A right circular cone of diameter k cm and height 12cm rests on the base of a rightcircular cylinder of radius k cm (their base lie in the same plane, as shown). Thecylinder is filled with water to a height of 12cm. If the cone is then removed, the heightto which the water will fall is
Question
A right circular cone of diameter k cm and height 12cm rests on the base of a rightcircular cylinder of radius k cm (their base lie in the same plane, as shown). Thecylinder is filled with water to a height of 12cm. If the cone is then removed, the heightto which the water will fall is
Solution
To find the height to which the water will fall when the cone is removed, we can use the concept of similar triangles.
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Let's start by finding the radius of the cone. Since the diameter of the cone is given as k cm, the radius will be half of that, which is k/2 cm.
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Now, let's consider the similar triangles formed by the cone and the cylinder. The height of the cone is given as 12 cm, and the height of the water in the cylinder is also 12 cm. Therefore, the ratio of the height of the cone to the height of the cylinder is 12/12, which simplifies to 1.
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Since the triangles are similar, the ratio of the corresponding sides will be the same. This means that the ratio of the radius of the cone to the radius of the cylinder is also 1.
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Now, when the cone is removed, the water level will drop until it reaches the new height. Let's denote this height as h cm.
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Since the ratio of the radius of the cone to the radius of the cylinder is 1, the ratio of the height of the cone to the new height of the water will also be 1.
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Therefore, we can set up the following proportion: 12 cm / h cm = 1 / 1.
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Solving this proportion, we find that h = 12 cm.
So, when the cone is removed, the water will fall to a height of 12 cm.
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