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Problem 358

Geometry Difficulty 2.5 Multiple choice CEMC Cayley · Canada · 2024

In the diagram, quadrilateral $ABCD$ has $AB =
20 $,$ BC = 12 $, and$ CD = 15 $. Also,$ AB $and$ CD $are perpendicular to$ BC$.

The perimeter of quadrilateral $ABCD$ is

Pick one

Official solution

Draw a perpendicular from $D$
to $F$ on $AB$ .

Since quadrilateral $FDCB$ has right
angles at $F$ , $C$ and $B$ , then it must be a rectangle.

This means that $FB = DC = 15$ and
$FD = BC = 12$ .

Further, $AF = AB - FB = 20 - 15 =
5$.

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Now, $\triangle AFD$ is
right-angled at $F$ .

By the Pythagorean Theorem, $AD^2 = AF^2 +
FD^2 = 5^2 + 12^2 = 25 + 144 = 169$.

Since $AD > 0$ , then $AD = 13$ . (Some might recognize the
Pythagorean triple $5$ - $12$ - $13$ directly.)

Thus, the perimeter of $ABCD$ is
$20 + 12 + 15 + 13 = 60$ .

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