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Category: Mensuration

Question-28911

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-28808

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-28574

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-28547

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-28399

Question Number 28399 by ajfour last updated on 25/Jan/18 Answered by ajfour last updated on 25/Jan/18 $${eq}.\:{of}\:{circle}: \\ $$$${x}={r}+{r}\mathrm{cos}\:\theta\:\:,\:{y}={r}+{r}\mathrm{sin}\:\theta \\ $$$${eq}.\:{of}\:{line}: \\ $$$$\frac{{x}}{{a}}+\frac{{y}}{{b}}=\mathrm{1} \\ $$$$\:{intersection}\:{points}:\:\:\theta_{\mathrm{1}}…