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MathsGee Android Q&A

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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.
in Mathematics by Diamond (66,975 points) | 98 views

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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)

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MathsGee Android Q&A

MathsGee Android Q&A