Library / /299 of 387

, 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 $. [[IMAGE0]] Now,$ $\triangle$ 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

Figure for this problem 5 $-$ 12 $-$ 13 $directly.) Thus, the perimeter of$ ABCD $is$ 20 + 12 + 15 + 13 = 60$.

Want a route through all this instead of an archive? The track puts 2,604 problems in a working order, from Junior Challenge level to the IMO shortlist.

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