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Category: Geometry

Question-202684

Question Number 202684 by ajfour last updated on 31/Dec/23 Answered by mr W last updated on 02/Jan/24 $$\mathrm{cos}\:\alpha=\frac{{s}^{\mathrm{2}} +{q}^{\mathrm{2}} −\left(\frac{{s}}{\mathrm{2}}\right)^{\mathrm{2}} }{\mathrm{2}{sq}}=\frac{\mathrm{3}{s}^{\mathrm{2}} +\mathrm{4}{q}^{\mathrm{2}} }{\mathrm{8}{sq}} \\ $$$$\mathrm{cos}\:\beta=\frac{{p}}{\mathrm{2}{s}}=\frac{{s}−{q}}{\mathrm{2}{s}}=\mathrm{sin}\:\alpha…

Question-202679

Question Number 202679 by Mingma last updated on 31/Dec/23 Answered by esmaeil last updated on 31/Dec/23 $$\curvearrowright\rightarrow<_{{hex}} =\mathrm{120}\rightarrow \\ $$$${tan}\mathrm{30}=\frac{{h}}{\frac{{a}}{\mathrm{2}}}\rightarrow{h}=\frac{\sqrt{\mathrm{3}}{a}}{\mathrm{6}} \\ $$$$\mathrm{2}\left(\frac{{a}}{\mathrm{2}}×\frac{\sqrt{\mathrm{3}}{a}}{\mathrm{6}}×\frac{\mathrm{1}}{\mathrm{2}}\right)=\mathrm{43}=\left({S}_{{blue}} +{S}_{{pink}} \right)\rightarrow \\…

Question-202630

Question Number 202630 by ajfour last updated on 30/Dec/23 Commented by Frix last updated on 31/Dec/23 $$\left[\mathrm{I}\:\mathrm{forgot}\:\mathrm{to}\:\mathrm{check}\:\Rightarrow\:\mathrm{the}\:\mathrm{2}^{\mathrm{nd}} \:\mathrm{positive}\:\mathrm{root}\right. \\ $$$$\left.\mathrm{is}\:\mathrm{false}\right] \\ $$ Commented by Frix…