vessel so that the depth of water will be 10 31. In the figure, the right pyramid $$ V A B C D $$ is cut by the planes $$ F G H E $$ an $$ K L I J $$ parallel to the base into three parts. The areas of $$ A B C D, F G H E $$ and $$ K L I J $$ are in the ratio $$ 49: 36: 16 $$. (a) Find $$ A F $$ : $$ F K: K V $$. (b) If the volume of frustum $$ E F G H I J K L $$ is $$ 30.4 \mathrm{~cm}^{3} $$, find the volume of frustum $$ A B C D E F G H $$. (c) Are the frustums $$ A B C D E F G H $$ and $$ E F G H I J K L $$ similar? Explain your answer.

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We begin by noting that the pyramid $$VABCD$$ is right and any cross‐section parallel to the base is similar to the base. Hence if the distance from the vertex $$V$$ along a lateral edge is proportional to the similarity ratio, then the area of a cross‐section is proportional to the square of that ratio.

Assume the total length of $$VA$$ (from vertex $$V$$ to the base vertex $$A$$) is $$h$$. Since the cross–sections are parallel to the base, the ratio of a cross–section’s linear dimensions to those of the base is the same as the ratio of its distance from $$V$$ to $$h$$. In other words, if a cross–section is cut at a distance $$x$$ from $$V$$, then its area is
$$
\left(\frac{x}{h}\right)^2\quad \text{times the area of the base.}
$$

In our problem the areas of the base and the two cross–sections are in the ratio
$$
49 : 36 : 16.
$$

We take the base area (of $$ABCD$$) to correspond to $$49$$ (units). Then if the cross–section at some level has area $$A$$, we can write
$$
\left(\frac{x}{h}\right)^2 =\frac{A}{49}.
$$

Because $$49>36>16$$ the base is the “largest cross section.” Working from the vertex $$V$$ downward along, say, edge $$VA$$ the first (upper) plane meets the edge at a point (which we label $$K$$) and the second (lower) plane meets it at a point (which we label $$F$$); the labeling on the figure is such that on $$VA$$ the points occur in the order
$$
A\;(\text{base}),\;F,\;K,\;V\;(\text{vertex}).
$$

Since the area of the cross–section through $$KLIJ$$ is $$16$$, we have
$$
\left(\frac{VK}{h}\right)^2=\frac{16}{49}\quad\Longrightarrow\quad VK=\frac{4}{7}h.
$$
Also, since the area of the cross–section through $$FGHE$$ is $$36$$,
$$
\left(\frac{VF}{h}\right)^2=\frac{36}{49}\quad\Longrightarrow\quad VF=\frac{6}{7}h.
$$

The points on segment $$VA$$ (from the base $$A$$ up to $$V$$) are arranged as:
- $$A$$ is at distance $$h$$ from $$V$$.
- $$F$$ is at distance $$VF=\frac{6}{7}h$$ from $$V$$ so that
  $$
  AF = VA - VF = h - \frac{6}{7}h = \frac{1}{7}h.
  $$
- $$K$$ is at distance $$VK=\frac{4}{7}h$$ from $$V$$ so that
  $$
  AK = h - \frac{4}{7}h = \frac{3}{7}h.
  $$
  Then
  $$
  FK = AK - AF = \frac{3}{7}h - \frac{1}{7}h = \frac{2}{7}h.
  $$

Thus, the segments along $$VA$$ are:
$$
AF=\frac{1}{7}h,\quad FK=\frac{2}{7}h,\quad KV=\frac{4}{7}h.
$$
It is customary to write the ratio starting at $$A$$ (the base) going upward:
\[
AF:FK:KV=1:2:4.
\
$$ AF : FK : KV = 1 : 2 : 4 $$.

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