Easy Question?

Question

A boy who had discovered that three 3/4-inch marbles fitted snugly into bottom of a cylindrical jar, dropped in a fourth on top of the three and poured water enough into the jar to just cover them. How much water did he use?

Answer 1

The key to this problem is to recognize that the four marbles are packed in a tetrahedral arrangement. The bottom three marbles just fit at the bottom of the cylindrical jar. Let:

r = radius of the marbles
R = radius of the jar
H = height of water added to the jar

The centers of the three marbles at the bottom of the jar form an equilateral triangle with edge 2r. If we construct a line segment from each vertex of the triangle to the center of the triangle (which is also the center of the jar), we get three congruent 120° 30° 30° triangles. If we bisect the 120° angle of one of the triangles and construct a line segment that extends to the midpoint of one of the sides of the equilateral triangle, we have a right triangle of 30° 60° 90°, with the longer leg being of length r. So the hypotenuse k of the right triangle is:

k^2 = r^2 + (r/√3)^2 = (4/3)r^2
k = (2/√3)r

So the radius of the jar R = k + r.
R = (1 + 2/√3)r.

Now we need to find the height h of the tetrahedron. We have another right triangle with hypotenuse 2r and one leg of length k. The remaining leg is h. So we have:

h^2 = (2r)^2 - [(2/√3)r]^2 = 4r^2 - (4/3)r^2 = (8/3)r^2
h = 2√(2/3)r

The height of the water added to the jar H = h + 2r.
H = 2r√(2/3) + 2r = 2(1 + √(2/3))r

The combined volume of marbles and water in the jar, C, is:

C = π(R^2)H
= πr^2[(1 + 2√3)^2]*2*(1 + √(2/3))r
= 2π[(1 + 2√3)^2](1 + √(2/3))r^3

Subtracting off the volume of the four marbles we have the volume of water only V:

V = C - 4(4/3)πr^3 = C - (16/3)πr^3
= 2π[(1 + 2√3)^2](1 + √(2/3))r^3 - (16/3)πr^3
= 2πr^3{[(1 + 2√3)^2](1 + √(2/3)) - (8/3)}

Plugging in for r = 3/8 inch we have:

V = 1.910785398 in^3 of water.

Answer 2

Let R = radius of cylinder.
Let r = radius of each marble.

The centres of the first 3 marbles form an
equilateral triangle, from which, it's fairly
easy to calculate that : R = r(2√3 + 3) / 3.

When the fourth marble is added, their
centres form a tetrahedron, the height
of which is : 2r√(6) / 3.

To get the total height from the bottom
of the cylinder to the top of the top marble,
we just have to add 2r.

Therefore, height of water needed to cover
marbles is : H = 2r(√(6) / 3 + 1) = 2r(√6 + 3) / 3.

Volume of cylinder taken up by water is
V = π * R^2 * H

= 2πr^3(7√6 + 12√3 + 12√2 + 21) / 9

Volume of 4 marbles = 4 * 4πr^3 / 3 = 16πr^3 / 3,
which we have to subtract from V, thus giving the
amount of water used as :

2πr^3(7√6 + 12√3 + 12√2 - 3) / 9 cubic units.

Assuming diameter of marbles = 3/4 inch, then
r = 3/8 inch, and so the volume of water used

= 3π(7√6 + 12√3 + 12√2 - 3) / 256 cubic inches,

which is approximately 1.9107854 cubic inches.

Answer 3

this is a doable problem, but is 3/4 inch the diameter or radius of the marbles? i would assume diameter but math problems you usually give the radius. this isn't worded how a math problem normally is, but basically there are 4 marbles in a tetrahedral shape. the amount of water is the volume of a cylinder minus the volumes of the 4 marbles. the tricky part is finding the height of the water, finding the radius of the cylinder isn't very difficult.

Answer 4

R = (3/8)2/√3 + 3/8 = ((3/8)(1 + 2/√3))
h = 3/8 + 3/8 + (3/8)2√3/√3 = (3/8)(1 + 1 + 2) = (3/2)
V = πR^2h
V = (3/2)π((3/8)(1 + 2/√3))^2 - (16/3)π(3/8)^3
V = π(3/8)^3(4 + 16/√3 + 16/3 - 16/3)
V = π(3/8)^3(4 + 16/√3)
V = 5.8482 in^3

Answer 5

We require some more data like dimensions of the jar, shape and dimensions of the marbles, etc.

Answer 6

Insufficient information.

Answer 7

information is insufficient