Given:
Three corner points of a triangular garden are (0,3),(3,0), and (4,3).
To find:
The area of the garden.
Solution:
We know that, area of a triangle is
[tex]A=\dfrac{1}{2}|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)|[/tex]
Let a denote the area of the garden, in square units.
Three vertices of the triangular garden are (0,3),(3,0), and (4,3). So, area of the triangular garden is
[tex]a=\dfrac{1}{2}|0(0-3)+3(3-3)+4(3-0)|[/tex]
[tex]a=\dfrac{1}{2}|0(-3)+3(0)+4(3)|[/tex]
[tex]a=\dfrac{1}{2}|0+0+12|[/tex]
[tex]a=\dfrac{1}{2}\times 12[/tex]
[tex]a=6[/tex]
Therefore, the area of the garden is a=6 square units.
