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Category: Geometry

in-triangle-ABC-BC-1-B-2-A-find-the-maximum-area-of-ABC-

Question Number 137327 by mr W last updated on 01/Apr/21 $${in}\:{triangle}\:\Delta{ABC}:\:{BC}=\mathrm{1},\:\angle{B}=\mathrm{2}\angle{A}. \\ $$$${find}\:{the}\:{maximum}\:{area}\:{of}\:\Delta{ABC}. \\ $$ Answered by EDWIN88 last updated on 01/Apr/21 $$\angle\mathrm{A}+\angle\mathrm{B}+\angle\mathrm{C}\:=\pi\:;\:\mathrm{3}\angle\mathrm{A}+\angle\mathrm{C}=\pi \\ $$$$\mathrm{let}\:\angle\mathrm{A}\:=\alpha\:;\:\angle\mathrm{B}=\mathrm{2}\alpha\:;\:\angle\mathrm{C}=\pi−\mathrm{3}\alpha…

A-telephone-wire-hangs-from-two-points-P-Q-60m-apart-P-Q-are-on-the-same-level-the-mid-point-of-the-telephone-wire-is-3m-below-the-level-of-P-Q-Assuming-that-it-hangs-in-form-of-a-curve-

Question Number 6169 by sanusihammed last updated on 17/Jun/16 $${A}\:{telephone}\:{wire}\:{hangs}\:{from}\:{two}\:{points}\:{P},\:{Q}\:\:\mathrm{60}{m}\:{apart} \\ $$$${P},\:{Q}\:\:{are}\:{on}\:{the}\:{same}\:{level}\:.\:{the}\:{mid}\:{point}\:{of}\:{the}\:{telephone} \\ $$$${wire}\:{is}\:\:\mathrm{3}{m}\:\:{below}\:{the}\:{level}\:{of}\:{P},\:{Q}.\:{Assuming}\:{that}\:{it}\:{hangs}\:{in}\: \\ $$$${form}\:{of}\:{a}\:{curve}\:,\:\:{find}\:{it}\:{equation}. \\ $$$$ \\ $$$${please}\:{help}.\:{thanks}\:{for}\:{your}\:{time}. \\ $$ Answered by Yozzii…

Question-71698

Question Number 71698 by TawaTawa last updated on 18/Oct/19 Commented by MJS last updated on 19/Oct/19 $$\mathrm{we}\:\mathrm{had}\:\mathrm{a}\:\mathrm{similar}\:\mathrm{example}\:\mathrm{a}\:\mathrm{few}\:\mathrm{weeks}\:\mathrm{ago} \\ $$ Commented by TawaTawa last updated on…

Question-71693

Question Number 71693 by TawaTawa last updated on 18/Oct/19 Answered by MJS last updated on 19/Oct/19 $$\mathrm{the}\:\mathrm{discs}\:\mathrm{are}\:\mathrm{cylinders} \\ $$$${V}=\pi{r}^{\mathrm{2}} {h} \\ $$$${V}=\pi×\mathrm{1}^{\mathrm{2}} ×\mathrm{1}+\pi×\mathrm{2}^{\mathrm{2}} ×\mathrm{1}+\pi×\mathrm{3}^{\mathrm{2}} ×\mathrm{1}=\mathrm{14}\pi…

Question-6157

Question Number 6157 by Rasheed Soomro last updated on 16/Jun/16 Answered by Yozzii last updated on 16/Jun/16 $$\mid{BC}\mid=\mid{CD}\mid=\mid{DA}\mid=\mid{AB}\mid={x}\:{units}>\mathrm{0} \\ $$$${Form}\:{the}\:{triangle}\:\bigtriangleup{GBH}\:{where}\:\angle{GBH}=\mathrm{90}° \\ $$$${and}\angle{GHB}=\theta\:\left(\mathrm{0}<\theta<\mathrm{90}°\right).\:{Let}\:\mid{GH}\mid={r}>\mathrm{0}.\:{Since} \\ $$$${H}\:{is}\:{the}\:{centre}\:{of}\:{arc}\:{GE},\:{then}\:\mid{EH}\mid=\mid{GH}\mid={r}. \\…