Question

An inverted regular square pyramid has an altitude of 12m and with top
edge of 3m is fully–filled with water.
1. What is the volume of the water in m3?
A portion of water is to be removed so that water will be 7m deep.
2. What is the area at the surface of the water when it is 7m deep?
3. How much water is to be removed so that water is 7m deep?
4. How much water retained when it is 7m deep?

Answer 1

height, h = 12m Base edge, B=12m

1. $V=\frac{1}{3}Ah$

A=area of the base = B² = 3² = 9m²

$V=\frac{1}{3}×9×12=36m³$

2. $\frac{H}{h} = \frac{B}{b}$

H = initial height of pyramid, h = final height of pyramid, B = initial base edge, b = final base edge

H=12m, h = 7m, B = 3m, b=?

$\frac{12}{7}=\frac{3}{b} => b=\frac{3×7}{12}=1.75m$

Area = b² = 1.75² = 3.0625m²

3. Volume initial Pyramid – Volume of final pyramid

Volume of initial pyramid = $\frac{1}{3}×9×12=36m³$

Volume of final pyramid=$\frac{1}{3}×3.0625×7=7.15m³$

36–7.15 = 28.85m³

4. Volume of new/final pyramid = $\frac{1}{3}×3.0625×7=7.15m³$