A room with a square base and height 10 ft occupies 25,000 ft3 volume. The room has cubicles of base area 25 ft2 in four corners. The walls and ceiling of the area of the room which is not occupied by cubicles is to be painted. How much area is to be painted?
4000 ft²
Volume of the room \(= \text{base area} \times \text{height}\), so base area \(= \dfrac{25000}{10} = 2500\) ft\(^2\).
Since the base is square, side \(= \sqrt{2500} = 50\) ft.
Each cubicle has base area 25 ft\(^2\), so each cubicle is a \(5 \times 5\) ft square reaching the full height of 10 ft, placed in the four corners.
Total wall area of the room \(= 4 \times (\text{side} \times \text{height}) = 4 \times (50 \times 10) = 2000\) ft\(^2\).
At each corner the cubicle covers two wall strips of \(5 \times 10 = 50\) ft\(^2\) each, i.e. \(100\) ft\(^2\) per cubicle, so wall area covered \(= 4 \times 100 = 400\) ft\(^2\); paintable wall area \(= 2000 - 400 = 1600\) ft\(^2\).
Ceiling area of the room \(= 2500\) ft\(^2\); the four cubicle tops cover \(4 \times 25 = 100\) ft\(^2\), so paintable ceiling area \(= 2500 - 100 = 2400\) ft\(^2\).
Total area to be painted \(= 1600 + 2400 = 4000\) ft\(^2\).
Hence, the area to be painted is 4000 ft\(^2\).
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Take $\pi = \frac{22}{7}$
