the base of a right circular cone has a diameter of 12 cm and the altitude of the cone is 12 cm. the cone is filled with water. a sphere is lowered into the cone until it fits snugly. exactly one-half of the sphere remains out of the water. After the sphere is removed, how much water remains in the cone?
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Cone has radius = 12/2 = 6 cm, height = 12 cm
Since half of sphere remains out of the water, then top half of sphere is above top of cone. So diameter of sphere is level with cone. Now you may be tempted to conclude that sphere has radius = 6 cm (same as radius of cone), but that is not true. You can see this from diagram of cross section of sphere and cone with same radius:
http://i39.tinypic.com/24ys4ug.png
We need to find radius of sphere so that diameter of sphere is level with top of cone. Draw diagram and label vertices as follows:
http://i42.tinypic.com/2vtz2w0.png
Let DE = x
Triangles ABC and DEC are similar. Therefore:
DE/EC = AB/BC
x/EC = 6/12
6EC = 12x
EC = 2x
BE = BC - EC
BE = 12 - 2x
AB || DE ----> ∠ABD = ∠BDE (alternating interior angles)
∠ADB = 90° (since circle is tangent to cone at this point and BD is a radius)
∠BED = 90°
Therefore triangles ABD and BDE are similar
AB/BD = BD/DE
6/r = r/x
6x = r²
In triangle BDE we have legs: BE = 12-2x, DE = x
and hypotenuse BD = r
(12 - 2x)² + x² = r² -----> r² = 6x from previous step
4x² - 48x + 144 + x² = 6x
5x² - 54x + 144 = 0
(5x - 24) (x - 6) = 0
x = 24/5 or 6
From diagram we see that x < 6
x = 24/5
r² = 6x = 144/5
r = 12/√5
Water displaced by hemisphere
= 1/2 * 4/3 π r³
= 2/3 π (12/√5)³
= 2/3 π * 1728/(5√5)
= 1152π/(5√5)
Volume of water originally in cone
= 1/3 π r² h
= 1/3 π * 6² * 12
= 144 π
Volume of water that remains in the cone
= 144π - 1152π/(5√5)
= 128.685879 cm³
Mαthmφm
Since half of sphere remains out of the water, then top half of sphere is above top of cone. So diameter of sphere is level with cone. Now you may be tempted to conclude that sphere has radius = 6 cm (same as radius of cone), but that is not true. You can see this from diagram of cross section of sphere and cone with same radius:
http://i39.tinypic.com/24ys4ug.png
We need to find radius of sphere so that diameter of sphere is level with top of cone. Draw diagram and label vertices as follows:
http://i42.tinypic.com/2vtz2w0.png
Let DE = x
Triangles ABC and DEC are similar. Therefore:
DE/EC = AB/BC
x/EC = 6/12
6EC = 12x
EC = 2x
BE = BC - EC
BE = 12 - 2x
AB || DE ----> ∠ABD = ∠BDE (alternating interior angles)
∠ADB = 90° (since circle is tangent to cone at this point and BD is a radius)
∠BED = 90°
Therefore triangles ABD and BDE are similar
AB/BD = BD/DE
6/r = r/x
6x = r²
In triangle BDE we have legs: BE = 12-2x, DE = x
and hypotenuse BD = r
(12 - 2x)² + x² = r² -----> r² = 6x from previous step
4x² - 48x + 144 + x² = 6x
5x² - 54x + 144 = 0
(5x - 24) (x - 6) = 0
x = 24/5 or 6
From diagram we see that x < 6
x = 24/5
r² = 6x = 144/5
r = 12/√5
Water displaced by hemisphere
= 1/2 * 4/3 π r³
= 2/3 π (12/√5)³
= 2/3 π * 1728/(5√5)
= 1152π/(5√5)
Volume of water originally in cone
= 1/3 π r² h
= 1/3 π * 6² * 12
= 144 π
Volume of water that remains in the cone
= 144π - 1152π/(5√5)
= 128.685879 cm³
Mαthmφm
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it's not really a problem about finding the volume of the cone, it is asking you what size sphere would fit in exactly half way. You can look at a cross section and try to inscribe the largest half circle in the triangle.
"Show Work!" Isn't that what they always say?
"Show Work!" Isn't that what they always say?
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Find the volume of the cone then the volume of the sphere. Hint: Both have a diameter of 12. The subtact half of the volume of the sphere ( The half that is in the cone) from the volume of the cone.