In the diagram, $\triangle P Q R$ is equilateral and has side length $6 \mathrm{~cm}$. Each of the shaded triangles is equilateral and has side length $2 \mathrm{~cm}$. What fraction of the area of $\triangle P Q R$ is shaded?
(A) $\frac{3}{7}$
(B) $\frac{1}{3}$
(C) $\frac{1}{2}$
(D) $\frac{3}{5}$
(E) $\frac{2}{3}$

![](https://cdn.mathpix.com/cropped/2024_04_20_6027bc27089ed4fc493cg-028.jpg?height=322&width=355&top_left_y=1601&top_left_x=1297)
Label points $S, T, U, V, W$, as shown in Figure 1.

Each shaded triangle is equilateral, $\triangle P Q R$ is equilateral, and so $\angle V S U=\angle V T W=\angle S P T=60^{\circ}$.

Therefore, $\angle P S V=180^{\circ}-\angle V S U=120^{\circ}$ (since $P S U$ is a straight angle), and similarly, $\angle P T V=120^{\circ}$.

In quadrilateral $P S V T, \angle S V T=360^{\circ}-\angle P S V-\angle S P T-\angle P T V$ or $\angle S V T=360^{\circ}-120^{\circ}-60^{\circ}-120^{\circ}=60^{\circ}$.

Therefore, $P S V T$ is a parallelogram, and since $S V=T V=2$, then $P S=P T=2$ (opposite sides of a parallelogram are equal in length). Join $S$ to $T$, as shown in Figure 2.

Since $S V=T V$, then $\angle V S T=\angle V T S=\frac{1}{2}\left(180^{\circ}-60^{\circ}\right)=60^{\circ}$.

That is, $\triangle S V T$ is equilateral with side length 2 , and is therefore congruent to each of the shaded triangles.

Similarly, $\angle S P T=60^{\circ}, P S=P T=2$, and so $\triangle P S T$ is congruent to each of the shaded triangles.

We may also join $U$ to $X$ and $W$ to $Y$ (as in Figure 3), and similarly show that $\triangle U Q X, \triangle U X V, \triangle W R Y$, and $\triangle W Y V$ are also congruent to the shaded triangles.

Thus, $\triangle P Q R$ can be divided into 9 congruent triangles.

Since 3 of these 9 triangles are shaded, the fraction of the area of $\triangle P Q R$ that is shaded is $\frac{3}{9}$ or $\frac{1}{3}$.
![](https://cdn.mathpix.com/cropped/2024_04_20_748fdafcde2a47412608g-142.jpg?height=1038&width=352&top_left_y=220&top_left_x=1559)

ANswer: (B)