Find the perimeter of the polygon

Mathematics

1 answer:

Zielflug [23.3K]2 years ago

4 0

Answer:

100

Step-by-step explanation:

Okay so I'm not 100% sure but I think it's 100 because:

Angle B and D are the same, so the 12.5 areas will be the same (I honestly suck at explaining I'm really sorry.) But, if you add the sides from angle B and D it will be 50 because 12.5 + 12.5 + 12.5 + 12.5 (or 12.5 * 4). Meanwhile, For angle A and C, both outsides are the same as the other side.

I'm so sorry I can't explain the problem well but once you add them:

11.5 + 11.5 + 12.5 + 12.5 + 12.5 + 12.5 + 13.5 + 13.5 = 100

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the coordinates of the vertices of a rectangle in order are (5,2), (-1,4), (-2,1) and (4,-1). what is the perimeter of the recta

Anna [14]

Answer:

19 units

Step-by-step explanation:

Plot points A(5,2), B(-1,4), C(-2,1) and D(4,-1) on the coordinate plane. Find lengths of all sides as distances between two points:

$AB=\sqrt{(-1-5)^2+(4-2)^2}=\sqrt{6^2+2^2}=\sqrt{36+4}=\sqrt{40}=2\sqrt{10}\ units\\ \\BC=\sqrt{(-2-(-1))^2+(1-4)^2}=\sqrt{1^2+3^2}=\sqrt{1+9}=\sqrt{10}\ units\\ \\CD=\sqrt{(4-(-2))^2+(-1-1)^2}=\sqrt{6^2+2^2}=\sqrt{36+4}=\sqrt{40}=2\sqrt{10}\ units\\ \\DA=\sqrt{(5-4)^2+(2-(-1))^2}=\sqrt{1^2+3^2}=\sqrt{1+9}=\sqrt{10}\ units$