ANSWER :
The answer is 30 cm/sec
EXPLANATION :
We have an inverted square pyramid with a square side of 7 cm and a height is 14 cm.
We need to find the area of the square at 2 cm from below.
Using similar triangles, we will express the side view as 2D :
We need to find the side of the square at 2 cm level.
The ratio of the sides of the smaller triangle and bigger triangle must be the same :
[tex]\begin{gathered} \frac{\text{ smaller}}{\text{ bigger}}=\frac{x}{7}=\frac{2}{14} \\ \\ \text{ Solve for x :} \\ \text{ Cross multiply :} \\ 14x=7(2) \\ 14x=14 \\ x=\frac{14}{14}=1 \end{gathered}[/tex]
So the value of x is 1, then the side of the square at 2 cm level is 1 cm
The area of that square is :
A = 1 x 1 = 1 cm^2
The inverted pyramid is filled with water at a constant rate of Q = 30 cm^3 per second.
And we are asked to find the rate when the water level is 2 cm or when the area of the square is 1 cm^2 from the result we calculated above.
Recall the formula of rate :
[tex]\begin{gathered} Q=AV \\ \text{ where :} \\ Q\text{ = constant rate in }\frac{cm^3}{sec} \\ \\ A\text{ = Area of the section in }cm^2 \\ \\ V\text{ = Velocity or rate in }\frac{cm}{sec} \end{gathered}[/tex]
We have the following :
[tex]\begin{gathered} Q=30\text{ }\frac{cm^3}{sec} \\ \\ A=1\text{ }cm^2 \end{gathered}[/tex]
Using the formula above, the rate is :
[tex]\begin{gathered} Q=AV \\ V=\frac{Q}{A} \\ \\ V=\frac{30\text{ }\frac{cm^{\cancel{3}}}{sec}}{1\text{ }\cancel{cm^2}} \\ \\ V=30\text{ }\frac{cm}{sec} \end{gathered}[/tex]
