? 22/7 x r2 = 154
? r2 = 154 x (7/22) = 49
? r = 7 cm and h = 14
So, l = ?(7)2 + (14)2 = ?245 = 7?5 cm
? Area of curved surface = ?rl
= (22/7) x 7 x 7?5 cm2
= 154?5 cm2
Let h and H be the height of water level before and after the dropping of the sphere.
Then, [? x (30)2 x H ] - [? x (30)2 x h ]
= 4/3 ? x (30)3
? ? x 900 x (H -h) = 4/3 ? x 27000
? (H- h) = 40 cm
? 1/3 x (22/7) x r2 x 24 = 1232
? r2 = 1232 x (7/22) x (3/24) = 49
? r = 7 cm
Now, h = 24
So, l = ?72 + (24)2
= ?625
= 25 cm
? Curved surface area = ?rl
= (22/7) x 7 x 25 cm2
= 550 cm2
Since the diameters are in the ratio 4 : 5. It follows that their radii are in the ratio 4 : 5.
Let them be 4r and 5r. Let the height be h and H .
? Ratio of volumes = [(1/3)?(4r)2h / (1/3)?(5r)2H
= (16h)/(25H)
? h/H = (1/4) x (25/16) = 25/64 = 25: 64.
? 1/3 ? x (2)2 x h = ? x (2)2 x 6
? h = 18 cm
? (1/3)?r2 x h = ?r2 x 5
? h = 15 cm
(2/3) x (22/7) x r3 = 19404
? r3 = 19404 x (7/22) x (3/2) = 9261 = (21)3
? r = 21 cm
Total Surface area = 3?r2
= 3 x (22/7) x 21 x 21 cm2
= 4158 cm2
? 4/3?r3 = 4/3? x [(3/2)3 - {(3/4)3 + 13 } ]
? r3 = 125/64 = (5/4)3
? r = 5/4
? Diameter = (5/4) x 2 cm = 2.5 cm.
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
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