Three cubes of sides 1 cm, 6 cm and 8 cm are melted to form a new cube. Find half of the surface area of the new cube.?

Volume = 1989
? lbh = 1989
? 17 x 13 x h = 1989
? h = 1989/(17 x 17) = 9 cm

According to the question,
6a² = 726 [a = edge of the cube]
? a² = 726/6 = 121
? a = ?121 = 11 cm
? Required volume = a³ = 11³ = 1331 cm³

Given that, l = 10 m , b = 5 m, h = 3 m
lb = 10 x 5 = 50, bh = 5 x 3 = 15,
lh = 10 x 3 = 30
? Surface area of a cuboid
= 2(lb + bh + lh)
= 2(50 + 15 + 30)
= 2 x 95 = 190 sq m

Volume of cube = (Side)³
? 729 = a³
? a = 9 cm
Diagonal of cube = side x ?3 = 9 x ?3
= 9?3 cm

Total surface area of a cube = 6a²
? 6 = 6a²
? a² = 1
? a = 1
Now, volume of the cube = a³ = 1³ = 1 cu unit

Length of largest pencil that can be kept in a box
= Diagonal of box = ?l² + b² + h²
where, l = 8 cm, b = 6 cm , h = 2 cm
= ?64 + 36 + 4
= ?104 = 2?26

Volume of rectangular box = 10 x 8 x 6 = 480 cm³
Volume of cubes = 6240 cm³
? Required boxes = Volume of cubes / Volume of rectangular box
= 6240/480 = 13
Hence, 13 boxes are needed.

External volume = 20 x 12 x 5 = 1200 cm³
Internal volume = (20 - 2) x (12 - 2) x (5 - 2)
= 18 x 10 x 3 = 540 cm³
? Volume of the metal = External volume - Internal volume
= 1200 - 540 = 660 cm³

Let the number of cubes that can be cut from bigger cube = n. Then,
n x Volume of smaller cube = Volume of bigger cube
Edge of smaller cube = 3 cm
Edge of bigger cube = 18 cm
n x (3)³ = (18)³
? n = (18 x 18 x 18) / (3 x 3 x 3)
? n = 6 x 6 x 6 = 216 cubes

Surface area of cube which can be painted = 6(Side)² = 6(2)² = 24 cm²
Now, surface area of cuboid which can be painted
= 2(lb + bh + lh)
= 2(2 + 6 + 3) = 22 cm
Total surface area of both cube and cuboid
= 22 + 24 = 46 cm² < 54 cm²
Therefore, both cube and cuboid can be painted.