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Question 1198557: The frustum of a cone of revolution is 25 cm high, and the radii of its bases are 8 cm and 2 cm, respectively. Find the height, in cm, of an equivalent right circular cylinder whose base is equal in area to the section of the frustum made by a plane parallel to its base and equidistant from the bases.
Answer by onyulee(41)   (Show Source):
You can put this solution on YOUR website! Certainly, let's find the height of the equivalent cylinder.
**1. Find the Radius of the Midsection**
* Since the cutting plane is equidistant from the bases, it divides the frustum's height equally.
* Height of each portion = 25 cm / 2 = 12.5 cm
* We can use similar triangles to find the radius (r) of the midsection:
* (r - 2) / 12.5 = (8 - 2) / 25
* (r - 2) / 12.5 = 6 / 25
* r - 2 = (6 * 12.5) / 25
* r - 2 = 3
* r = 5 cm
**2. Find the Area of the Midsection**
* Area of the midsection (A) = π * r²
* A = π * (5 cm)²
* A = 25π cm²
**3. Find the Volume of the Frustum**
* Volume of Frustum (V) = (1/3) * π * h * (R² + r² + Rr)
* Where:
* h = height of frustum (25 cm)
* R = radius of larger base (8 cm)
* r = radius of smaller base (2 cm)
* V = (1/3) * π * 25 * (8² + 2² + 8 * 2)
* V = (1/3) * π * 25 * (64 + 4 + 16)
* V = (1/3) * π * 25 * 84
* V = 700π cm³
**4. Find the Volume of the Equivalent Cylinder**
* Volume of Cylinder (V) = Area of Base * Height
* 700π cm³ = 25π cm² * Height
* Height = 700π cm³ / 25π cm²
* Height = 28 cm
**Therefore, the height of the equivalent right circular cylinder is 28 cm.**
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