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Geometry Difficulty 3.7 AMC 10/12 Find the answer Canada

Figure PQRSTPQRST is shown below.Figure 0In the figure, $PQR=QRS=TPQ=60°\$\angle PQR=\angle QRS=\angle TPQ=60\degree.Also,. Also, PTisparallelto is parallel to SRand and TSisparallelto is parallel to QR.If. If PQ=10PQ=10\text{} cm}and and TS=6TS=6\text{} cm},theperimeteroffigure, the perimeter of figure PQRST$ is

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Solution

We begin by extending PTPT to point UU on QRQR, as shown. [[IMAGE0]] Since PTPT is parallel to SRSR, then TUTU is parallel to SRSR. Similarly, since TSTS is parallel to QRQR, then TSTS is parallel to URUR. Since TUTU is parallel to SRSR and TSTS is parallel to URUR, then TURSTURS is a parallelogram, and so TU=SRTU=SR and UR=TS=6UR=TS=6 cm. In PQU\triangle PQU, the sum of the measures of the three angles is 180°180\degree, and so PUQ=180°60°60°=60°\angle PUQ=180\degree-60\degree-60\degree=60\degree.

Thus, PQU\triangle PQU is an equilateral triangle, and so QU=PU=PQ=10QU=PU=PQ=10 cm. The perimeter of PQRSTPQRST is PQ+QR+SR+TS+PT=PQ+QU+UR+SR+TS+PT  (since QR=QU+UR)=PQ+QU+UR+TU+TS+PT  (since SR=TU)=PQ+QU+UR+PT+TU+TS  (reordering some sides)=PQ+QU+UR+PU+TS  (since PT+TU=PU)=10 cm+10 cm+6 cm+10 cm+6 cm=42 cm\begin{align*} PQ+QR+SR+TS+PT&=PQ+QU+UR+SR+TS+PT \ \ \text{(since }QR=QU+UR)\\ &=PQ+QU+UR+TU+TS+PT \ \ \text{(since }SR=TU)\\ &=PQ+QU+UR+PT+TU+TS \ \ \text{(reordering some sides)}\\ &=PQ+QU+UR+PU+TS \ \ \text{(since }PT+TU=PU)\\ &=10\text{ cm}+10\text{ cm}+6\text{ cm}+10\text{ cm}+6\text{ cm}\\ &=42\text{ cm}\end{align*}

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Source: CEMC, University of Waterloo, licensed CC-BY-NC-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.