A plane parallel to the base divides a cone of height 40 cm into two parts. If the volumeof the upper smaller cone formed is \(\frac{1}{64}\) of the volume of the original cone, calculate the height of the plane from the base of the cone.
30 cm
To solve this problem, we need to apply the concept of similar cones and the formula for the volume of a cone. Let's break down the solution step by step:
The volume \((V)\) of a cone is given by the formula:
\(V = \frac{1}{3} \pi r^2 h\)
where \(r\) is the base radius and \(h\) is the height of the cone.
Let's consider the original cone with a height of 40 cm, and let the base radius be \(R\).
A plane parallel to the base divides the cone into two parts: an upper smaller cone and a remaining frustum.
It is given that the volume of the upper smaller cone is \(\frac{1}{64}\) of the original cone's volume.
Let's assume the height of the smaller cone is \(h_1\). The volume of the smaller cone is:
\(V_{small} = \frac{1}{3} \pi \left(\frac{r_1}{R}\right)^2 h_1\right)\)
We are given, \(V_{small} = \frac{1}{64} \times V\).
Also, due to the similarity of the cones, \(\left(\frac{r_1}{R}\right) = \frac{h_1}{h}\).
Thus, the ratio of volumes condition becomes:
\(\left(\frac{h_1}{40}\right)^3 = \frac{1}{64}\)
Taking cube roots on both sides:
\(\frac{h_1}{40} = \frac{1}{4}\)
Solving for \(h_1\) gives:
\(h_1 = \frac{1}{4} \times 40 = 10 \, \text{cm}\)
The height of the plane from the base of the cone is therefore:
\(40 \, \text{cm} - 10 \, \text{cm} = 30 \, \text{cm}\)
Hence, the height of the plane from the base of the cone is 30 cm.
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