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Geometry Difficulty 3.7 AMC 10/12 Find the answer Canada

In the diagram, $PQRS$ is a square with $RS=30~\text{cm}$ . $TUVW$ is a square with $U$ and $V$ on $PS$ so that $PU=UV=VS$ .Point $X$ is on $QR$ so that $TX$ is parallel to $WR$ . The area of parallelogram $TXRW$ is

Pick one

Solution

$PQRS$ is a square with $PS=RS=30\text{ cm}$ . Since $PU+UV+VS=PS=30\text{ cm}$ and $PU=UV=VS$ , then $UV=\dfrac{30\text{ cm}}{3}=10\text{ cm}$ .

$TUVW$ is a square with $UT=TW=VW=UV=10\text{ cm}$ . $TXRW$ is a parallelogram, and so $XR=TW=10\text{ cm}$ . The height of $TXRW$ is the vertical distance between its two parallel and horizontal sides $TW$ and $XR$ . This vertical height is equal to $\$RS+VW=30\text{ cm}+10\text{ cm}=40\text{}$
cm} $. The area of parallelogram$ TXRW is the product of its base and its height, which is equal to

Figure for this problem $10\text{ cm}\times40\text{ cm}=400\text{}$
cm}^2$.

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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.