Question

A hall is 15 m long and 12 m broad. If the sum of the areas of the floor and the ceiling is equal to the sum of the areas of four walls, the volume of the hall in m$^3$ is:

• A. 720
• B. 900
• C. 1200
• D. 1800

Hint:

Whenever the sum of the floor and ceiling areas equals the area of the four walls, the height \(H\) can be quickly calculated as:
\[ H = \frac{L \cdot B}{L + B} \]
Here, \(H = \frac{15 \times 12}{15 + 12} = \frac{180}{27} = \frac{20}{3}\text{ m}\).

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
A rectangular hall represents a cuboid.
Solving this problem involves setting up equations for the area of different faces of the cuboid to determine the unknown height, which is then used to find the volume.
Key Formula or Approach:
For a cuboid of length \(L\), breadth \(B\), and height \(H\):
- Area of floor = \(L \times B\)
- Area of ceiling = \(L \times B\)
- Sum of areas of floor and ceiling = \(2 \cdot L \cdot B\)
- Area of four walls = \(2H(L + B)\)
- Volume of the cuboid = \(L \times B \times H\)

Step 2: Detailed Explanation:
Let us substitute the given values into our equations:
- Given: Length, \(L = 15\text{ m}\), and Breadth, \(B = 12\text{ m}\).
- Let the height of the hall be \(H\).
- Area of floor + ceiling:
\[ \text{Area}_{\text{floor+ceiling}} = 2 \times 15 \times 12 = 360\text{ m}^2 \]
- Area of four walls:
\[ \text{Area}_{\text{four walls}} = 2H(L + B) = 2H(15 + 12) = 2H(27) = 54H \]
- According to the problem, these two quantities are equal:
\[ 54H = 360 \]
\[ H = \frac{360}{54} = \frac{20}{3}\text{ m} \]
- Now, we compute the volume of the hall:
\[ \text{Volume} = L \times B \times H \]
\[ \text{Volume} = 15 \times 12 \times \frac{20}{3} \]
\[ \text{Volume} = 5 \times 12 \times 20 = 1200\text{ m}^3 \]

Step 3: Final Answer:
The volume of the hall is 1200 m$^3$ (Option C).