Given :
Vertices of triangle (−4, 0), (−3, 1) , (0, −2), and (−1, −3).
To Find :
The area of a rectangle.
Solution :
Breadth of triangle :
[tex]B = \sqrt{(-4-(-3))^2+(0-1)^2}\\\\B = \sqrt{1^2+1^2}\\\\B = \sqrt{2}\ units[/tex]
Length of triangle :
[tex]L = \sqrt{(-4-0)^2+(0-(-2))^2}\\\\L = \sqrt{16+4}\\\\L = 2\sqrt{5}\ units[/tex]
Now, area is given by :
A = L×B
A = √2×2√5
A = 2√10 units
Therefore, the area is 2√10 units.
Hence, this is the required solution.
