In the figure, triangles $A B C$ and $D E F$ are equilateral with sides $14 \mathrm{~cm}$ and $13 \mathrm{~cm}$, respectively, and sides $B C$ and $E F$ are parallel.

![](https://cdn.mathpix.com/cropped/2024_05_01_984603b876c0bc28095bg-13.jpg?height=388&width=392&top_left_y=1528&top_left_x=932)

a) Calculate the measure of angle EÛT.

b) Calculate the perimeter of polygon $P Q R S T U$.

c) If segment $P Q$ measures $6 \mathrm{~cm}$, what is the measure of segment $S T$?
Solution

a) Since $BC$ and $EF$ are parallel, the angles $EUT$ and $ACB$ are alternate interior angles, hence $E \hat{U} T = A \hat{C} B = 60^{\circ}$. b) From item a) we can conclude that all triangles in the figure are equilateral. Thus, we have $QP = FP$, $UT = UE$, $TS = CS$, and $RQ = RB$. Therefore, the perimeter of $PQRSTU$ is

\[
\begin{gathered}
QP + PU + UT + TS + SR + RQ = (FP + PU + UE) + (CS + SR + RB) \\
= FE + CB = 13 \text{ cm} + 14 \text{ cm} = 27 \text{ cm}
\end{gathered}
\]

c) From $PQ = 6 \text{ cm}$ it follows that $FP = 6 \text{ cm}$, since triangle $QFP$ is equilateral, and we conclude that $PE = FE - EP = 13 \text{ cm} - 6 \text{ cm} = 7 \text{ cm}$. Since $BC$ is parallel to $EF$ and $AB$ is parallel to $DE$, the quadrilateral $PESB$ is a parallelogram, hence $BS = PE = 7 \text{ cm}$. Finally, we have $SC = BC - BS = 14 \text{ cm} - 7 \text{ cm} = 7 \text{ cm}$; thus $ST = SC = 7 \text{ cm}$, since triangle $TCS$ is equilateral.

![](https://cdn.mathpix.com/cropped/2024_05_01_984603b876c0bc28095bg-38.jpg?height=471&width=488&top_left_y=1609&top_left_x=687)

A similar solution can be given starting from the parallelogram QDTA.