1.18. The radius of the circle circumscribed around a right triangle is to the radius of the inscribed circle as 5:2. Find the area of the triangle if one of its legs is equal to $a$.
1.18. Let \( r \) and \( R \) be the radii of the inscribed and circumscribed circles, \( BC = a \) (Fig. 1.16). Let \( BD = x \); then \( BL = x \) (as tangents drawn from the same point), \( LA = AK = 2R - x \) (since \( \triangle ABC \) is a right triangle, then \( AB = 2R \)). We have \( AC^2 + BC^2 = AB^2 \).

or

\[
(r + 2R - x)^2 + a^2 = 4R^2
\]

But \( R = \frac{5r}{2}, x = a - r \) and the last equation becomes

\[
(7r - a)^2 + a^2 = 25r^2
\]

or

\[
12r^2 - 7ar + a^2 = 0
\]

from which \( r_1 = \frac{a}{3}, r_2 = \frac{a}{4} \). These roots correspond to the values \( (AC)_1 = \frac{4a}{3} \),

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-422.jpg?height=457&width=462&top_left_y=108&top_left_x=690)

Fig. 1.16

\( (AC)_2 = \frac{3a}{4} \), As a result, we get two solutions:

\( S_1 = \frac{1}{2} a \cdot \frac{4a}{3} = \frac{2a^2}{3}, S_2 = \frac{1}{2} a \cdot \frac{3a}{4} = \frac{3a^2}{8} \).

Answer: \( \frac{2a^2}{3} \) and \( \frac{3a^2}{8} \).