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CEMC Gauss (Grade 7), 2025

Geometry Difficulty 3.7 AMC 10/12 Find the answer Canada

Figure PQRSTPQRST is shown
below.

Figure 0

In the figure, ∠PQR=∠QRS=∠TPQ=60°\angle PQR=\angle QRS=\angle TPQ=60\degree. Also, PTPT is parallel to SRSR and TSTS is parallel to QRQR. If PQ=10 cmPQ=10\text{ cm} and TS=6 cmTS=6\text{ cm}, the perimeter of figure
PQRSTPQRST is

Pick one

Solution

We begin by extending PTPT to
point UU on QRQR, as shown.

Figure 1

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.