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

TheMathCoder  Apr 27, 2018

1+0 Answers


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. 

Melody  Apr 27, 2018

18 Online Users

New Privacy Policy (May 2018)
We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  Privacy Policy