Points A(0,0), B(9,6) and C(6,12) are vertices of triangle ABC. Point D is on segment AB such that 3(AD) = DB, point E is on segment BC such that 3(BE) = EC and point F is on segment CA such that 4(CF) = FA. What is the ratio of the area of triangle DEF to the area of triangle ABC? Express your answer as a common fraction.

Guest May 31, 2021

#1**+1 **

Best Answer

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I found the same question here: https://web2.0calc.com/questions/2-problems_1

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Varxaax May 31, 2021

#3**+3 **

Points A(0,0), B(9,6) and C(6,12) are vertices of triangle ABC. Point D is on segment AB such that 3(AD) = DB, point E is on segment BC such that 3(BE) = EC, and point F is on segment CA such that 4(CF) = FA. What is the ratio of the area of triangle DEF to the area of triangle ABC? Express your answer as a common fraction.

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Using the shoelace method we find that [ABC] = 36, and [DEF] = 16.65.

[DEF] / [ABC] = 37 / 80

jugoslav May 31, 2021