check the picture below.
for the parallelogram, we can simply get the area of those 2 blue triangles and the yellow rectangle, sum them up and that's the area of the parallelogram.
for the triangle on the right, we know the right-angle is at vertex A(4,2), so the base will be the distance from (4,2) to (1,4), and its height is the distance form (4,2) to (6,5)
[tex]\bf \stackrel{\textit{area of the parallelogram}}{\stackrel{triangle}{\cfrac{1}{2}(2)(2)}~~~~+~~~~\stackrel{rectangle}{2\cdot 1}~~~~+~~~~\stackrel{triangle}{\cfrac{1}{2}(2)(2)}}\implies 2+2+2\implies 6[/tex]
now for the area of the triangle.
[tex]\bf ~~~~~~~~~~~~\textit{distance between 2 points}
\\\\
(\stackrel{x_1}{4}~,~\stackrel{y_1}{2})\qquad
(\stackrel{x_2}{1}~,~\stackrel{y_2}{4})\qquad \qquad
% distance value
d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2}
\\\\\\
base=\sqrt{(1-4)^2+(4-2)^2}\implies base=\sqrt{(-3)^2+2^2}
\\\\\\
base=\sqrt{9+4}\implies \boxed{base=\sqrt{13}}[/tex]
[tex]\bf (\stackrel{x_1}{4}~,~\stackrel{y_1}{2})\qquad
(\stackrel{x_2}{6}~,~\stackrel{y_2}{5})\qquad \qquad height=\sqrt{(6-4)^2+(5-2)^2}
\\\\\\
height=\sqrt{2^2+3^2}\implies height=\sqrt{4+9}\implies \boxed{height=\sqrt{13}}\\\\
-------------------------------\\\\
\stackrel{\textit{area of the triangle}}{\cfrac{1}{2}(\sqrt{13})(\sqrt{13})}\implies \cfrac{1}{2}(\sqrt{13})^2\implies \cfrac{1}{2}\cdot 13\implies \cfrac{13}{2}\implies 6.5[/tex]
so, the triangle is 6.5 square units and the parallelogram has 6 square units, surely you know which statement is true.
