Answer:
The area of the triangle is 7.5 sq units.
Step-by-step explanation:
The coordinates of the vertices of triangle ABC is,
A = (2, 1)
B = (-1, 4)
C = (2, 6)
D = (2, 4)
So,
\(AC=\sqrt{(2-2)^2+(6-1)^2}=\sqrt{0^2+5^2}=\sqrt{5^2}=5\)
\(BD=\sqrt{(-1-2)^2+(4-4)^2}=\sqrt{(-3)^2+0^2}=\sqrt{9}=3\)
we know that,
\(\text{Area}=\dfrac{1}{2}\times \text{Base}\times \text{Height}\)
\(=\dfrac{1}{2}\times \text{AC}\times \text{BD}\)
\(=\dfrac{1}{2}\times 5\times 3\)
\(=7.5\)
