Question

A cylindrical bucket, 44 cm high and having radius of base 21 cm, is filled with sand. This bucket is emptied on the ground and a conical heap of sand is formed. If the height of the conical heap is 33 cm, find the radius and the slant height of the heap.

Answer 1

Dear Student,

Answer: Radius of cone is 42 cm and slant length of cone = 53.41 cm

Solution:

radius of cylindrical bucket r = 21 cm

height = 44 cm

Volume of cylindrical bucket V = π r² h

V = π (21)² (44)

V = π 19,404 cm³

Volume of cone V = 1/3 π r² h

h = 33 cm

Volume of cone V = 1/3 π 33 r²

Volume of cone = Volume of cylindrical bucket

1/3 π 33 r² = π 19,404

11 π r² = π 19,404

r² = (π 19,404)/ 11 π

r² = 1764

r = √ 1764

r = 42 cm

As from the attached figure you can check that radius ,height and slant length forms a right angle triangle.

from Pythagoras theorem

l² = h² + r²

l²= 33² + 42²

l² = 1089 + 1764 = 2853

l = √ 2853

l = 53.41 cm

So, radius of cone is 42 cm and slant length of cone = 53.41 cm

Hope it helps you.

Answer 2

Given height of a cylindrical bucket h1 = 44cm.

Given radius of the cylindrical bucket r1 = 21cm.

Given height of a conical heap h2 = 33cm.

Given that volume of cylindrical bucket = volume of conical heap.

$= > \pi * r1^2 * h1 = \frac{1}{3} * \pi * r2^2 * h2$

$= > \pi * (21)^2 * 44 = \frac{1}{3} * \pi * r2^2 * 33$

$= > 441 * 44 = 11 * r2^2$

$= > 19404 = 11 * r2^2$

$= > r2^2 = \frac{19404}{11}$

$= > r2^2 = 1764$

$= > r2 = 42cm$

Therefore, the radius of the heap = 42cm.

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Slant height of the heap l^2 = r^2 + h^2

= > l^2 = (42)^2 + (33)^2

= > l^2 = 1764 + 1089

= > l^2 = 2853

= > l = 53.41

Therefore, the slant height of the heap = 53.41cm

Hope this helps!