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Areas Related To Circles
EXAMPLES
CBSE_NCERT_TEXTBOOK
Grade 10

Question:

The decorative block shown in Fig. 12.7 is made of two solids — a cube and a hemisphere. The base of the block is a cube with edge 5 cm, and the hemisphere fixed on the top has a diameter of 4.2 cm. Find the total surface area of the block. (Take  = 22 7 )

Step-by-Step Solution

Key Concept: Find the exposed area of the cube (all faces except the part covered by the hemisphere) and add the curved surface area of the hemisphere. The circular base of the hemisphere is not exposed.
1. Data given
- Edge of cube, \(a = 5\) cm ⇒ each face area = \(a^2 = 5^2 = 25\) cm².
- Diameter of hemisphere = 4.2 cm ⇒ radius \(r = \frac{4.2}{2}=2.1\) cm = \(\frac{21}{10}\) cm.
- \(π = \frac{22}{7}\).

2. Area of the cube that remains exposed
- The cube has 6 faces. The bottom and the four side faces are completely exposed:
\[5\text{ faces} \times 25\text{ cm}^2 = 125\text{ cm}^2.\]
- The top face is partially covered by the circular base of the hemisphere. The area of the circular base is
\[\text{Area}_{\text{circle}} = πr^2 = \frac{22}{7}\times\left(\frac{21}{10}\right)^2 = \frac{22}{7}\times\frac{441}{100}=\frac{9702}{700}=13.86\text{ cm}^2.\]
- Exposed part of the top face = square area – circular area
\[25 - 13.86 = 11.14\text{ cm}^2.\]
- Total exposed area of the cube = \(125 + 11.14 = 136.14\) cm².

3. Curved surface area of the hemisphere
- Curved surface area of a hemisphere = \(2πr^2\).
- \[2πr^2 = 2\times\frac{22}{7}\times\frac{441}{100}=\frac{44\times441}{700}=\frac{19404}{700}=27.72\text{ cm}^2.\]

4. Total surface area of the block
\[\text{Total SA}= \text{exposed area of cube} + \text{curved area of hemisphere}
= 136.14 + 27.72 = 163.86\text{ cm}^2.\]
- Rounding to the nearest whole number, the total surface area ≈ 164 cm².

5. Answer
The total surface area of the decorative block is \(\boxed{163.86\text{ cm}^2 \;(\approx 164\text{ cm}^2)}\).

Correct Answer: 163.86 cm² (≈ 164 cm²)
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