Question

A cone fits perfectly (height-wise) inside a 6 cm × 8 cm × 10 cm cuboid such that the entire base of the cone rests on one of the faces of the cuboid. Also, the circumference of the base of the cone just touches one of the pairs of the opposite sides of the face of the cuboid which is beneath it. The volume of the cone is \( 32 \pi \, \text{cm}^3 \).

• A. Always
• B. Sometimes
• C. Never
• D.
• E.

Hint:

To solve volume problems involving cones inside a cuboid, always check the relationship between the cone's base radius and the cuboid's dimensions.

Answer 1

The Correct Option is B

Step 1: We are given a cone fitting perfectly inside a cuboid with dimensions 6 cm × 8 cm × 10 cm. The base of the cone rests on one of the cuboid faces, and the circumference of the base touches one of the pairs of opposite sides of the cuboid. Step 2: The formula for the volume of a cone is: \[ V = \frac{1}{3} \pi r^2 h \] where \(r\) is the radius of the base and \(h\) is the height of the cone. Step 3: The radius \(r\) can be calculated using the circumference condition. The circumference of the cone’s base is equal to 8 cm (the length of one side of the cuboid face). \[ 2\pi r = 8 \quad \Rightarrow \quad r = \frac{8}{2\pi} = \frac{4}{\pi} \, \text{cm} \] Step 4: Now, substitute \(r = \frac{4}{\pi}\) and height \(h = 6\) cm into the volume formula: \[ V = \frac{1}{3} \pi \left( \frac{4}{\pi} \right)^2 \times 6 = \frac{1}{3} \pi \times \frac{16}{\pi^2} \times 6 = \frac{96}{3\pi} = \frac{32}{\pi} \, \text{cm}^3 \] Thus, the volume of the cone is \( \frac{32}{\pi} \, \text{cm}^3 \), not \(32 \pi \, \text{cm}^3\). Step 5: Therefore, the volume is only \( 32 \pi \, \text{cm}^3 \) under specific conditions, so the correct answer is sometimes.