In convex quadrialteral ABCD, the area of triangle DAB is 1, that of triangle ABC is 6, and that of triangle CDA is 2. Find the area of triangle BCD.
Since area of DAB, ABC, and CDA are 1, 6, and 2, respectively, we have
\(a+b=1\),
\(b+c=6\), and
\(a+d=2\).
We need to find the area of BCD, that is \(c+d\).
Add equations two and three to get
\(a+b+c+d=8\).
Subtract equation one to get
\(c+d=8-1=7\)
In convex quadrialteral ABCD, the area of triangle DAB is 1, that of triangle ABC is 6, and that of triangle CDA is 2.
Find the area of triangle BCD.
\(\begin{array}{|rcll|} \hline \mathbf{CDA + ABC} &=& \mathbf{DAB + BCD} \\ 2 + 6 &=& 1 + BCD \\ 8 &=& 1 + BCD \\ BCD &=& 8-1 \\ \mathbf{BCD} &=& \mathbf{7} \\ \hline \end{array}\)