A cone of radius 8 cm and height 12 cm is divided into two parts by a plane through the mid-point of its axis parallel to its base. Find the ratio of the volumes of the two parts.
1 : 7
Step 1: Understand the cone dimensions.
The original cone has:
Step 2: Where the plane cuts the cone.
The plane passes through the mid-point of the axis, so it divides the height into two equal parts:
Step 3: Find radius of the small cone at the top.
Because the plane is parallel to the base, the small cone formed at the top is similar to the whole cone.
Ratio of heights = \(\dfrac{6}{12} = \dfrac{1}{2}\).
So, its radius = \(\dfrac{1}{2} \times 8 = 4\,\text{cm}\).
Step 4: Volume of the full cone.
Formula: \(V = \tfrac{1}{3}\pi r^2 h\).
\(V_{\text{big}} = \tfrac{1}{3} \pi (8^2)(12) = \tfrac{1}{3} \pi (64)(12) = 256\pi\,\text{cm}^3\).
Step 5: Volume of the small top cone.
\(V_{\text{small}} = \tfrac{1}{3} \pi (4^2)(6) = \tfrac{1}{3} \pi (16)(6) = 32\pi\,\text{cm}^3\).
Step 6: Volume of the frustum (bottom part).
Frustum volume = (whole cone volume) − (small cone volume).
\(V_{\text{frustum}} = 256\pi - 32\pi = 224\pi\,\text{cm}^3\).
Step 7: Ratio of volumes.
\(V_{\text{small}} : V_{\text{frustum}} = 32\pi : 224\pi = 1 : 7\).
Final Answer: The ratio of the volumes of the two parts is 1 : 7.