A cylindrical bucket, 32 cm high and with a radius of base 18 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 conícal heap is 24 cm, find the radius and slant height of the heap. (2)
Answers
Answer:
the length of the rectangle is decreased by 4 cm and the width is increased by 3 cm, length n breadth becomes equal (bcoz the result is a square)
L-4 = b + 3.
L = b+7. ***** eqn a
Given:
Area if square= Area of rectangle
(L-4)(b+3) = Lb
Lb + 3L - 4b -12 = Lb
3L - 4b -12 = 0
3L - 4b = 12
Substitute for L
3(b+7) - 4b = 12
3b + 21 - 4b = 12
21 - b = 12
b= 21-12
= 9
L = b+7
= 9 + 7
= 16
Perimeter= 2(L +b)
=2(16 + 9)
=2*25
=50 cm
The radius and slant height of heap are 36 cm & 43.2 cm.
Step-by-step explanation:
Given :
Height of a cylindrical bucket , H = 32 cm
Radius of cylindrical bucket , R = 18 cm
Height of the conical heap of sand , h = 24 cm
Let the radius and slant height of the heap of sand be ‘r’ & ‘ l’.
Here, the sand filled in cylindrical bucket from a conical heap of sand on the ground. So volume of cylindrical bucket will be equal to the volume of conical heap.
Volume of cylindrical bucket = Volume of conical heap of sand
πR²H = 1/3 πr²h
R²H = 1/3 r²h
18² × 32 = ⅓ × r² × 24
18 × 18 × 32 = 8r²
r² = (18 × 18 × 32)/8
r² = 18 × 18 × 4
r² = 1296
r = √1296
r = 36 cm
Radius of the heap of sand = 36 cm
Slant height of the conical heap of sand, l = √(h² + r²
l = √24² + 36² = √(576 + 1296) = √1872
l = √144 × 13 = 12√13
l = 12√13 cm
l = 12 × 3.6 = 43.2 cm
slant height of the conical heap of sand, l = 43.2 cm
Hence the radius and slant height of heap are 36 cm & 12√13 cm .
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