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Areas Related To Circles
EXERCISE 12.1
CBSE_NCERT_TEXTBOOK
Grade 10

Question:

A toy is in the form of a cone of radius 3.5 cm mounted on a hemisphere of same radius. The total height of the toy is 15.5 cm. Find the total surface area of the toy.

Step-by-Step Solution

Key Concept: Use the curved surface area formulas: \(\text{CSA of cone}=\pi r l\) where \(l=\sqrt{r^{2}+h^{2}}\), and \(\text{CSA of hemisphere}=2\pi r^{2}\). The base of the cone and the flat face of the hemisphere are not exposed, so they are not included in the total surface area.
1. Given data
\[ r = 3.5\ \text{cm}, \quad \text{total height}=15.5\ \text{cm} \]
2. Height of the cone
The hemisphere contributes a height equal to its radius \(r\).
\[ h_{\text{cone}} = 15.5 - 3.5 = 12\ \text{cm} \]
3. Slant height of the cone
\[ l = \sqrt{r^{2}+h_{\text{cone}}^{2}} = \sqrt{(3.5)^{2} + (12)^{2}} = \sqrt{12.25 + 144} = \sqrt{156.25} = 12.5\ \text{cm} \]
4. Curved surface area of the cone
\[ \text{CSA}_{\text{cone}} = \pi r l = \pi \times 3.5 \times 12.5 = 43.75\pi \ \text{cm}^{2} \]
5. Curved surface area of the hemisphere
\[ \text{CSA}_{\text{hemisphere}} = 2\pi r^{2} = 2\pi \times (3.5)^{2} = 2\pi \times 12.25 = 24.5\pi \ \text{cm}^{2} \]
6. Total surface area of the toy (only the exposed curved surfaces)
\[ \text{Total SA} = \text{CSA}_{\text{cone}} + \text{CSA}_{\text{hemisphere}} \]
\[ = (43.75\pi + 24.5\pi) \ \text{cm}^{2} = 68.25\pi \ \text{cm}^{2} \]
7. Numerical value (using \(\pi = \frac{22}{7}\))
\[ 68.25\pi = 68.25 \times \frac{22}{7} = \frac{1501.5}{7} = 214.5 \ \text{cm}^{2} \]
Thus, the total surface area of the toy is \(68.25\pi\ \text{cm}^{2}\) or \(214.5\ \text{cm}^{2}\) (to 1 decimal place).

Correct Answer: Total surface area = $68.25\pi\ \text{cm}^2$ \(= 214.5\ \text{cm}^2\) (using $\pi=\frac{22}{7}$).
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