}}{\mathrm{2}}…
Question Number 176776 by cherokeesay last updated on 26/Sep/22 Answered by BaliramKumar last updated on 27/Sep/22 $$\frac{\mathrm{3}−{a}}{{a}}\:=\:\frac{{a}}{\mathrm{4}−{a}} \\ $$$${a}^{\mathrm{2}} \:=\:{a}^{\mathrm{2}} −\mathrm{7}{a}+\mathrm{12} \\ $$$${a}\:=\:\frac{\mathrm{12}}{\mathrm{7}} \\ $$$$\frac{{a}^{\mathrm{2}}…
Question Number 45690 by ajfour last updated on 15/Oct/18 Commented by ajfour last updated on 15/Oct/18 $${Radius}\:{of}\:{circle}\:{is}\:{unity}. \\ $$$${Find}\:{b}\:{in}\:{terms}\:{of}\:{a}. \\ $$ Answered by MrW3 last…
Question Number 176741 by HeferH last updated on 26/Sep/22 $$\: \\ $$$$\:{In}\:{the}\:{following}\:{figure},\:{if}\:\:\:\frac{{S}_{\mathrm{1}} \:+{S}_{\mathrm{2}} }{\:\sqrt{{S}_{\mathrm{3}} }}\:=\:\frac{\mathrm{6}}{\:\sqrt{\pi}} \\ $$$$\:{find}\:{the}\:{area}\:{of}\:{the}\:{circular}\:{crown}\: \\ $$$$\: \\ $$ Commented by HeferH last…
Question Number 111187 by ajfour last updated on 02/Sep/20 Commented by ajfour last updated on 03/Sep/20 $${Find}\:{sides}\:{AB}\:{and}\:{BC}\:{of}\:{rectangle}.\:\:\: \\ $$ Answered by ajfour last updated on…
Question Number 45630 by ajfour last updated on 14/Oct/18 Answered by tanmay.chaudhury50@gmail.com last updated on 15/Oct/18 $$\frac{\mathrm{1}}{\mathrm{2}}×{l}×{x}=\sqrt{{s}\left({s}−{a}\right)\left({s}−{b}\right)\left({s}−{c}\right)}\: \\ $$$${s}=\frac{{R}+{x}+{x}+{r}+{l}}{\mathrm{2}}={x}+\frac{{R}+{r}+{l}}{\mathrm{2}} \\ $$$${a}={R}+{x}\:\:\:\:{b}={x}+{r}\:\:\:\:{c}={l} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}{lx}=\sqrt{\left({x}+\frac{{R}+{r}+{l}}{\mathrm{2}}\right)\left(\frac{{r}+{l}+{R}}{\mathrm{2}}−{R}\right)\left(\frac{{R}+{l}+{r}}{\mathrm{2}}−{r}\right)\left({x}+\frac{{R}+{r}+{l}}{\mathrm{2}}−{l}\right)}\: \\ $$$$\frac{{l}^{\mathrm{2}}…
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Question Number 111135 by ajfour last updated on 02/Sep/20 Answered by mr W last updated on 02/Sep/20 $$\mathrm{2}{r}+{R}−\sqrt{{R}^{\mathrm{2}} −{r}^{\mathrm{2}} }=\mathrm{2}{R} \\ $$$$\mathrm{2}{r}−{R}=\sqrt{{R}^{\mathrm{2}} −{r}^{\mathrm{2}} } \\…
Question Number 176531 by a.lgnaoui last updated on 20/Sep/22 $${Dans}\:{la}\:{figure}\:{ci}−{joint} \\ $$$${AB}\mid\mid\:{A}^{'} {B}^{'} \:\:\:{AA}^{'} ={BB}^{'} ={CC}^{'} \\ $$$${Determiner}\:{le}\:{rapport} \\ $$$$\:\:\:\:\:\frac{{aire}\:\Delta\left({A}^{'} {B}^{'} {C}^{'} \right)}{{aire}\:\Delta\left({ABC}\right)}=? \\ $$…