Question

Find the perimeter of the polygon with the vertices A (−1, 1), B (4, 1), C (4,−2), and D(−1,−2).

The perimeter of ABCD is ___ units.


Could someone explain how I can figure this out?

Answer 1

Given:

The vertices of a polygon are A (−1, 1), B (4, 1), C (4,−2), and D(−1,−2).

To find:

The perimeter of ABCD.

Solution:

Distance formula:

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Using distance formula, we get

$AB=\sqrt{(4-(-1))^2+(1-1)^2}$

$AB=\sqrt{(4+1)^2+(0)^2}$

$AB=\sqrt{(5)^2}$

$AB=5$

Similarly,

$BC=\sqrt{(4-4)^2+(-2-1)^2}=3$

$CD=\sqrt{(-1-4)^2+(-2-(-2))^2}=5$

$AD=\sqrt{(-1-(-1))^2+(-2-1)^2}=3$

Now, perimeter of ABCD is

$P=AB+BC+CD+AD$

$P=5+3+5+3$

$P=16$

Therefore, the perimeter of ABCD is 16 units.

Answer 2

The perimeter of polygon ABCD is 16 units

The coordinates are given as:

$\mathbf{A = (-1,1)}$

$\mathbf{B = (4,1)}$

$\mathbf{C = (4,-2)}$

$\mathbf{D = (-1,-2)}$

Start by calculating the distance between the coordinates using the following distance formula

$\mathbf{d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}}$

So, we have:

$\mathbf{AB = \sqrt{(-1 - 4)^2 + (1 - 1)^2} = 5}$

$\mathbf{BC = \sqrt{(4 - 4)^2 + (-2 - 1)^2} = 3}$

$\mathbf{CD = \sqrt{(4 - -1)^2 + (-2 - -2)^2} = 5}$

$\mathbf{AD = \sqrt{( -1- -1)^2 + (-2 - 1)^2} = 3}$

The perimeter is then calculated as:

$\mathbf{Perimeter = AB + BC + CD + AD}$

Substitute values for AB, BC, CD and DA

$\mathbf{Perimeter = 5+ 3 + 5 + 3}$

$\mathbf{Perimeter = 16}$

Hence, the perimeter of the polygon is 16 units

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