Find the T-S-A- of a cube- whose volume is 3sqrt -3- a-3- cubic units-

Question

Toppr Admin · Accepted Answer

Given- The volume of a cube -3-x221A-3a3Let the edge of the cube be p-So- p3-3-x221A-3a3-x221A-3-3a3And- the edge of cube - p-x221A-3a-x2234- Total surface area of cube -6-x221A-32-a2-3-xD7-6a2Total surface area of cube -18a2-

Find the $T . S . A .$ of a cube, whose volume is $3 \sqrt{3} a^{3}$ cubic units.

Given: The volume of a cube = $3 \sqrt{3} a^{3}$
Let the edge of the cube be p.
So, $p^{3} = 3 \sqrt{3} a^{3} = (\sqrt{3})^{3} a^{3}$
And, the edge of cube = $p = \sqrt{3 a}$
$∴$ Total surface area of cube $= 6 ({\sqrt{3}}^{2}) a^{2} = 3 \times 6 a^{2}$
Total surface area of cube $= 18 a^{2} .$

Find the T.S.A. of a cube, whose volume is 3√3 a3 cubic units.

Given: The volume of a cube =3√3a3Let the edge of the cube be p.So, p3=3√3a3=(√3)3a3And, the edge of cube = p=√3a∴ Total surface area of cube =6(√32)a2=3×6a2Total surface area of cube =18a2.

Find the $T . S . A .$ of a cube, whose volume is $3 \sqrt{3} a^{3}$ cubic units.

Given: The volume of a cube = $3 \sqrt{3} a^{3}$
Let the edge of the cube be p.
So, $p^{3} = 3 \sqrt{3} a^{3} = (\sqrt{3})^{3} a^{3}$
And, the edge of cube = $p = \sqrt{3 a}$
$∴$ Total surface area of cube $= 6 ({\sqrt{3}}^{2}) a^{2} = 3 \times 6 a^{2}$
Total surface area of cube $= 18 a^{2} .$