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Geometry Difficulty 2.3 Junior Find the answer Canada

$PQRS$ is a square with side length 8. Points $T$ and $U$ are on $PS$ and $QR$ respectively with $QU=TS=1$ .

The length of $TU$ is closest to

Pick one

Solution

We draw a line through $T$ to point $W$ on $QR$ so that $TW$ is perpendicular to $QR$ .

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Since $TWRS$ has three right angles (at $W$ , $R$ and $S$ ), then it must be a rectangle.

Therefore, $WR=TS=1$ and $TW=SR=8$ .

Since $QU=1$ , then $UW = QR - QU - WR = 8 - 1- 1 = 6$ .

Now, $\triangle TWU$ is right-angled at $W$ .

By the Pythagorean Theorem, we have $TU^2 = TW^2 + UW^2$ .

Thus, $TU^2 = 8^2 + 6^2 = 64+36=100$ .

Since $TU>0$ , then $TU = \sqrt{100}=10$ .

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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.