The height of a monument is $20$ m. A horizontal cross-section at a distance x meters from the top is an equilateral triangle with side $\frac{x}{4}$ meters. Find the volume of the monument.
A) $\frac{125 \sqrt{3}}{3} \mathrm{~m}^{3}$
B) $\frac{125 \sqrt{2}}{3} \mathrm{~m}^{3}$
C) $\frac{\sqrt{3}}{3} \mathrm{~m}^{3}$
D) $\frac{125 \sqrt{2}}{2} \mathrm{~m}^{3}$
E) $125 \sqrt{3} \mathrm{~m}^{3}$

Question

Quizplus · Accepted Answer

Let's divide the monument into small horizontal slices of thickness $\Delta x$, starting from the base. 
Consider a slice at a height $h$ from the base. The horizontal cross-section of this slice is an equilateral triangle with side length $s = \frac{1}{4}(20-h)$.
The area of an equilateral triangle with side length $s$ is $\frac{\sqrt{3}}{4}s^2$. Therefore, the area of the cross-section of this slice is $$A(h) = \frac{\sqrt{3}}{4}\left(\frac{1}{4}(20-h)ight)^2$$
The volume of this slice is $A(h)\Delta x$, so the total volume of the monument is given by the integral: $$\int_0^{20} A(h)\Delta x = \int_0^{20} \frac{\sqrt{3}}{4}\left(\frac{1}{4}(20-h)ight)^2 \Delta x$$
Taking the limit as $\Delta x ightarrow 0$ and replacing the integral by the corresponding definite integral, we get:
\begin{align*}
V &= \int_0^{20} \frac{\sqrt{3}}{4}\left(\frac{1}{4}(20-h)ight)^2 dh\
&= \frac{\sqrt{3}}{256} \int_0^{20} (20-h)^2 dh\
&= \frac{\sqrt{3}}{256} \cdot \frac{1}{3} \cdot (20)^3\
&= \frac{125\sqrt{3}}{3} \mathrm{~m}^{3}
\end{align*}

The Height of a Monument Is 202020 M a Horizontal Cross-Section at a Distance X Meters from Cross-Section

The Height of a Monument Is $20$ M a Horizontal Cross-Section at a Distance X Meters from Cross-Section