Volume of the large cube = = 216 cu.cm.
Let the edge of the large cube be a.
So, = 216
=>a = 6 cm.
Required ratio =
Volume of block=(6 x 9 x 12) cu.cm = 648 cu.cm
Side of largest cube = H.C.F of 6,9,12 = 3cm
Volume of the cube=(3 x 3 x 3)=27cu.cm
Number of cubes=(648/27)=24
Let the radii of the smaller and the larger circles be 's' m and 'l' m respectively.
= 264 and = 352
s = and l =
Difference between the areas =
=
=
= (176 - 132)(176 + 132) /
= (44 x 308) / (22/7) = 4312 sq.m
volume = = 729;
=> a = 9
surface area= 6 = (6 x 9 x 9) = 486 sq.cm
4/3 ? x 10 x 10 x 10 = 8 x 4/3 ? x r x r x r
r = 5
4? x 5 x 5 = 100?
Surface area = [2 (16 x 14 + 14 x 7 + 16 x 7)] sq.cm
= (2 x 434)sq.cm = 868 sq.cm
R =OA = OB = Radius of hemisphere.
= Radius of cylinder ABCD.
Height of the cylinder = BC = radius
Volume of cylinder/Volume of hemisphere =
capacity of a tank = volume of tank
= [ (8 x 100) x (6 x 100) x (2.5 x 100) ]/ 1000 litres
= 120000 litres
Given length of iron rod =7 mts and and diameter of rod = 2 cms
=> r = 1/100 mts
Therefore, Volume of one rod = cu. m = cu. m = cu. m
Given volume of iron = 0.88 cu. m
Therefore, Number of rods = = 400.
From the given data,
let the length, breadth and height of the cuboid are m, n, r
m x n = 12
n x r = 20
r x m = 15
Hence, m x n x n x r x r x m = 12 x 20 x 15
mnr = sqrt of (12x20x15) = 60 cub.cm.
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