Question

The coordinates of polygon ABCD are A(-4,-1), B(-2,3), C(2,2), and D(4,-3). Use these coordinates to complete the sentences below. The perimeter of polygon ABCD, to the nearest thousandth, is units. If point D were translated 2 units up and 1 unit to the left to create point D', the perimeter of polygon ABCD', to the nearest thousandth, would be units.

Answer 1

Answer:

The perimeter of polygon ABCD, to the nearest thousandth units is

22.227 units

The perimeter of polygon ABCD', to the nearest thousandth, would be 20.980 units

The area of polygon ABCD' is 19.5 units²

Step-by-step explanation:

The coordinates forming the polygon are

A (-4,-1), B(-2,3), C(2,2), and D(4,-3)

The perimeter then is given by the sum of the length of the sides as follows;

Length of line between two X and Y points distance, xi and yj apart is

length XY =  $\sqrt{x^2+y^2}$

Therefore, the length between points AB is

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

= $\sqrt{(-4 +2)^2+(-4)^2} = \sqrt{-2^2+-4^2} = \sqrt{20}$  = 4.472 units

Similarly, length BC is given by

Length BC =  $\sqrt{(-4)^2+1^2} = \sqrt{17}$ = 4.123 units

Length CD = $\sqrt{(-2)^2+5^2} = \sqrt{29}$ = 5.385 units

Length DA = $\sqrt{(8)^2+(-2)^2} = \sqrt{68}$ = 8.246 units

The perimeter is equal to;

length AB + Length BC + Length CD + Length DA

= 4.472 + 4.123 + 5.385 + 8.246 = 22.227 units

If the point D is moved up 2 units and left 1 unit we have

D' = x-1, y+2 where x and y are the coordinates of point D

D(4, -3) → D'((4-1), (-3+2)) = D'(3, -1)

The length D'A =  $\sqrt{(7)^2+(0)^2} = \sqrt{49}$ =7 units

The perimeter of polygon ABCD'

length AB + Length BC + Length CD + Length D'A

= 4.472 + 4.123 + 5.385 + 7 = 20.980 units.

The area is given by the determinant of the 3 by 3 matrix using Cramer's Rule as follows

 $S_{\bigtriangleup}$ = $(\frac{1}{2})|x_1y_2+x_2y_3+x_3y_1-x_1y_3-x_2y_1-x_3y_2|$

= $(\frac{1}{2})|x_1(y_2-y_3)+x_2(y_3-y_1) +x_3(y_1-y_2)|$

Triangle ABC we have

A(-4,-1)

B(-2,3)

C(2,2)

$S_{{\bigtriangleup}ABC}$ = $(\frac{1}{2})|x_1(y_2-y_3)+x_2(y_3-y_1) +x_3(y_1-y_2)|$

$=(\frac{1}{2})|(-4)(3-2)+(-2)(2-(-1)) +2((-1)-3)|$

= $=(\frac{1}{2})|-4-6 -8|= 9 units^2$

Triangle ACD'

A(-4,-1)

C(2,2)

D'(3, -1)

$S_{{\bigtriangleup}ACD'}$ = $(\frac{1}{2})|x_1(y_2-y_3)+x_2(y_3-y_1) +x_3(y_1-y_2)|$

$=(\frac{1}{2})|(-4)(2+1)+(2)(-1+1) +3((-1)-2)| = \frac{21}{2}$ = 10.5 units²

Area of polygon =   $S_{{\bigtriangleup}ABCD'}$ = $S_{{\bigtriangleup}ABC}$ + $S_{{\bigtriangleup}ACD'}$ = (9 + 10.5) units²

Area of polygon ABCD' = 19.5 units².