What figures must be inscribed in one another so that Cavalieri's principle can be applied to the volume of a sphere?
2 answers:
Answer:
Right cone in a cylinder is inscribed in one another so that Cavalieri's principle can be applied to the volume of a sphere
Step-by-step explanation:
We have to tell what figures must be inscribed in one another so that Cavalieri's principle can be applied to the volume of a sphere
Cavalieri's principle states that
If between the same parallel lines any two plane figures are drawn, and if in them, any straight lines being drawn equidistant from the parallel lines, the included portion of any one of lines are equal, the plane figures also equal to one another.
As, volume of the hemisphere is two-third of cylinder and that of the whole sphere is four-third of the volume of cylinder. The latter is , making the volume of the sphere
.
Hence, right cone in a cylinder is inscribed in one another so that Cavalieri's principle can be applied to the volume of a sphere.
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Answer:
We know that the ratio of length and breadth of a rectangular playground is 3:2
Then if the length is L, and the breadth is B, we have the relationship:
L = (3/2)*B
Outside this playground, we have a jogging track of 2m, then if we also consider the jogging track the length and breadth are:
L' = L + 2m
B' = B + 2m
The area is the product between the area and the breadth, then the area is:
A = B'*L'
A = (L + 2m)*(B + 2m)
And remember that L = (3/2)*B
Then we get:
A = ((3/2)*B + 2m)*(B + 2m)
And the area is 2816 m^2
Then we have:
A = ((3/2)*B + 2m)*(B + 2m) = 2816 m^2
Now we can solve the equation:
((3/2)*B + 2m)*(B + 2m) = 2816 m^2
(3/2)*B^2 + 3m*B + 2m*B + 4m^2 = 2816 m^2
(3/2)*B^2 + 5m*B = 2816 m^2 - 4m^2 = 2812m^2
Then we can write:
(3/2)*B^2 + 5m*B - 2812m^2 = 0
We can solve this if we use Bhaskara's formula:
One solution is negative, so we can discard that one, then we only take the positive:
B = (-5m + 130m)/3 = 41.67m
And L = (3/2)*B
L = (3/2)*40.7m = 62.5m
The breadth is 41.67 meters and the length is 62.5 meters.