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Solid Mensuration

The document contains two examples of calculating the volume of water needed to cover balls arranged in a container. In the first example, it finds that 7.5 cm of water is needed to cover two balls of different diameters (15 cm and 10 cm) placed in a 20 cm diameter cylindrical jar. In the second example, it determines that 36.33 mm of water is required to cover four 20 mm marbles arranged in a regular tetrahedron formation at the bottom of a cylindrical jar with a 21.55 mm radius.

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113 views3 pages

Solid Mensuration

The document contains two examples of calculating the volume of water needed to cover balls arranged in a container. In the first example, it finds that 7.5 cm of water is needed to cover two balls of different diameters (15 cm and 10 cm) placed in a 20 cm diameter cylindrical jar. In the second example, it determines that 36.33 mm of water is required to cover four 20 mm marbles arranged in a regular tetrahedron formation at the bottom of a cylindrical jar with a 21.55 mm radius.

Uploaded by

Jovani Hard Core
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© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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Example 015

Two balls, one 15 cm in diameter and the other 10 cm in diameter, are placed in a
cylindrical jar 20 cm in diameter, as shown in Figure 014. Find the volume of water
necessary to cover them.

x=2057.5x=2057.5
x=7.5 cmx=7.5 cm
h2+x2=(5+7.5)2h2+x2=(5+7.5)2
h2+7.52=12.52h2+7.52=12.52
h=10 cmh=10 cm
Depth of water

H=7.5+h+5H=7.5+h+5
H=7.5+10+5H=7.5+10+5

H=22.5 cmH=22.5 cm
Volume of water

Vwater=VHVballsVwater=VHVballs
Vwater=14(202)(22.5)43(7.53+53)Vwater=14(202)(22.5)43(7.53+53)
Vwater=4777.84 cm3Vwater=4777.84 cm3

answer

Example 014
A boy who had discovered that 20 mm marbles fitted snugly into the 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?
The lines connecting the center of marbles will form into a regular tetrahedron of edge
20 cm.

From triangle ABC

cos30=10AEcos30=10AE
AE=10cos30=1032AE=10cos30=1032
AE=11.55 mmAE=11.55 mm
CE=AECE=AE
CE=11.55 mmCE=11.55 mm
Radius of cylinder

R=10+AE=10+11.55R=10+AE=10+11.55
R=21.55 mmR=21.55 mm
Solving for h from right triangle CED

h2+CE2=CD2h2+CE2=CD2
h2+11.552=202h2+11.552=202
h=16.33 mmh=16.33 mm
Depth of water

H=20+h=20+16.33H=20+h=20+16.33
H=36.33 mm
H=36.33 mm
Volume of water

Vwater=Vcylinder4VmarbleVwater=Vcylinder4Vmarble
Vwater=(21.552)(36.33)4[43(103)]Vwater=(21.552)(36.33)4[43(103)]
Vwater=36248.98 mm3Vwater=36248.98 mm3

answer

http://www.mathalino.com/reviewer/solid-mensuration-solid-geometry/011-equalspheres-piled-form-pyramid

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