CEMC Cayley, 2024

Geometry Difficulty 2.5 Junior Find the answer Canada

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

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

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 $\text{AD gt; 0}$ , then $AD = 13$ . (Some might recognize the
Pythagorean triple $5$ - $12$ - $13$ directly.)

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

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