Answer:
11. Perimeter: 17.7 units
11. Area: 14 square units
12. Perimeter: 24.7 units
12. Area: 34 square units
Step-by-step explanation:
Question 11
Given vertices of the polygon:
- A = (-2, 1)
- B = (2, 5)
- C = (5, 1)
Plot the given points in the coordinate plane (see attachment 1).
From inspection, we can see that the polygon is a triangle with the following dimensions:
- Base = 7 units
- Height = 4 units
[tex]\boxed{\begin{minipage}{7.4 cm}\underline{Distance between two points}\\\\$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$\\\\\\where $(x_1,y_1)$ and $(x_2,y_2)$ are the two points.\\\end{minipage}}[/tex]
The length of AC is 7 units.
To find the length of AB and BC, use the distance formula.
[tex]\begin{aligned}AB&=\sqrt{(x_B-x_A)^2+(y_B-y_A)^2}\\&=\sqrt{(2-(-2))^2+(5-1)^2}\\&=\sqrt{(4)^2+(4)^2}\\&=\sqrt{16+16}\\&=\sqrt{32}\\&=4\sqrt{2}\;\sf units\end{aligned}[/tex]
[tex]\begin{aligned}BC&=\sqrt{(x_C-x_B)^2+(y_C-y_B)^2}\\&=\sqrt{(5-2)^2+(1-5)^2}\\&=\sqrt{(3)^2+(-4)^2}\\&=\sqrt{9+16}\\&=\sqrt{25}\\&=5\;\sf units\end{aligned}[/tex]
Therefore:
[tex]\begin{aligned}\textsf{Perimeter}&=AB+BC+AC\\&=4\sqrt{2}+5+7\\&=12+4\sqrt{2}\\&=17.6568542...\\&=17.7\;\sf units\end{aligned}[/tex]
[tex]\begin{aligned}\textsf{Area}&=\dfrac{1}{2}\cdot 7\cdot 4\\\\&=\dfrac{7}{2} \cdot 4\\\\&=\dfrac{28}{2}\\\\&=14\;\sf square\;units\end{aligned}[/tex]
Question 12
Given vertices of the polygon:
- A = (-3, 5)
- B = (1, 6)
- C = (3, -2)
- D = (-1, -3)
Plot the given points in the coordinate plane (see attachment 2).
From inspection, we can see that the polygon is a rectangle.
[tex]\boxed{\begin{minipage}{7.4 cm}\underline{Distance between two points}\\\\$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$\\\\\\where $(x_1,y_1)$ and $(x_2,y_2)$ are the two points.\\\end{minipage}}[/tex]
Since AB = DC and BC = AD, find the length of AB and BC using the distance formula.
[tex]\begin{aligned}AB&=\sqrt{(x_B-x_A)^2+(y_B-y_A)^2}\\&=\sqrt{(1-(-3))^2+(6-5)^2}\\&=\sqrt{(4)^2+(1)^2}\\&=\sqrt{16+1}\\&=\sqrt{17}\;\sf units\end{aligned}[/tex]
[tex]\begin{aligned}BC&=\sqrt{(x_C-x_B)^2+(y_C-y_B)^2}\\&=\sqrt{(3-1)^2+(-2-6)^2}\\&=\sqrt{(2)^2+(-8)^2}\\&=\sqrt{4+64}\\&=\sqrt{68}\\&=2\sqrt{17}\;\sf units\end{aligned}[/tex]
Therefore:
[tex]\begin{aligned}\textsf{Perimeter}&=AB+BC+CD+AD\\&=2(AB+BC)\\&=2(\sqrt{17}+2\sqrt{17})\\&=2(3\sqrt{17})\\&=6\sqrt{17}\\&=24.7386337...\\&=24.7\;\sf units\end{aligned}[/tex]
[tex]\begin{aligned}\textsf{Area}&=AB\cdot BC\\&=\sqrt{17} \cdot 2\sqrt{17}\\&=34\;\sf square\;units\end{aligned}[/tex]
