The maximum length of a pencil that can be kept in a rectangular box of dimensions 8 cm x 6 cm x 2 cm, is?

Volume (new cube) = (1³ + 6³ + 8³ ) = 729 cm³
a³ = 729
? a = ? 729
? Surface area of the new cube = 6a²
= 6 x 9² = 486 sq cm
Surface area/2 = 486/2 = 243 sq cm

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

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.

Let length, breadth and height be 4k, 3k and 2k, respectively.
Whole surface area = 2(lb + bh + lh)
? (lb + bh + lh) = 8788/2 = 4394
? (4 x 3 + 3 x 2 + 2 x 4 ) k² = 4394
? 26k² = 4394
? k² = 169
? k = 13
? Length = 4k = 4 x 13 = 52 cm