find the area of the figure with the given vertices. (5, 5), (2, 1), (11, 1), (8, 5)?

Answer 1

It's a trapezoid. The base is 9 (11-2) and the top is 3 (8-5). The height is 4 (5-1).

The area is the average of the bases x the height.

So that is (9+3)/2 x 4

= 6 x 4
= 24 sq. units.

Answer 2

(5,5) and (2,1)
D = sqrt((2 - 5)^2 + (1 - 5)^2)
D = sqrt((-3)^2 + (-4)^2)
D = sqrt(9 + 16)
D = sqrt(25)
d= 5

(2,1) and (11,1)
D = sqrt((11 - 2)^2 + (1 - 1)^2)
D = sqrt(9^2)
D = 9

(11,1) and (8,5)
D = sqrt((8 - 11)^2 + (5 - 1)^2)
D = sqrt((-3)^2 +4^2)
D = sqrt(9 + 16)
D = sqrt(25)
d = 5

(5,5) and (8,5)
D = sqrt((8 - 5)^2 - (5 - 5)^2)
D = sqrt(3^2)
D = 3

Doing this, and plotting it, i find that the shape is of a Trapezoid

Area = (h(a + c))/2

h^2 + 3^2 = 5^2
h^2 + 9 = 25
h^2 = 16
h = 4

Area = (4(3 + 9))/2
A = 2(12)
A = 24

ANS : 24

Answer 3

Use cross product.
A: (5, 5)
B: (2, 1)
C: (11, 1)
D (8, 5)

Vector BA (3, 4)
Vector BC (9, 0)

| i j k |
| 3 4 0 |
| 9 0 0 |

(3 - 36)i - (0)j + (-36)k

Area Vector = -33i - 36k


Area = √(33^2 + 36^2) = 48.83 square units

Answer 4

The points give you the shape of a rectangle when you put it on a grid. Then, the formula for area is length times width or the distance between the x values times the distance between the y values. Good luck on your homework.