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mrs_skeptik [129]
4 years ago
13

The tip of a solid metal cone was placed into a cube that has 10 inch edges, as shown.

Mathematics
1 answer:
Licemer1 [7]4 years ago
7 0

Answer:

r=\sqrt{\dfrac{75}{\pi}}\approx 4.89\ in

Step-by-step explanation:

If the water in the cube rose from 6 inches to 8 inches when the cone was placed in the cube, then the difference in volumes is the volume of the submerged portion of the cone.

Initially, 10 in by 10 in by 6 in:

V_{initial}=10\cdot 10\cdot 6=600\ in^2.

At he end, 10 in by 10 in by 8 in:

V_{final}=10\cdot 10\cdot 8=800\ in^3.

Thus,

V_{submerged \ cone}=800-600=200\ in^3.

Use cone's volume formula

V_{cone}=\dfrac{1}{3}\pi r^2 \cdot h,

where r is the radius of the cone's base, h is the height of the cone.

From the diagram, h=8 in, then

200=\dfrac{1}{3}\cdot \pi\cdot r^2\cdot 8\\ \\\pi r^2=75\\ \\r^2=\dfrac{75}{\pi}\\ \\r=\sqrt{\dfrac{75}{\pi}}\approx 4.89\ in

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