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Areas Related To Circles
EXAMPLES
CBSE_NCERT_TEXTBOOK
Grade 10

Question:

Mayank made a bird-bath for his garden in the shape of a cylinder with a hemispherical depression at one end (see Fig. 12.9). The height of the cylinder is 1.45 m and its radius is 30 cm. Find the total surface area of the bird-bath. (Take  = 22 7 )

Step-by-Step Solution

Key Concept: Total surface area = curved surface area of the cylinder + curved surface area of the hemispherical depression + area of the circular base. Use 2πrh for the curved surface of a cylinder, 2πr² for a hemisphere (half of the sphere’s surface area 4πr²), and πr² for the base.
1. Convert all dimensions to the same unit\
Radius \(r = 30\,\text{cm}=0.30\,\text{m}\)\
Height of the cylindrical part \(h = 1.45\,\text{m}\)\
2. Curved surface area of the cylinder\
$$\text{CSA}_{\text{cyl}} = 2\pi r h = 2 \times \frac{22}{7} \times 0.30 \times 1.45$$\
$$= 2 \times \frac{22}{7} \times 0.435 = \frac{44}{7} \times 0.435 = 2.734\,\text{m}^2$$\
3. Curved surface area of the hemispherical depression\
The surface area of a full sphere is \(4\pi r^2\); half of it (hemisphere) is \(2\pi r^2\).\
$$\text{CSA}_{\text{hem}} = 2\pi r^2 = 2 \times \frac{22}{7} \times (0.30)^2$$\
$$= 2 \times \frac{22}{7} \times 0.09 = \frac{44}{7} \times 0.09 = 0.566\,\text{m}^2$$\
4. Area of the circular base\
$$\text{Base area} = \pi r^2 = \frac{22}{7} \times (0.30)^2 = \frac{22}{7} \times 0.09 = 0.283\,\text{m}^2$$\
5. Total surface area\
$$\text{Total SA} = \text{CSA}_{\text{cyl}} + \text{CSA}_{\text{hem}} + \text{Base area}$$\
$$= 2.734 + 0.566 + 0.283 = 3.583\,\text{m}^2$$\
(Rounded to two decimal places, \(\approx 3.58\,\text{m}^2\)).

Correct Answer: Approximately \(3.58\,\text{m}^2\)
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