Rasheed got a playing top (lattu) as his birthday present, which surprisingly had no colour on it. He wanted to colour it with his crayons. The top is shaped like a cone surmounted by a hemisphere (see Fig 12.6). The entire top is 5 cm in height and the diameter of the top is 3.5 cm. Find the area he has to colour. (Take = 22 7 )
Step-by-Step Solution
Key Concept: The coloured area is the sum of the curved surface area of the cone and the curved surface area of the hemisphere. Use the 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 cone height is obtained by subtracting the hemisphere radius from the total height.
1. Given data
- Total height of the top, \(H = 5\,\text{cm}\)
- Diameter = 3.5 cm \(\Rightarrow\) radius \(r = \dfrac{3.5}{2}=1.75\,\text{cm}\)
- Height of a hemisphere = its radius \(= r = 1.75\,\text{cm}\)
2. Find the height of the cone
\[ h_{\text{cone}} = H - r = 5 - 1.75 = 3.25\,\text{cm} \]
3. Slant height of the cone
\[ l = \sqrt{r^{2}+h_{\text{cone}}^{2}} = \sqrt{(1.75)^{2}+(3.25)^{2}} \]
\[ (1.75)^{2}=3.0625,\;(3.25)^{2}=10.5625 \]
\[ l = \sqrt{13.625}\approx 3.69\,\text{cm} \]
4. Curved surface area of the cone
\[ \text{CSA}_{\text{cone}} = \pi r l = \frac{22}{7}\times 1.75 \times 3.69 \]
\[ = \frac{22}{7}\times \frac{7}{4}\times 3.69 = \frac{22}{4}\times 3.69 = 5.5\times 3.69 \approx 20.30\,\text{cm}^{2} \]
5. Curved surface area of the hemisphere
\[ \text{CSA}_{\text{hemisphere}} = 2\pi r^{2} = 2\times \frac{22}{7}\times (1.75)^{2} \]
\[ = \frac{44}{7}\times 3.0625 = \frac{134.75}{7} \approx 19.25\,\text{cm}^{2} \]
6. Total area to be coloured
\[ \text{Total area}= \text{CSA}_{\text{cone}}+\text{CSA}_{\text{hemisphere}} \]
\[ \approx 20.30 + 19.25 = 39.55\,\text{cm}^{2} \]
Rounding to one decimal place, \(\boxed{39.5\,\text{cm}^{2}}\).
Correct Answer: ≈ 39.5 cm²