I think it is c, I could be wrong though
Answer:
Perimeter
units. Area 12 square units.
Step-by-step explanation:
Perimeter: total distance around the figure.
Distance Formula: the distance between points
is
![d=\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}](https://tex.z-dn.net/?f=d%3D%5Csqrt%7B%5Cleft%28x_2-x_1%5Cright%29%5E2%2B%5Cleft%28y_2-y_1%5Cright%29%5E2%7D)
![AB=\sqrt{(6-1)^2+(2-1)^2}=\sqrt{25+1}=\sqrt{26}](https://tex.z-dn.net/?f=AB%3D%5Csqrt%7B%286-1%29%5E2%2B%282-1%29%5E2%7D%3D%5Csqrt%7B25%2B1%7D%3D%5Csqrt%7B26%7D)
![BC=\sqrt{(5-6)^2+(4-2)^2}=\sqrt{1+4}=\sqrt{5}](https://tex.z-dn.net/?f=BC%3D%5Csqrt%7B%285-6%29%5E2%2B%284-2%29%5E2%7D%3D%5Csqrt%7B1%2B4%7D%3D%5Csqrt%7B5%7D)
![CD=\sqrt{(2-5)^2+(5-4)^2}=\sqrt{9+1}=\sqrt{10}](https://tex.z-dn.net/?f=CD%3D%5Csqrt%7B%282-5%29%5E2%2B%285-4%29%5E2%7D%3D%5Csqrt%7B9%2B1%7D%3D%5Csqrt%7B10%7D)
![DA=\sqrt{(1-2)^2+(1-5)^2}=\sqrt{1+16}=\sqrt{17}](https://tex.z-dn.net/?f=DA%3D%5Csqrt%7B%281-2%29%5E2%2B%281-5%29%5E2%7D%3D%5Csqrt%7B1%2B16%7D%3D%5Csqrt%7B17%7D)
The perimeter is the sum of all those segment lengths.
One way to find the area of the figure is to surround it with a rectangle, insert some lines so that the areas you do not want can be found and subtracted from the rectangle's area. (See attached image.)
The area of the large rectangle around the figure is 5 x 4 = 20 square units.
The triangles have areas 1/2 (base) (height):
A. (1/2)(1)(4) = 2 square units
B. (1/2)(3)(1) = 1.5 square units
D. (1/2)(1)(2) = 1 square unit
E. (1/2)(5)(1) = 2.5 square units
Square C. (1)(1) = 1 square unit
Total of all the area you don't want to include:
2 + 1.5 + 1 + 2.5 + 1 = 8 square units
Subtract 8 from the surrounding rectangle's area of 20, and you get the area of the figure is 20 - 8 = 12 square units.
I believe the answer is C 1,300
What are the ordered pairs
Answer:
141°
Step-by-step explanation:
It's an octagon so the Sum of interior angles is: (8-2)×180=1080
5x+133+16+126+162-12+30=5x+455=1080
5x=625
x=125
x+16=141