The area of the triangle ABC for the considered triangle plotted in the considered image is 66.5 sq. units approximately.
What is the distance between two points ( p,q) and (x,y)?
The shortest distance(length of the straight line segment's length connecting both given points) between points ( p,q) and (x,y) is:
[tex]D = \sqrt{(x-p)^2 + (y-q)^2} \: \rm units.[/tex]
The coordinates of the points A, B, C, and D in the given figure are:
- A(0, 7)
- B( 7,-2)
- C(-3, -4)
- D(2, -3)
Finding the length of the line segments AD and CB, which will be the distance between A and D, and C and B respectively.
Thus, we get:
Length of line segment AD = |AD| = distance between A and D = [tex]\sqrt{(0-7)^2 + (7-(-2))^2} = \sqrt{7^2 + 9^2} = \sqrt{130} \: \rm units.[/tex]
Similarly, we get:
|CB| = [tex]\sqrt{(-3-7)^2 + (-4-(-2))^2} = \sqrt{10^2 + 6^2} = \sqrt{136} \: \rm units.[/tex]
If we take CB as the base of ABC triangle, then as AD is perpendicular on CB, and touches the peak of the triangle ABC from its base, so AD is height of the triangle.
Thus, as we know know that:
- Height's measurement of ABC = |AD|= [tex]\sqrt{130} \: \rm units[/tex]
- Base length of ABC = |BC| = [tex]\sqrt{136} \: \rm units[/tex]
Thus, the area of the triangle ABC is:
[tex]A = \dfrac{base \times height}{2} = \dfrac{\sqrt{130} \times \sqrt{136}}{2} \approx 66.5 \: \rm unit^2[/tex]
Thus, the area of the triangle ABC for the considered triangle plotted in the considered image is 66.5 sq. units approximately.
Learn more about distance between two points here:
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