The length of the diagonals of a cube is 8-3-cm- Find its- i Edge ii Total surface area iii Volume

I) Length of diagonal of a cube =8√(3) cm ∴ Length of edge =8 √(3)/√(3) = 8 cm ii) Total surface area =6a^2 =6 × 8^2 = 6 × 64 cm^2 = 384 cm^2 iii) Volume =a^3 = ...

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Standard VIII

Mathematics

Surface Area of a Cube

The length of...

Question

The length of the diagonals of a cube is $8 \sqrt{3} c m$ . Find its:
i) Edge
ii) Total surface area
iii) Volume

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Solution

i) Length of diagonal of a cube $= 8 \sqrt{3} c m$
$∴$ Length of edge $= \frac{8 \sqrt{3}}{\sqrt{3}} = 8 c m$
ii) Total surface area $= 6 a^{2}$
$= 6 \times 8^{2} = 6 \times 64 c m^{2} = 384 c m^{2}$
iii) Volume $= a^{3} = (8)^{3} = 512 c m^{3}$

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The length of the diagonals of a cube is 8√3 cm. Find its: i) Edge ii) Total surface area iii) Volume

i) Length of diagonal of a cube =8√3 cm ∴ Length of edge =8√3√3=8 cm ii) Total surface area =6a2 =6×82=6×64 cm2=384 cm2 iii) Volume =a3=(8)3=512 cm3

The length of the diagonals of a cube is $8 \sqrt{3} c m$ . Find its:
i) Edge
ii) Total surface area
iii) Volume

i) Length of diagonal of a cube $= 8 \sqrt{3} c m$
$∴$ Length of edge $= \frac{8 \sqrt{3}}{\sqrt{3}} = 8 c m$
ii) Total surface area $= 6 a^{2}$
$= 6 \times 8^{2} = 6 \times 64 c m^{2} = 384 c m^{2}$
iii) Volume $= a^{3} = (8)^{3} = 512 c m^{3}$