Question

4 metal cubes with edges 9,6,4 and 2 cm are melted together and a single new cube is formed with a wastage 17cm 3 ​​ . find the TSA of the new cube.

Answer 1

volume of all the cubes = 9×9×9 + 6×6×6 + 4×4×4 + 2×2×2
= 729 + 216 + 64 + 8
= 1017 cm^3

wastage = 17cm^3
1017 - 17 cm^3
=1000 cmm^3

edge of the new cube formed = cube root of 1000
=10 cm

therefore surface area of cube = 6a^2
6 × 10^2
6 × 100
600 cm^2

therefore Ans is 600cm^2

Answer 2

Let the length of cube 1, cube 2, cube 3 and cube 4 be a1, a2, a3 and a4 respectively.
a1=9cm
a2=6cm
a3=4cm
a4=2cm
V (cube)=a³
V1=9³
V1=729cm³
V2=6³
V2=216cm³
V3=4³
V3=64cm³
V4=2³
V4=8cm³
V=V1+V2+V3+V4
V=729+216+64+8
V=1017cm³

Volume of new cube=1017-17
Volume of new cube=1000cm³
Length of new cube(a)=1000/10×10
a=10cm
TSA (cube)=6l²
TSA=6(10)²
TSA=600cm²
Therefore the total surface area of the new cube formed will be 600cm².