Let ABC be a triangle with AB = 18, BC = 24, and CA = 20. D is placed on AB such that AD = 15. E is placed on BC such that EC = 20. Call the intersection of lines AE and DC point F. Compute (\frac{\text{area of DEF}}{7^3} \cdot \left( \frac{6}{5} \right)^3).

To solve the problem, we first need to calculate the area of triangle \( \triangle DEF \) and then evaluate the expression given.

### Step 1: Compute the area of \( \triangle ABC \)

The side lengths of the triangle \( ABC \) are \( AB = 18 \), \( BC = 24 \), and \( CA = 20 \). We can use Heron's formula to find the area of \( \triangle ABC \).

First, we compute the semi-perimeter \( s \):

\[
s = \frac{AB + BC + CA}{2} = \frac{18 + 24 + 20}{2} = 31
\]

Now, apply Heron's formula:

\[
\text{Area of } \triangle ABC = \sqrt{s(s-a)(s-b)(s-c)}
\]
where \( a = 18 \), \( b = 24 \), and \( c = 20 \):

\[
\text{Area} = \sqrt{31(31-18)(31-24)(31-20)} = \sqrt{31 \times 13 \times 7 \times 11}
\]

\[
\text{Area} = \sqrt{31 \times 1001} = \sqrt{31031} \approx 176.1
\]

For simplicity and precision in calculations, let’s use the exact radical expression for the area:
\[
\text{Area} = \sqrt{31031}
\]

### Step 2: Use mass point geometry or area ratios to find the area of \( \triangle DEF \)

Using mass point geometry or similar triangles (based on the ratios of side lengths), the area of \( \triangle DEF \) can be derived as follows:

Given:
- \( D \) is on \( AB \) such that \( AD = 15 \), hence \( DB = 3 \).
- \( E \) is on \( BC \) such that \( EC = 20 \), hence \( BE = 4 \).

The area of \( \triangle DEF \) is related to the area of \( \triangle ABC \) by the product of the ratios:

\[
\text{Area of } \triangle DEF = \frac{15}{18} \times \frac{4}{24} \times \text{Area of } \triangle ABC = \frac{5}{6} \times \frac{1}{6} \times \text{Area of } \triangle ABC = \frac{5}{36} \times \sqrt{31031}
\]

### Step 3: Calculate the given expression

Finally, we need to compute:

\[
\frac{\text{Area of } \triangle DEF}{7^3} \times \left( \frac{6}{5} \right)^3
\]

Given \( 7^3 = 343 \):

\[
\text{Expression} = \frac{\frac{5}{36} \times \sqrt{31031}}{343} \times \left(\frac{216}{125}\right)
\]

\[
\text{Expression} = \frac{5 \times \sqrt{31031}}{12348} \times \frac{216}{125} = \frac{1080 \times \sqrt{31031}}{1543500}
\]

This simplifies to:

\[
\frac{216 \times \sqrt{31031}}{125 \times 343 \times 36}
\]

Finally:

\[
\frac{216 \times \sqrt{31031}}{1543500} = \frac{\sqrt{31031}}{7140.28}
\]

Hence, the exact value depends on further simplification but approximately can be calculated if needed.

Would you like a deeper explanation or details on simplifying or approximating further? Here are some questions for further practice:

1. What is Heron's formula, and how is it applied in triangle area calculations?
2. How do mass points help in solving geometric problems?
3. What properties of similar triangles aid in solving for area ratios?
4. How can exact radicals be approximated in simpler forms?
5. What other methods can be used to solve for triangle area besides Heron's formula?

**Tip**: For complicated expressions, break down the problem step-by-step and use exact values for accuracy before approximating for a final answer.

Learn how to compute the area of triangle DEF formed by points D, E, and F in triangle ABC. Use Heron's formula and area ratios to solve for the area, then evaluate an expression involving cube roots and ratios. Suitable for high school students studying advanced geometry.

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