Library / /357 of 387

, 2025

Geometry Difficulty 3.2 AMC 10/12 Find the answer Canada

In the diagram, point $D$ is on side $AB$ of $\triangle ABC$ and points $E$ and $F$ are on $BC$ .For some positive real number $t$ , the area of $\triangle DBE$ is $t$ , the area of $\triangle DEF$ is $t$ , the area of $\triangle DFC$ is $2t$ , and the area of $\triangle DAC$ is $4t$ . If the area of $\triangle DEC$ is $63$ , what is the area of $\triangle ABC$ ?

Pick one

Solution

The area of $\triangle DEC$ is equal to the sum of the areas of $\$\triangle$
DEF $and$ $\triangle$ DFC $, and so$ t+2t=63 $or$ 3t=63 $or$ t=21 $. The area of$ $\triangle$ ABC is equal to the sum of the areas of

Figure for this problem $\triangle$ ADC $,$ $\triangle$ DBE $,$ $\triangle$ DEF $and$ $\triangle$ DFC $, which is equal to$ 4t+t+t+2t=8t=8(21)=168$.

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.