0 like 0 dislike
98 views
Points $A(0,0), B(9,6)$ and $C(6,12)$ are vertices of triangle $A B C$. Point $D$ is on segment $A B$ such that $2(A D)=D B$, point $E$ is on segment $B C$ such that $2(B E)=E C$ and point $F$ is on segment $C A$ such that $2(C F)=F A$. What is the ratio of the area of triangle $D E F$ to the area of triangle $A B C$ ? Express your answer as a common fraction.
| 98 views

0 like 0 dislike
The area of triangle $A B C$ is $A_{A B C}=\frac{1}{2}(9)(6)=27$. The area of triangle $D E F$ is $A_{D E F}=\frac{1}{2}(3)(6)=9$. Therefore, the ratio of the area of triangle $D E F$ to the area of triangle $A B C$ is $\frac{A_{D E F}}{A_{A B C}}=\frac{9}{27}=\frac{1}{3}$.

Final Answer: The final answer is $\frac{1}{3}$.
by Diamond (66,975 points)

0 like 0 dislike
0 like 0 dislike
0 like 0 dislike
0 like 0 dislike
0 like 0 dislike
0 like 0 dislike
0 like 0 dislike
0 like 0 dislike
0 like 0 dislike