The six triangles in the figure are right-angled and their angles with vertex at point $A$ are equal. In addition, $A B=24 \mathrm{~cm}$ and $A C=54 \mathrm{~cm}$. What is the length of $A D$?

![](https://cdn.mathpix.com/cropped/2024_05_01_c051a4aaadd76ab6970bg-13.jpg?height=306&width=485&top_left_y=438&top_left_x=700)
A) $30 \mathrm{~cm}$
B) $34 \mathrm{~cm}$
C) $36 \mathrm{~cm}$
D) $38 \mathrm{~cm}$
E) $39 \mathrm{~cm}$

#
Solution

## ALTERNATIVE C

Let's denote the measures, in centimeters, of the hypotenuses of the right triangles that appear in the figure by $a, b, x, d$, and $c$, as shown in the figure below. Our goal is to find $x = AD$.

The six right triangles are similar because they share the angle at vertex $A$. Therefore,

$$
\frac{24}{a}=\frac{a}{b}=\frac{b}{x}=\frac{x}{c}=\frac{c}{d}=\frac{d}{54}
$$

Multiplying the first three terms above and, separately, the last three, we get $\frac{24}{x}=\frac{x}{54}$. Thus, $x^{2}=24 \times 54=2^{3} \times 3 \times 2 \times 3^{3}=2^{4} \times 3^{4}=4^{2} \times 9^{2}=36^{2}$, from which $x=36$.

![](https://cdn.mathpix.com/cropped/2024_05_01_c051a4aaadd76ab6970bg-41.jpg?height=300&width=503&top_left_y=1392&top_left_x=685)

Alternatively, let $\lambda=\frac{24}{a}$. Multiplying the six terms of the sequence of equalities above, we get $\lambda^{6}=\frac{24}{54}=\frac{4}{9}=\left(\frac{2}{3}\right)^{2}$, from which $\lambda^{3}=\frac{2}{3}$. On the other hand, $\lambda^{3}=\frac{24}{a} \times \frac{a}{b} \times \frac{b}{x}=\frac{24}{x}$ and we get $\frac{24}{x}=\frac{2}{3}$, from which $x=36$.