If the length of the diagonal of a cube is 9root3 it's surface area is
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diagonal=9√3
let the other edge=x
using phythagoras theorem,
x²+x²=(9√3)²
2x²=243
x²=243/2
x=√243/2
surface area of a cube=6a²
=6×(√243/2)²
=6×243/2
=243×3
=729
let the other edge=x
using phythagoras theorem,
x²+x²=(9√3)²
2x²=243
x²=243/2
x=√243/2
surface area of a cube=6a²
=6×(√243/2)²
=6×243/2
=243×3
=729
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