# 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.

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= 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

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.