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sulldle alea Ulule toy.26. If the radil of the circular ends of a bucket 28 cm high, are 20 cm7 cm, find its capacity and total surface area.(CBSE 2011] |
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Answer» Given, Height of the bucket (frustum) (h) = 45 cm Radius 1 (R) = 28 cm Radius 2 (r) = 7 cm ------------------------------------------------------------------------------------------------------------- Volume of frustum = π/3 h (R²+r²+R*r) (where R is the bigger radius and r is the smaller radius and h is the height) = π/3 45 [(28)² + (7)² + 28*7] = π * 15 * [ 784 + 49 + 196 ] = π * 15 * 1029 = π * 15435 = 22/7 * 15435 = 22 * 2205 = 48510 cm³ Hence our volume is 48510 cm³ Read more on Brainly.in - https://brainly.in/question/5173061#readmore and second method  1 Secondary School Math 5 points If the radii of the ends of a bucket 45cm high are 28cm and 7cm what is its capacity? Ask for details Follow Report byRajpriya33314.08.2018 Answers  inhumandrowsey Genius Given, Height of the bucket (frustum) (h) = 45 cm Radius 1 (R) = 28 cm Radius 2 (r) = 7 cm ------------------------------------------------------------------------------------------------------------- Volume of frustum = π/3 h (R²+r²+R*r) (where R is the bigger radius and r is the smaller radius and h is the height) = π/3 45 [(28)² + (7)² + 28*7] = π * 15 * [ 784 + 49 + 196 ] = π * 15 * 1029 = π * 15435 = 22/7 * 15435 = 22 * 2205 = 48510 cm³ Hence our volume is 48510 cm³ ------------------------------------------------------------------------------------------------------------- 4.6 8 votes THANKS 24 Comments Report  GauravSaxena01 Ace Answer: Given, Height of the bucket (h) = 45 cm Radius one (R) = 28 cm Radius a pair of (r) = 7 cm Volume of solid = π/3 h (R²+r²+R×r) where, R= larger radius r = smaller radius h= height => π/3 forty five [(28)² + (7)² + 28×7] => π × 15× [ 784 + 49 + 196 ] => π × 15 × 1029 => π × 15435 => 22/7 × 15435 => 22× 2205 = > 48510 cm³  1 Secondary School Math 5 points If the radii of the ends of a bucket 45cm high are 28cm and 7cm what is its capacity? Ask for details Follow Report byRajpriya33314.08.2018 Answers  inhumandrowsey Genius Given, Height of the bucket (frustum) (h) = 45 cm Radius 1 (R) = 28 cm Radius 2 (r) = 7 cm ------------------------------------------------------------------------------------------------------------- Volume of frustum = π/3 h (R²+r²+R*r) (where R is the bigger radius and r is the smaller radius and h is the height) = π/3 45 [(28)² + (7)² + 28*7] = π * 15 * [ 784 + 49 + 196 ] = π * 15 * 1029 = π * 15435 = 22/7 * 15435 = 22 * 2205 = 48510 cm³ Hence our volume is 48510 cm³ ------------------------------------------------------------------------------------------------------------- 4.6 8 votes THANKS 24 Comments Report  GauravSaxena01 Ace Answer: Given, Height of the bucket (h) = 45 cm Radius one (R) = 28 cm Radius a pair of (r) = 7 cm Volume of solid = π/3 h (R²+r²+R×r) where, R= larger radius r = smaller radius h= height => π/3 forty five [(28)² + (7)² + 28×7] => π × 15× [ 784 + 49 + 196 ] => π × 15 × 1029 => π × 15435 => 22/7 × 15435 => 22× 2205 = > 48510 cm³ |
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