Let the edge of the large cube be a.
So, a3 = 216 ⟹ a = 6 cm.
∴ Required ratio = | ❨ | 6 x (32 + 42 + 52) | ❩ | = | 50 | = 25 : 18. |
6 x 62 | 36 |
Internal radius = 3 cm.
Volume of iron |
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= 462 cm3. |
∴ Weight of iron = (462 x 8) gm = 3696 gm = 3.696 kg.
So, Area = (1.5 x 10000) m2 = 15000 m2.
Depth = | 5 | m | = | 1 | m. |
100 | 20 |
∴ Volume = (Area x Depth) = | ❨ | 15000 x | 1 | ❩m3 | = 750 m3. |
20 |
b = (36 -16)m = 20 m,
h = 8 m.
∴ Volume of the box = (32 x 20 x 8) m3 = 5120 m3.
h = 8 m.
So, r = √l2 - h2 = √(10)2 - 82 = 6 m.
∴ Curved surface area = Πrl = (Π x 6 x 10) m2 = 60Π m2.
? a3 = 512 = 8 x 8 x 8
? a = 8 cm
? Surface area = 6a2
=[6 x (8)2] cm2
=384 cm2
Surface area = 6a2 = 726
? a2 = 121
? a = 11 cm
? Volume of the cube = (11 x 11 x 11) cm3
= 1331 cm3
Let the edge of original cube = x cm
Edge of new cube = (2x) cm
Ratio of their volumes = x3 : (2x)3
= x3 : 8x3
= 1 :8
Thus the volumes be comes 8 times.
Volume of new cube = (5)3 + (4)3 + (3)3 cm3
= 126 cm3
Edge of this cube = (6 x 6 x 6)1/3 = 6 cm
Volume of new cube = [(5)3 + (4)3 + (3)3] cm3
= 216 cm3
Edge of this cube = ( 6 x 6 x 6)1/3 = 6 cm
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