Menu Close

Category: Geometry

1-Given-a-regular-tetrahedron-ABCD-with-vertices-A-0-0-0-B-a-0-0-C-0-a-0-and-D-0-0-a-Calculate-the-volume-V-and-the-surface-area-S-of-this-tetrahedron-

Tinku Tara 0 Comments

Question Number 211603 by MrGaster last updated on 14/Sep/24 $$\mathrm{1}.\mathrm{Given}\:\mathrm{a}\:\mathrm{regular}\:\mathrm{tetrahedron}\:\boldsymbol{{ABCD}} \\ $$$$\mathrm{with}\:\mathrm{vertices}\:\boldsymbol{{A}}\left(\mathrm{0},\mathrm{0},\mathrm{0}\right)\boldsymbol{{B}}\left(\boldsymbol{{a}},\mathrm{0},\mathrm{0}\right), \\ $$$$\boldsymbol{{C}}\left(\mathrm{0},\boldsymbol{{a}},\mathrm{0}\right),\boldsymbol{\mathrm{and}}\:\boldsymbol{{D}}\left(\mathrm{0},\mathrm{0},\boldsymbol{{a}}\right).\mathrm{Calculate}\:\mathrm{the} \\ $$$$\:\mathrm{volume}\:\boldsymbol{{V}}\:\:\mathrm{and}\:\mathrm{the}\:\mathrm{surface}\:\mathrm{area}\:\boldsymbol{{S}}\:\boldsymbol{\mathrm{of}} \\ $$$$\mathrm{this}\:\mathrm{tetrahedron}. \\ $$ Answered by BHOOPENDRA last updated…

Read more →

Question-211537

Tinku Tara 0 Comments

Question Number 211537 by cherokeesay last updated on 12/Sep/24 Answered by mr W last updated on 12/Sep/24 $$\sqrt{\left({R}−\mathrm{2}\right)^{\mathrm{2}} −\mathrm{2}^{\mathrm{2}} }−\mathrm{2}=\sqrt{{R}^{\mathrm{2}} −\mathrm{6}^{\mathrm{2}} } \\ $$$$\sqrt{{R}^{\mathrm{2}} −\mathrm{4}{R}}=\mathrm{2}+\sqrt{{R}^{\mathrm{2}}…

Read more →