Helpp nowww 25 points

Surface area = 197 in²

Step-by-step explanation:

To determine the surface area of the solid shown in the diagram, we would first determine the area of the different faces.

1) It has 2 trapezoidal faces at the front and opposite end, the formula for determining the area of a trapezoid is expressed as

Area = (a + b)h/2

a and b are the length of the bases while h is the height of the trapezoid.

Area of trapezoid = (5 + 9)5/2 = 35 in²

Area of the trapezoid at the opposite end = 35 in²

2) Area of base = 9 × 5 = 45 in²

3) Area of top = 5 × 5 = 25 in²

4) Area of slant face = 6.4 × 5 = 32 in²

5) Area of back = 5 × 5 = 25in²

Surface area of the design is the sum of the area of its faces. It becomes

35 + 35 + 45 + 25 + 32 + 25 = 197in²

there are lots of ways.   pick's theorem tells us if the vertices are lattice points the area a is

a = b/2 + i - 1

where b is the number of boundary lattice points and i is the number of interior lattice points.

in the figure i count b=16, i=50 so

a = 8 + 50 - 1 = 57

let's try the shoelace formula.   the area is half the sum of the cross products of the sides.  

a(-2,6)   -2(2)-6(-5) = 26

b(-5,2)   -5(-3) - 2(-2) = 19

c(-2,-3) -2(-3) - (-3)(4) = 18

d(4,-3)   4(3) - (-3)(5) = 27

e(5,3)   5(3) - 3(1) = 12

f(1,3)     1(6) - 3(-2) = 12

a(-2,6)

each of the cross products is twice the area of the triangle with the associated side and one vertex the origin. so the area of abo=26/2=13.

area = (1/2) (26 + 19 +   18 + 27 + 12 + 12) = 57

that checks!

for the friend request.

when a figure, such as triangle (xyz) is reflected over the line y = x to create x'y'z' the x-coordinates and y-coordinates of the original image (pre-image) will be reversed for the image. (e.g., (x, y) > (y, x)) this means that the pre-image position of the x- and y-coordinates switch places which in turn provides a mirror like image of the original xyz.  

i notice that when i draw a line segment from x to x', that the points are parallel from each other -- they are exactly across from each other and do not intersect with any other points.  

yes, think i would see the same characteristic if you drew the line segment connecting y with the reflecting line and then y' with the reflecting line because of the essence of reflections. because it is a mirror image, each point was effected the same way which leads me to conclude that they same would apply to y and y'.