The triangle $\triangle A B C$ is equilateral, with sides measuring $3 \mathrm{~cm}$, and the triangle $\triangle C B D$ is right-angled, with sides measuring 3, 4 and 5 $\mathrm{cm}$, as shown in the given figure. Calculate the distance between points $A$ and $D$.

![](https://cdn.mathpix.com/cropped/2024_05_01_30c9a294a58a6e4b8190g-026.jpg?height=317&width=574&top_left_y=253&top_left_x=1255)
Solution 1: Note that the angle $A \widehat{B} D$ is known. Indeed, the triangle $\triangle A B C$ is equilateral, so $A \widehat{B} C=60^{\circ}$ and, since $C \widehat{B} D=90^{\circ}$, we get $A \widehat{B} D=150^{\circ}$. Thus, applying the Law of Cosines to the triangle $\triangle A B D$, we have

$$
A D^{2}=3^{2}+4^{2}-2 \cdot 3 \cdot 4 \cos 150^{\circ}=25+24 \cos 30^{\circ}=25+12 \sqrt{3}
$$

It follows that $A D=\sqrt{25+12 \sqrt{3}} \mathrm{~cm}$.

Solution 2: Let $E$ be the point on the line $B D$ such that the triangle $\triangle A E B$ is a right triangle at vertex $E$ (see figure). In this triangle, we have

$$
\frac{\sqrt{3}}{2}=\cos 30^{\circ}=\frac{E B}{A B}=\frac{E B}{3} \quad \text { and } \quad \frac{1}{2}=\sin 30^{\circ}=\frac{A E}{A B}=\frac{A E}{3}
$$

therefore,

$$
E B=\frac{3 \sqrt{3}}{2} \quad \text { and } \quad A E=\frac{3}{2}
$$

in particular, $E D=4+\frac{3 \sqrt{3}}{2}$.

![](https://cdn.mathpix.com/cropped/2024_05_01_30c9a294a58a6e4b8190g-091.jpg?height=389&width=714&top_left_y=645&top_left_x=751)

Applying the Pythagorean Theorem to the triangle $\triangle A E D$, we get

$$
A D^{2}=A E^{2}+E D^{2}=\left(\frac{3}{2}\right)^{2}+\left(\frac{8+3 \sqrt{3}}{2}\right)^{2}=25+12 \sqrt{3}
$$

It follows that $A D=\sqrt{25+12 \sqrt{3}} \mathrm{~cm}$.