In triangle ABC, angle C is a right angle. Point D lies on side AB, and point E lies on side BC. The line DE is parallel to side AC. If the length of side AB is 39 units, the length of DE is 8 units, and the length of side AC is greater than the length of side BC, and the area of triangle ABC is 270 square units, what is the length of side BE, in units?

Given that triangle ABC is a right triangle with angle C being the right angle, we can denote the lengths of the sides as follows:

• $AB = c = 39$ units (hypotenuse)
• $AC = b$ units
• $BC = a$ units

We are also given that the area of triangle ABC is 270 square units. The area of a right triangle can be calculated using the formula: $\text{Area} = \frac{1}{2} \times a \times b$ Thus, we have: $\frac{1}{2} \times a \times b = 270$ $a \times b = 540$

Since DE is parallel to AC, triangle ADE is similar to triangle ABC by the AA (Angle-Angle) similarity criterion. The ratio of the sides of similar triangles is equal. Let $k$ be the ratio of similarity. Then: $\frac{DE}{AC} = k$ $\frac{8}{b} = k$

Since DE is parallel to AC, the ratio of the areas of triangles ADE and ABC is $k^2$. The area of triangle ADE is: $\text{Area of } \triangle ADE = k^2 \times \text{Area of } \triangle ABC$ $\text{Area of } \triangle ADE = \left(\frac{8}{b}\right)^2 \times 270$

Now, we need to find the length of BE. Since DE is parallel to AC, the length of BE can be found using the properties of similar triangles. The length of BE is: $BE = a - k \times a$ $BE = a - \left(\frac{8}{b}\right) \times a$ $BE = a \left(1 - \frac{8}{b}\right)$

We need to find the values of $a$ and $b$. Using the Pythagorean theorem in triangle ABC: $a^2 + b^2 = c^2$ $a^2 + b^2 = 39^2$ $a^2 + b^2 = 1521$

We also have: $a \times b = 540$

We can solve these two equations simultaneously. Let's solve for $b$ in terms of $a$: $b = \frac{540}{a}$

Substitute $b$ into the Pythagorean theorem equation: $a^2 + \left(\frac{540}{a}\right)^2 = 1521$ $a^2 + \frac{291600}{a^2} = 1521$ $a^4 - 1521a^2 + 291600 = 0$

Let $x = a^2$: $x^2 - 1521x + 291600 = 0$

Solve this quadratic equation using the quadratic formula: $x = \frac{1521 \pm \sqrt{1521^2 - 4 \times 291600}}{2}$ $x = \frac{1521 \pm \sqrt{2311441 - 1166400}}{2}$ $x = \frac{1521 \pm \sqrt{1145041}}{2}$ $x = \frac{1521 \pm 1069}{2}$

This gives us two solutions for $x$: $x = \frac{1521 + 1069}{2} = 1295$ $x = \frac{1521 - 1069}{2} = 226$

Thus, $a^2 = 1295$ or $a^2 = 226$. Since $a$ and $b$ are positive, we take the positive square roots: $a = \sqrt{1295} \approx 36$ $b = \frac{540}{a} \approx 15$

Since $b > a$, we have: $b = \sqrt{1295} \approx 36$ $a = \frac{540}{b} \approx 15$

Now, we can find the length of BE: $BE = a \left(1 - \frac{8}{b}\right)$ $BE = 15 \left(1 - \frac{8}{36}\right)$ $BE = 15 \left(1 - \frac{2}{9}\right)$ $BE = 15 \left(\frac{7}{9}\right)$ $BE = \frac{105}{9}$ $BE = 11.67$

Thus, the length of BE is approximately 11.67 units.