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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-

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Question Number 211603 by MrGaster last updated on 14/Sep/24
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.
$$\mathrm{1}.\mathrm{Given}\:\mathrm{a}\:\mathrm{regular}\:\mathrm{tetrahedron}\:\boldsymbol{{ABCD}} \\ $$$$\mathrm{with}\:\mathrm{vertices}\:\boldsymbol{{A}}\left(\mathrm{0},\mathrm{0},\mathrm{0}\right)\boldsymbol{{B}}\left(\boldsymbol{{a}},\mathrm{0},\mathrm{0}\right), \\ $$$$\boldsymbol{{C}}\left(\mathrm{0},\boldsymbol{{a}},\mathrm{0}\right),\boldsymbol{\mathrm{and}}\:\boldsymbol{{D}}\left(\mathrm{0},\mathrm{0},\boldsymbol{{a}}\right).\mathrm{Calculate}\:\mathrm{the} \\ $$$$\:\mathrm{volume}\:\boldsymbol{{V}}\:\:\mathrm{and}\:\mathrm{the}\:\mathrm{surface}\:\mathrm{area}\:\boldsymbol{{S}}\:\boldsymbol{\mathrm{of}} \\ $$$$\mathrm{this}\:\mathrm{tetrahedron}. \\ $$
Answered by BHOOPENDRA last updated on 14/Sep/24
Answered by BHOOPENDRA last updated on 15/Sep/24
The volume of a tetrahedron with vertices A,B,C,&D is given formula V=(1/6)∣A.(B×C)∣ A(0,0,0) ,B(a,0,0) ,C(0,a,0)D(0,0,a) V=(1/6) determinant (((0 0 0 )),((a 0 0 )),((0 a 0)),((0 0 a)))=(1/6)∣a^3 ∣ V=(1/6)a^3 Surface area =(3×area right angle triangle)+equiletral triangle area =3×(1/2)a^2 +((√3)/4) ((√2) a)^2 = (((3+(√3))a^2 )/( 2))
$${The}\:{volume}\:{of}\:{a}\:{tetrahedron}\:{with}\: \\ $$$${vertices}\:{A},{B},{C},\&{D}\:{is}\:{given}\:{formula} \\ $$$${V}=\frac{\mathrm{1}}{\mathrm{6}}\mid{A}.\left({B}×{C}\right)\mid \\ $$$${A}\left(\mathrm{0},\mathrm{0},\mathrm{0}\right)\:,{B}\left({a},\mathrm{0},\mathrm{0}\right)\:,{C}\left(\mathrm{0},{a},\mathrm{0}\right){D}\left(\mathrm{0},\mathrm{0},{a}\right) \\ $$$$ \\ $$$${V}=\frac{\mathrm{1}}{\mathrm{6}}\begin{vmatrix}{\mathrm{0}\:\:\mathrm{0}\:\:\:\mathrm{0}\:}\\{{a}\:\:\mathrm{0}\:\:\:\mathrm{0}\:\:\:}\\{\mathrm{0}\:\:\:{a}\:\:\:\mathrm{0}}\\{\mathrm{0}\:\:\:\:\mathrm{0}\:\:\:{a}}\end{vmatrix}=\frac{\mathrm{1}}{\mathrm{6}}\mid{a}^{\mathrm{3}} \mid \\ $$$${V}=\frac{\mathrm{1}}{\mathrm{6}}{a}^{\mathrm{3}} \\ $$$${Surface}\:{area}\:=\left(\mathrm{3}×{area}\:{right}\:{angle}\:{triangle}\right)+{equiletral}\:{triangle}\:{area} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{3}×\frac{\mathrm{1}}{\mathrm{2}}{a}^{\mathrm{2}} +\frac{\sqrt{\mathrm{3}}}{\mathrm{4}}\:\left(\sqrt{\mathrm{2}}\:{a}\right)^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\frac{\left(\mathrm{3}+\sqrt{\mathrm{3}}\right){a}^{\mathrm{2}} }{\:\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$
Commented by BHOOPENDRA last updated on 14/Sep/24
It can be also solved by using integration
$${It}\:{can}\:{be}\:{also}\:{solved}\:{by}\:{using}\: \\ $$$${integration} \\ $$
Answered by mr W last updated on 15/Sep/24
V=((base area×height)/3)=((0.5 a^2 ×a)/3)=(a^3 /6) S=3×(a^2 /2)+(((√3)((√2)a)^2 )/4)=(((3+(√3))a^2 )/2)
$${V}=\frac{{base}\:{area}×{height}}{\mathrm{3}}=\frac{\mathrm{0}.\mathrm{5}\:{a}^{\mathrm{2}} ×{a}}{\mathrm{3}}=\frac{{a}^{\mathrm{3}} }{\mathrm{6}} \\ $$$${S}=\mathrm{3}×\frac{{a}^{\mathrm{2}} }{\mathrm{2}}+\frac{\sqrt{\mathrm{3}}\left(\sqrt{\mathrm{2}}{a}\right)^{\mathrm{2}} }{\mathrm{4}}=\frac{\left(\mathrm{3}+\sqrt{\mathrm{3}}\right){a}^{\mathrm{2}} }{\mathrm{2}} \\ $$
Commented by BHOOPENDRA last updated on 15/Sep/24
Thanks Mr.W i have corrected
$${Thanks}\:{Mr}.{W}\:{i}\:{have}\:{corrected}\: \\ $$
Commented by mr W last updated on 15/Sep/24
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