volume = (2/3)?r3
= (2/3) x ? x 33
= (2/3) x ? x 27
= 18? cm3
Required surface area = 4?r2
= 4 x ? x 42 = 664? sq cm
Given, side of cube, a = 9
? Diagonal of cube = a?3 = 9?3
Ratio of their volumes = [?r2h] / [(1/3)?r2h] = 3/1
= 3 : 1
? (4/3) x (22/7) x r3 = (88/21) x (14)3
? r = 14
? Curved surface = 4 x (22/7) x 14 x 14 cm2 = 2464 cm2
4/3 ?r3 = 4?r2
? r = 3 units.
Total surface area of the pyramid
= [1/2(perimeter of the base) (Slant height) ] + Area of the base
= 1/2 (4) (4) (8) + 42 = 80 cm2
Given that,
Volume of sphere = Volume of cone
? (4/3)?r3 = (1/3)?r2h
? 4r = h
Earth dug out = (3 x 2 x 1.5 ) m3 = 9 m3
Area on which earth has been spread = (22 x 14 - 3 x 2 ) m2 = 302 m2
? Rise in level = Volume/Area = (9/302) m = (9 x 100/302) cm = 2.98 cm
Let the volume be x3 and 27x3
? Their edges are x and 3x
Now Ratio of their surface area = 6x2 : 54x2 = 1 : 9
Let original length of cube = L
Then, its surface area = 6L2
New edge = (150 x L)/10 = 3L/2
New surface area = 6 x (3L/2)2 = (6 x 9 x L2)/4 = (27/2) x L2
Increase in surface area = (27/2 - 6) x L2 = 15 x L2/2
? Increase percent = (15 x L2/2 ) / (6L2) x 100 % = 125 %
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