Library / /430 of 481

, 2026

Geometry Difficulty 3.5 AMC 10/12 Find the answer Canada

Points $E$ and $F$ are positioned on the sides of rectangle $ABCD$ so that $BF = ED = DC = AB= 8$ and $EF = 10$ , as shown.The area of rectangle $ABCD$
is

Pick one

Solution

We begin by constructing line segment $EG$ perpendicular to $BC$ , as shown. [[IMAGE0]] Then $EG=AB=8$ and $\triangle EFG$ is right-angled at $G$ . By the Pythagorean theorem, $FG^2=EF^2-EG^2=10^2-8^2=36$ , and so $FG=6$ (since $FG>0$ ). Since $EGCD$ is a square, then $GC=ED=8$ . Thus, $BC=BF+FG+GC=8+6+8=22$ , and so the area of $ABCD$ is $AB\times BC=8\times22=176$ .

Figure for this problem

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.