Not the exact question you were looking for?

Paste your question to our Mathbee AI Mentor below to get an instant step-by-step solution.

Areas Related To Circles
EXERCISE 12.1
CBSE_NCERT_TEXTBOOK
Grade 10

Question:

A wooden article was made by scooping out a hemisphere from each end of a solid cylinder, as shown in Fig. 12.11. If the height of the cylinder is 10 cm, and its base is of radius 3.5 cm, find the total surface area of the article.

Step-by-Step Solution

Key Concept: The flat circular faces of the cylinder are removed, leaving a curved cylindrical surface and two hemispherical cavities. The total surface area = curved surface area of the cylinder + surface area of the two hemispheres (which together form a complete sphere). Use formulas: Curved surface area of cylinder = $2\pi r h$, Surface area of a sphere = $4\pi r^2$.
1. Identify the given data
- Radius of cylinder (and also radius of each hemisphere) $r = 3.5\,\text{cm}$
- Height of the cylinder $h = 10\,\text{cm}$

2. Curved surface area of the cylinder
$$\text{CSA}_{\text{cyl}} = 2\pi r h = 2\pi (3.5)(10) = 70\pi\ \text{cm}^2$$

3. Surface area contributed by the two hemispheres
- Two hemispheres together make a complete sphere of radius $r$.
- Surface area of a sphere: $\text{SA}_{\text{sphere}} = 4\pi r^2$
$$\text{SA}_{\text{hemis}} = 4\pi (3.5)^2 = 4\pi \times 12.25 = 49\pi\ \text{cm}^2$$

4. Total surface area of the wooden article
$$\text{Total SA} = \text{CSA}_{\text{cyl}} + \text{SA}_{\text{hemis}} = 70\pi + 49\pi = 119\pi\ \text{cm}^2$$

5. Numerical value (using $\pi \approx \frac{22}{7}$ or $3.14$)
- Using $\pi = \frac{22}{7}$:
$$119\pi = 119 \times \frac{22}{7} = 374\ \text{cm}^2$$
- Using $\pi \approx 3.14$:
$$119\pi \approx 119 \times 3.14 = 373.66\ \text{cm}^2$$

Hence, the total surface area of the article is $119\pi\,\text{cm}^2$ (approximately $374\,\text{cm}^2$).

Correct Answer: $119\pi\ \text{cm}^2 \;\text{(≈ 374 cm}^2\text{)}$
Mathbee AI Mentor (Free Demo)

Confused by the solution? Ask the AI to explain a specific step, tell you where you went wrong, or break down the key trap in this question.

Master Areas Related To Circles with Mathbee

Practice this topic under real exam conditions with strict timers, or ask our AI Mentor to explain the concepts step-by-step.

Start Practicing for Free