Introduction / Context:
This is a mensuration problem involving a rectangular hall where we are given a relationship between the areas of the floor, ceiling and walls. The hall is assumed to be a rectangular cuboid. The statement that the sum of the floor area and the ceiling area equals the sum of the wall areas allows us to find the height. Once the height is known, we can calculate the volume. Such problems test conceptual understanding of surface area formulas for cuboids and the ability to set up and solve simple equations.
Given Data / Assumptions:
- Length of the hall, l = 15 m.
- Breadth of the hall, b = 12 m.
- Height of the hall, h = unknown.
- Floor area = l * b.
- Ceiling area = l * b (same as floor).
- Wall areas: two walls of size l by h and two walls of size b by h.
- Condition: floor area + ceiling area = total area of four walls.
Concept / Approach:
For a rectangular hall:
- Floor area = l * b.
- Ceiling area = l * b.
- Total area of the four walls = 2 * l * h + 2 * b * h = 2 * h * (l + b).
The given condition is:
2 * l * b = 2 * h * (l + b)
From this we can solve for h. After finding h, volume V is:
V = l * b * h
Step-by-Step Solution:
Step 1: Write the floor and ceiling area.
Floor area = 15 * 12 = 180 square metres.
Ceiling area = 180 square metres.
Sum of floor and ceiling areas = 2 * 180 = 360 square metres.
Step 2: Express wall areas.
Total wall area = 2 * l * h + 2 * b * h = 2 * h * (15 + 12) = 54 * h.
Step 3: Use given condition.
360 = 54 * h
h = 360 / 54 = 20 / 3 metres.
Step 4: Compute volume of the hall.
V = l * b * h = 15 * 12 * 20 / 3.
V = 180 * 20 / 3 = 60 * 20 = 1200 cubic metres.
So the volume of the hall is 1200 cubic metres.
Verification / Alternative check:
We can recheck the condition using h = 20 / 3 metres. Wall area = 2 * 15 * 20 / 3 + 2 * 12 * 20 / 3 = 2 * 100 + 2 * 80 = 200 + 160 = 360 square metres. Floor plus ceiling area is 360 square metres as computed earlier. Hence the condition is satisfied, so the derived height and volume are correct.
Why Other Options Are Wrong:
Option A (720) corresponds to a smaller height that would not satisfy the wall condition. Option B (900) and Option D (1800) also lead to inconsistent wall area sums. Option E (600) assumes a height that is half of the correct height. Only 1200 cubic metres satisfies both the area relation and the volume formula.
Common Pitfalls:
Some learners mistakenly take only one of floor or ceiling areas in the equation instead of both. Others forget that there are two pairs of opposite walls leading to the factor 2 in the total wall area. Careful setup of the equality and correct arithmetic division are crucial to avoid errors.
Final Answer:
The volume of the hall is
1200 cubic metres.
Discussion & Comments