A shuttle cock used for playing badminton has the shape of a frustum of a cone mounted on a hemisphere. The two diameters of the frustum are 5 cm and 2 cm, the height of the entire shuttle cock is 6 cm. Find the external surface area.
63.68 cm 2
Solution:
Step 1: Identify Given Values
- Radius of lower base of frustum (r1) = 1 cm
- Radius of upper base of frustum (r2) = 2.5 cm
- Total height of shuttlecock = 6 cm
- Radius of hemisphere = 1 cm (same as r1)
- Height of frustum = 6 − 1 = 5 cm
Step 2: Curved Surface Area of Frustum
Formula: CSA = π (r1 + r2) × l
Slant height l = √[(r2 − r1)² + height²]
l = √[(2.5 − 1)² + 5²] = √[2.25 + 25] = √27.25 ≈ 5.22 cm
CSAfrustum = π × (1 + 2.5) × 5.22 ≈ π × 3.5 × 5.22 ≈ 3.1416 × 18.27 ≈ 57.40 cm²
Step 3: Curved Surface Area of Hemisphere
Formula: CSA = 2πr²
CSAhemisphere = 2 × π × 1² = 2π ≈ 6.28 cm²
Step 4: Total External Surface Area
Total = CSAfrustum + CSAhemisphere
= 57.40 + 6.28 = 63.68 cm²
✅ Final Answer: 63.68 cm²
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