Library / /362 of 428

, 2026

Geometry Difficulty 2.7 Junior Find the answer Canada

The line passing through the points $A$ , $B$ , $C$ , and $D$ , is parallel to the line passing
through the points $E$ , $F$ , and $G$ , as shown.

$ABFE$ is a parallelogram and $CDG$ is a triangle. If $AB = 8~\text{cm}$ and $CD = 12~\text{cm}$ , what is the ratio of the area of $ABFE$ to the area of $\triangle CDG$ ?

Pick one

Solution

The distance between two parallel lines is constant. Thus, the
height of parallelogram $ABFE$ is equal to the height of $\$\triangle$
CDG $. Suppose that height is$ h. In this case, the ratio of the area of

Figure for this problem ABFE $to the area of$ $\triangle$ CDG $is$ $AB\times h:\dfrac12\times CD\times$
h. Substituting and simplifying, this ratio is equal to

Figure for this problem $8\times h:\dfrac12\times12\times$
h=8:6=4:3$.

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.