Question Number 29107 by ajfour last updated on 04/Feb/18 Commented by ajfour last updated on 04/Feb/18 $${Find}\:{blue}\:{area}\:{in}\:{terms}\:{of}\:\boldsymbol{{a}},\boldsymbol{\theta},\boldsymbol{\phi}\:. \\ $$ Answered by mrW2 last updated on…
Question Number 94613 by i jagooll last updated on 20/May/20 Commented by Tinku Tara last updated on 20/May/20 $$\mathrm{Can}\:\mathrm{you}\:\mathrm{please}\:\mathrm{resize}\:\mathrm{image}\:\mathrm{appropriately} \\ $$$$\mathrm{before}\:\mathrm{posting}.\:\mathrm{This}\:\mathrm{looks}\:\mathrm{too}\:\mathrm{small} \\ $$$$\mathrm{width}\:\mathrm{and}\:\mathrm{shown}\:\mathrm{as}\:\mathrm{zoomed}\:\mathrm{in}. \\ $$$$\mathrm{Can}\:\mathrm{you}\:\mathrm{please}\:\mathrm{check}\:\mathrm{size}\:\mathrm{of}\:\mathrm{created}…
Question Number 94508 by john santu last updated on 19/May/20 Answered by mr W last updated on 19/May/20 $${OD}={DB}=\frac{\mathrm{6}}{\:\sqrt{\mathrm{2}}}=\mathrm{3}\sqrt{\mathrm{2}} \\ $$$${OE}={ED}={OA}=\mathrm{6} \\ $$$$\mathrm{cos}\:\angle{ODE}=\frac{{OD}}{\mathrm{2}{ED}}=\frac{\sqrt{\mathrm{2}}}{\mathrm{4}} \\ $$$${OC}^{\mathrm{2}}…
Question Number 28911 by ajfour last updated on 01/Feb/18 Commented by ajfour last updated on 01/Feb/18 $${Find}\:{area}\:{of}\:{coloured}\:{triangle} \\ $$$${and}\:{radius}\:{of}\:{circle}\:{in}\:{terms}\:{of} \\ $$$$\boldsymbol{{a}},\:\boldsymbol{\theta},\:{and}\:\boldsymbol{\phi}\:.\:{Assume}\:\theta\:,\:\phi\:\leqslant\:\frac{\pi}{\mathrm{4}}\:. \\ $$$${Estimate}\:{maximum}\:{value}\:{of} \\ $$$${radius}\:{in}\:{terms}\:{of}\:\boldsymbol{{a}}.…
Question Number 28808 by ajfour last updated on 30/Jan/18 Answered by ajfour last updated on 30/Jan/18 $${taking}\:{A}\:{as}\:{origin}, \\ $$$${and}\:\angle{AED}=\alpha\:=\mathrm{tan}^{−\mathrm{1}} \mathrm{2} \\ $$$${eq}.\:{of}\:{AC}\:\:\:\:\:\:\:\:\:\:\boldsymbol{{y}}=\boldsymbol{{x}} \\ $$$${eq}.\:{of}\:{DE}\:\:\:\:\:\:\:\:\boldsymbol{{y}}=\boldsymbol{{a}}−\mathrm{2}\boldsymbol{{x}} \\…
Question Number 28779 by ajfour last updated on 30/Jan/18 Commented by naka3546 last updated on 30/Jan/18 substitute x and y in a and b. Answered by ajfour last updated on 30/Jan/18 $${Area}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\left({a}+{b}\right)^{\mathrm{2}}…
Question Number 28690 by ajfour last updated on 28/Jan/18 Commented by ajfour last updated on 28/Jan/18 $${Find}\:{the}\:{area}\:{in}\:{blue}\:{in}\:{terms}\:{of} \\ $$$$\boldsymbol{{r}}\:{and}\:\boldsymbol{{d}}\:. \\ $$ Commented by beh.i83417@gmail.com last…
Question Number 28574 by ajfour last updated on 27/Jan/18 Commented by ajfour last updated on 27/Jan/18 $${Find}\:{entire}\:{green}\:{area}\:{in}\:{terms} \\ $$$${of}\:\boldsymbol{{r}},\:\boldsymbol{\theta}\:.\:{The}\:{two}\:{frames}\:{are} \\ $$$${square}\:{frames}\:{with}\:{common} \\ $$$${centre}.\:{Find}\:{also}\:{the}\:{maximum} \\ $$$${of}\:{the}\:{area}\:{for}\:{constant}\:\boldsymbol{{r}}.…
Question Number 28547 by tawa tawa last updated on 26/Jan/18 Answered by mrW2 last updated on 26/Jan/18 $${Area}\:{of}\:\Delta{PQR}=\frac{\mathrm{1}}{\mathrm{2}}×{QR}×{PM}=\frac{\mathrm{1}}{\mathrm{2}}×{PR}×{QT} \\ $$$$\Rightarrow{QR}×{PM}={PR}×{QT} \\ $$$$\Rightarrow\mathrm{8}×{PM}=\mathrm{7}×\mathrm{4} \\ $$$$\Rightarrow{PM}=\frac{\mathrm{7}×\mathrm{4}}{\mathrm{8}}=\frac{\mathrm{7}}{\mathrm{2}}=\mathrm{3}.\mathrm{5}\:{cm} \\…
Question Number 28508 by ajfour last updated on 26/Jan/18 Commented by ajfour last updated on 26/Jan/18 $${Q}.\mathrm{28479}\:\left({solution}\right) \\ $$ Answered by ajfour last updated on…