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ABCD is a square of side length 1.  E is on AB and F is on BC such that DE and DF trisect angle ADC.  Find the area of triangle DEF.

 May 15, 2020
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Since ABCD is a square, each of its corner angles are right angles, each = 90o.

Since DE and DF trisect angle(ADC), angle(ADE) = 30o, angle(EDF) = 30o, angle(FDC) = 30o.

 

Triangle(DCF) is a 30o - 60o - 90o right triangle.

 

Since DC = 1, CF = 1/sqrt(3) and DF = 2/sqrt(3).

Similarly, DE = 2/sqrt(3).

 

Using this formula to find the area of triangle(EDF):  area = ½·a·b·sin(C)

--->   area  =  ½ · 2/sqrt(3) · 2/sqrt(3) ·sin(30o)

--->   area  =  ½ · 2/sqrt(3) · 2/sqrt(3) · 0.50

--->   area  =  1/3

 May 15, 2020

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