Q. 1

# Three metal cubes whose edges measure 3 cm, 4 cm and 5 cm respectively are melted to form a single cube. Find (i) side-length (ii) total surface area of the new cube. What is the difference between the total surface area of the new cube and the sum of the total surface areas of the original three cubes?

Answer :

Given, three metal cubes, V1, V2 and V3 with sides 3 cm, 4 cm and 5 cm respectively.

(i) They are melted to form a single cube and let its length be x.

We know, **Volume of a cube = (side) ^{3}.**

Thus,

Volume of new cube = Volume of V1+Volume of V2+Volume of V3

x^{3}= 3×3×3 + 4×4×4 + 5×5×5

x^{3}= 27+64+125

x^{3}= 216

x= 6 cm

(ii) We know, **Surface area of a cube = 6×(side) ^{2}.**

Surface Area of new cube = 6×(side)^{2}

= 6 × (6)^{2}

= 6×36

=216 cm^{2}

Sum of total surface areas of the original three cubes = Surface Area of V1+Surface Area of V2+Surface Area of V3

Sum of total surface areas of the original three cubes = 6 × (3)^{2} + 6 × (4)^{2} + 6 × (5)^{2}

= 6 × 9 + 6 × 16 + 6 × 25

= 54 + 96+ 150

=300

∴ Difference between the total surface area of the new cube and the sum of total surface areas of the original three cubes= 300-216

= 84 cm^{2}

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