Question

Given: A right-angled triangle has sides of lengths 6 cm, 8 cm, and 10 cm. Quantity A: Volume of the cone formed by rotating the triangle about the side of length 6 cm. Quantity B: Volume of the cone formed by rotating the triangle about the side of length 8 cm.

• A. Quantity A is greater.
• B. Quantity B is greater.
• C. The two quantities are equal.
• D. The relationship cannot be determined from the information given.
• E.

Hint:

When rotating a right-angled triangle to form a cone, use the formula \( V = \frac{1}{3} \pi r^2 h \), where the base radius and height are determined by the two perpendicular sides of the triangle.

Answer 1

The Correct Option is B

We need to find the volumes of the cones formed by rotating the triangle about the sides of lengths 6 cm and 8 cm.
Step 1: When a right-angled triangle is rotated about one of its sides, the volume of the resulting cone can be found using the formula: \[ V = \frac{1}{3} \pi r^2 h \] Where \( r \) is the radius (which is one leg of the triangle) and \( h \) is the height (the other leg of the triangle).
Step 2: For Quantity A (rotation about side 6 cm), the radius \( r = 6 \) cm and the height \( h = 8 \) cm. The volume is: \[ V_A = \frac{1}{3} \pi (6)^2 (8) = \frac{1}{3} \pi \times 36 \times 8 = 96 \pi \, \text{cm}^3 \] Step 3: For Quantity B (rotation about side 8 cm), the radius \( r = 8 \) cm and the height \( h = 6 \) cm. The volume is: \[ V_B = \frac{1}{3} \pi (8)^2 (6) = \frac{1}{3} \pi \times 64 \times 6 = 128 \pi \, \text{cm}^3 \] Step 4: Clearly, \( V_B = 128 \pi \, \text{cm}^3 \) is greater than \( V_A = 96 \pi \, \text{cm}^3 \), so Quantity B is greater.