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In the figure, what is the area of triangle \(ABD\)? Express your answer as a common fraction.

 Apr 27, 2018
 #1
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The only way I can think to do this is with co-ordinate geometry.

 

 

 

Equation of AF is    y=(-4/3)x+4

Equation of EB is    y=(-2/7)x+2

Simultaneous solution of these gives the point  D which is \((\frac{21}{11},\frac{16}{11})\)

 

Distance AB can be found using Pythagoras's theoem.  It is  \(\sqrt{65}\)

 

Equation of AB is 

  \(y=\frac{-4x}{7}+4\\ 7y=-4x+28\\ 4x+7y-28=0 \)

 

Now I can use the perpendicular distance formula of a point to a line to find the perpendicular height between AB and D

 

\(d=\frac{|ax_1+by_1+c|}{\sqrt{a^2+b^2}}\\ d=\frac{|4*\frac{21}{11}+7*\frac{16}{11}-28|}{\sqrt{16+49}}\\ d=\frac{|4*\frac{21}{11}+7*\frac{16}{11}-28|}{\sqrt{16+49}}\\ d=\frac{112}{11}*\frac{1}{\sqrt{65}}\\ d=\frac{112\sqrt{65}}{11*65}\\ d=\frac{112\sqrt{65}}{715}\\ \)

 

Area of ABD = 

 

\(\frac{1}{2}*\sqrt{65}*\frac{112\sqrt{65}}{715}\\ =\frac{56*65}{715}\\ =\frac{56*13}{143}\\ =\frac{728}{143}\\ =5\frac{13}{143}u^2\)

 

 

That is assuming I have made not stupid errors.

You could check this using heron's formula but that would be a lot of work too. 

Since the area of ABC is 14 my answer does sound reasonable. 

 Apr 27, 2018

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