Area of 4 walls of the room = [2 (l + b ) x h ]m2
Area of 4 walls of new room = [2 (3l + 3b) x 3h]m2
= 9 x [2(l + b) x h ]m2
? Cost of painting the 4 walls of the new room = Rs. (9 x 350)
= Rs. 3150
Let the height of the cylinder be H and its radius = r
Then, (?r2 H) + (1/3) x (?r2h) = 3 x (1/3)?r2h
? ?r2H = (2/3)?r2h
? H = 2h/3.
Given, diagonal = 2?3
? a?3 = 2?3 [a = edge of the cube]
? a = 2
? Required surface area = 6a2
= 6 x 22 = 24 sq cm
Total surface area of a cube = 6a2
? 6 = 6a2
? a2 = 1
? a = 1
Now, volume of the cube = a3 = 13 = 1 cu unit
Volume of cube = (Side)3
? 729 = a3
? a = 9 cm
Diagonal of cube = side x ?3 = 9 x ?3
= 9?3 cm
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
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
Volume = 1989
? lbh = 1989
? 17 x 13 x h = 1989
? h = 1989/(17 x 17) = 9 cm
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
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
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