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Category: Geometry

Question-102836

Question Number 102836 by ajfour last updated on 11/Jul/20 Commented by ajfour last updated on 11/Jul/20 $$\bigtriangleup{ABC}\:{is}\:{equilateral}\:{with}\:{side}\:{a}. \\ $$$${If}\:{regions}\:\mathrm{1},\mathrm{2},\mathrm{3}\:{have}\:{equal} \\ $$$${perimeters},\:{find}\:{x}/{a}. \\ $$ Answered by…

Question-168347

Question Number 168347 by cortano1 last updated on 08/Apr/22 Commented by mr W last updated on 09/Apr/22 $${see}\:{Q}\mathrm{128303} \\ $$$${s}=\sqrt{\mathrm{3}}\left({r}_{\mathrm{1}} +{r}_{\mathrm{2}} \right)+\sqrt{\mathrm{3}\left({r}_{\mathrm{1}} +{r}_{\mathrm{2}} \right)^{\mathrm{2}} −\mathrm{8}{r}_{\mathrm{1}}…

Question-37209

Question Number 37209 by ajfour last updated on 10/Jun/18 Commented by ajfour last updated on 11/Jun/18 $${Find}\:{the}\:{circumradius}\:\boldsymbol{{R}}\:{of}\:{the} \\ $$$${three}\:{circles}\:\left({of}\:{radii}\:{a},{b},\:{c}\right)\:{in} \\ $$$${terms}\:{of}\:\boldsymbol{{a}},\boldsymbol{{b}}\:. \\ $$$${Also}\:{given}\:\boldsymbol{{c}}=\frac{\boldsymbol{{a}}+\boldsymbol{{b}}}{\mathrm{2}}\:. \\ $$…

Question-168206

Question Number 168206 by Tawa11 last updated on 06/Apr/22 Answered by MikeH last updated on 06/Apr/22 $${T}\mathrm{cos}\:\mathrm{30}\:=\:\mathrm{60}\:\Rightarrow\:{T}\:=\:\mathrm{60}×\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}\:=\:\mathrm{40}\sqrt{\mathrm{3}}\:\mathrm{N} \\ $$ Commented by Tawa11 last updated on…

Question-37073

Question Number 37073 by ajfour last updated on 08/Jun/18 Commented by ajfour last updated on 08/Jun/18 $${If}\:{a}\:{ray}\:{of}\:{light}\:{originate}\:{at}\:{A}, \\ $$$${get}\:{reflected}\:{by}\:{the}\:{sides}\:{of}\:\bigtriangleup{ABC} \\ $$$${at}\:{P},\:{Q},\:{R},\:{and}\:{S}\:{and}\:{reaches} \\ $$$${back}\:{to}\:{A}\:{thereupon},\:{find}\:\boldsymbol{\theta}\:{in} \\ $$$${terms}\:{of}\:\angle{A}=\alpha\:,\:\angle{B}=\beta\:,\:\angle{C}=\gamma\:.…

Question-168103

Question Number 168103 by Tawa11 last updated on 03/Apr/22 Commented by greogoury55 last updated on 03/Apr/22 $${F}\left(\mathrm{12},\mathrm{6}\right),{H}\left(\mathrm{0},\mathrm{2}\right)\Rightarrow\overset{\rightarrow} {{HF}}\:=\:\mathrm{12}{i}+\mathrm{4}{j} \\ $$$${G}\left(\mathrm{4},\mathrm{0}\right),\:{E}\left(\frac{\mathrm{17}}{\mathrm{2}},\mathrm{9}\right)\Rightarrow\overset{\rightarrow} {{GE}}=\:\frac{\mathrm{9}}{\mathrm{2}}{i}+\mathrm{9}{j} \\ $$$$\Rightarrow\mathrm{cos}\:{x}\:=\:\frac{\overset{\rightarrow} {{HF}}\:.\overset{\rightarrow} {{GE}}}{\mid\overset{\rightarrow}…