How do you find the area and perimeter of a figure by only using points?
For example:
(-5,3) (-1,2) (0,6)
I. How do you find the area of a figure by only using points?
For n points, the points must be oriented:
$$\boxed{2A= \sum \limits_{i=1}^n(y_i+y_{i+1})(x_i-x_{i+1}) \quad x_{n+1} = x_1 \quad y_{n+1} = y_1}$$ [Carl Friedrich Gauß]
For example (-5,3) (-1,2) (0,6), the points are oriented!
$$\begin{array}{|c|r|r|}
\hline
Point & x & y \\
\hline
1 &-5&3 \\
2 &-1&2 \\
3 &0&6 \\
\hline
\end{array}$$
$$\begin{array}{rcl}
2A & = & (y_1+y_2)(x_1-x_2)+(y_2+y_3)(x_2-x_3)+(y_3+y_1)(x_3-x_1) \\
2A & = & (3+2)(-5+1)+(2+6)(-1-0)+(6+3)(0+5)\\
2A & = & (5)(-4)+(8)(-1)+(9)(5)\\
2A & = & -20 - 8 + 45\\
2A & = & 17\\
A & = & 8.5
\end{array}$$
II. How do you find the Perimeter of a figure by only using points?
$$\boxed{Perimeter = \sum \limits_{i=1}^n
\sqrt{(x_{i+1}-x_i)^2+(y_{i+1}-y_i)^2}\quad x_{n+1} = x_1 \quad y_{n+1} = y_1}
}$$
Example:
$$\begin{array}{|c|r|r|}
\hline
Point & x & y \\
\hline
1 &-5&3 \\
2 &-1&2 \\
3 &0&6 \\
\hline
\end{array}$$
$$\begin{array}{rcl}
Perimeter & = & \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}+ \sqrt{(x_3-x_2)^2+(y_3-y_2)^2}+ \sqrt{(x_1-x_3)^2+(y_1-y_3)^2} \\
& = & \sqrt{(-1+5)^2+(2-3)^2}+ \sqrt{(0+1)^2+(6-2)^2}+ \sqrt{(-5-0)^2+(3-6)^2} \\
& = & \sqrt{(4)^2+(-1)^2}+ \sqrt{(1)^2+(4)^2}+ \sqrt{(-5)^2+(-3)^2} \\
& = & \sqrt{16+1}+ \sqrt{1+16}+ \sqrt{25+9} \\
& = & \sqrt{17}+ \sqrt{17}+ \sqrt{34} \\
Perimeter & = & = 14.0771631461
\end{array}$$
Find the distance between all three points using the distance formula.....
Add all these results for the perimeter
To find the area of this triangle ....let's use something called the "semi-perimeter" = S ...this is given by .....(A + B + C) / 2 where A, B and C are the side lengths
Then the area is given by....
√[S * (S - A) * (S - B) * (S - C)]
If you get stuck, let me know..........!!!
I. How do you find the area of a figure by only using points?
For n points, the points must be oriented:
$$\boxed{2A= \sum \limits_{i=1}^n(y_i+y_{i+1})(x_i-x_{i+1}) \quad x_{n+1} = x_1 \quad y_{n+1} = y_1}$$ [Carl Friedrich Gauß]
For example (-5,3) (-1,2) (0,6), the points are oriented!
$$\begin{array}{|c|r|r|}
\hline
Point & x & y \\
\hline
1 &-5&3 \\
2 &-1&2 \\
3 &0&6 \\
\hline
\end{array}$$
$$\begin{array}{rcl}
2A & = & (y_1+y_2)(x_1-x_2)+(y_2+y_3)(x_2-x_3)+(y_3+y_1)(x_3-x_1) \\
2A & = & (3+2)(-5+1)+(2+6)(-1-0)+(6+3)(0+5)\\
2A & = & (5)(-4)+(8)(-1)+(9)(5)\\
2A & = & -20 - 8 + 45\\
2A & = & 17\\
A & = & 8.5
\end{array}$$
II. How do you find the Perimeter of a figure by only using points?
$$\boxed{Perimeter = \sum \limits_{i=1}^n
\sqrt{(x_{i+1}-x_i)^2+(y_{i+1}-y_i)^2}\quad x_{n+1} = x_1 \quad y_{n+1} = y_1}
}$$
Example:
$$\begin{array}{|c|r|r|}
\hline
Point & x & y \\
\hline
1 &-5&3 \\
2 &-1&2 \\
3 &0&6 \\
\hline
\end{array}$$
$$\begin{array}{rcl}
Perimeter & = & \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}+ \sqrt{(x_3-x_2)^2+(y_3-y_2)^2}+ \sqrt{(x_1-x_3)^2+(y_1-y_3)^2} \\
& = & \sqrt{(-1+5)^2+(2-3)^2}+ \sqrt{(0+1)^2+(6-2)^2}+ \sqrt{(-5-0)^2+(3-6)^2} \\
& = & \sqrt{(4)^2+(-1)^2}+ \sqrt{(1)^2+(4)^2}+ \sqrt{(-5)^2+(-3)^2} \\
& = & \sqrt{16+1}+ \sqrt{1+16}+ \sqrt{25+9} \\
& = & \sqrt{17}+ \sqrt{17}+ \sqrt{34} \\
Perimeter & = & = 14.0771631461
\end{array}$$
I don't remember seeing that area formula before Heureka. I've got a lot to learn.
Your dispay is as brilliant as we have all come to expect. Thanks. :)
You know Chris I have seen you use that formula a number of times and it still has not sunk into my brain!
Thanks ( I think I may reserve that icon for when i am feeling more "blond" than usual)