A shuttle cock used for playing badminton has the shape of a frustum of a cone mounted hemisphere (as shown in the figure). The external diameters of the frustum are 5 cm and 2 cm, the height of the entire shuttle cock is 7 cm. Then its external surface area is

Options

A. `4.25 cm^(2)`
B. `74.26 cm^(2)`
C. `73.26 cm^(2)`
D. `74.36 cm^(2)`

Answer

To find the external surface area of the shuttlecock, which consists of a frustum of a cone and a hemisphere, we will follow these steps:

### Step 1: Identify the dimensions
- The external diameters of the frustum are given as:
- Diameter of the top base (D1) = 5 cm → Radius (R1) = D1/2 = 2.5 cm
- Diameter of the bottom base (D2) = 2 cm → Radius (R2) = D2/2 = 1 cm
- The total height of the shuttlecock (H) = 7 cm

### Step 2: Determine the height of the frustum
- The height of the hemisphere (h_hemisphere) = Radius of the bottom base (R2) = 1 cm
- Therefore, the height of the frustum (h_frustum) = Total height - Height of the hemisphere = 7 cm - 1 cm = 6 cm

### Step 3: Calculate the slant height of the frustum
- The slant height (L) of the frustum can be calculated using the formula:
\[ L = \sqrt{h_{frustum}^2 + (R1 - R2)^2} \]
- Substituting the values:
\[ L = \sqrt{6^2 + (2.5 - 1)^2} = \sqrt{36 + 1.5^2} = \sqrt{36 + 2.25} = \sqrt{38.25} \approx 6.18 \text{ cm} \]

### Step 4: Calculate the curved surface area of the frustum
- The curved surface area (CSA) of the frustum is given by:
\[ CSA_{frustum} = \pi (R1 + R2) L \]
- Substituting the values:
\[ CSA_{frustum} = \pi (2.5 + 1) \times 6.18 = \pi \times 3.5 \times 6.18 \]
- Using \(\pi \approx \frac{22}{7}\):
\[ CSA_{frustum} = \frac{22}{7} \times 3.5 \times 6.18 \approx 74.26 \text{ cm}^2 \]

### Step 5: Calculate the surface area of the hemisphere
- The surface area (SA) of the hemisphere is given by:
\[ SA_{hemisphere} = 2\pi R2^2 \]
- Substituting the values:
\[ SA_{hemisphere} = 2\pi (1^2) = 2\pi \approx 2 \times \frac{22}{7} = \frac{44}{7} \approx 6.29 \text{ cm}^2 \]

### Step 6: Calculate the total external surface area
- The total external surface area (TSA) of the shuttlecock is the sum of the curved surface area of the frustum and the surface area of the hemisphere:
\[ TSA = CSA_{frustum} + SA_{hemisphere} \]
- Substituting the values:
\[ TSA = 74.26 + 6.29 \approx 80.55 \text{ cm}^2 \]

### Final Answer
The external surface area of the shuttlecock is approximately **80.55 cm²**.

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