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Areas Related To Circles
EXAMPLES
CBSE_NCERT_TEXTBOOK
Grade 10

Question:

Shanta runs an industry in a shed which is in the shape of a cuboid surmounted by a half cylinder (see Fig. 12.12). If the base of the shed is of dimension 7 m × 15 m, and the height of the cuboidal portion is 8 m, find the volume of air that the shed can hold. Further, suppose the machinery in the shed occupies a total space of 300 m3, and there are 20 workers, each of whom occupy about 0.08 m3 space on an average. Then, how much air is in the shed? (Take  = 22 7 ) Fig. 12.12 Fig. 12.11

Step-by-Step Solution

Key Concept: The volume of the shed is the sum of the volume of the cuboidal part and the volume of the half‑cylinder placed on its top. Use V(cuboid)=l·b·h and V(half‑cylinder)=½·π·r²·L, where r is the radius of the cylinder (half of the shorter side of the base) and L is its length (the longer side of the base). After obtaining the total volume, subtract the space occupied by machinery and workers to get the volume of air remaining.
1. Identify dimensions\
- Base of cuboid: length \(l = 15\) m, breadth \(b = 7\) m.\
- Height of cuboidal part: \(h_c = 8\) m.\
- The half‑cylinder sits on the top of the cuboid. Its length equals the longer side of the base, i.e. \(L = 15\) m.\
- Diameter of the cylinder = breadth of the base = 7 m, therefore radius \(r = \dfrac{7}{2}=3.5\) m.

2. Volume of the cuboidal portion\
$$V_{cuboid}=l\times b\times h_c = 15\times 7\times 8 = 840\ \text{m}^3$$

3. Volume of the half‑cylinder\
- Full cylinder volume: $$V_{cyl}=\pi r^{2}L$$\
- Using \(\pi = \dfrac{22}{7}\):\
$$\pi r^{2}=\frac{22}{7}\times (3.5)^{2}=\frac{22}{7}\times 12.25=\frac{269.5}{7}=38.5$$\
- Hence, \(V_{cyl}=38.5\times 15=577.5\) m³.\
- Half‑cylinder volume: $$V_{half\,cyl}=\frac{1}{2}V_{cyl}=\frac{1}{2}\times 577.5=288.75\ \text{m}^3$$

4. Total volume of the shed (air space when empty)\
$$V_{total}=V_{cuboid}+V_{half\,cyl}=840+288.75=1128.75\ \text{m}^3$$

5. Space occupied by machinery\
$$V_{mach}=300\ \text{m}^3$$

6. Space occupied by workers\
- Each worker occupies \(0.08\) m³.\
- For 20 workers: $$V_{workers}=20\times 0.08=1.6\ \text{m}^3$$

7. Volume of air actually present\
$$V_{air}=V_{total}-V_{mach}-V_{workers}
=1128.75-300-1.6=827.15\ \text{m}^3$$

8. Answer (rounded to one decimal place)\
- Volume of shed = \(1128.8\) m³ (≈ \(1.13\times10^{3}\) m³).\
- Volume of air after accounting for machinery and workers = \(827.2\) m³.

Correct Answer: Total volume of the shed = 1128.75 m³. After subtracting the machinery (300 m³) and the workers (1.6 m³), the volume of air remaining = 827.15 m³ (≈ 827.2 m³).
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