please help me pleaseeeeeeeeeee?

The surface of the prism included 5 parts :
1 rectangular base, 2 isos. Δs with base 3 ft, and 2 isos. Δs with base 8 ft

Let H be the height of each isos. Δ with base 3ft, and h be the height of each isos. Δ with base 8 ft.

(8/2 ft), (H ft) and (5 ft) form a right-angled Δ with hypotenuse (H ft).
By Pythagorean theorem :
H² = (8/2)² + 5²
H = √41

(3/2 ft), (h ft) and (5 ft) form a right-angled Δ with hypotenuse (h ft).
By Pythagorean theorem :
h² = (3/2)² + 5²
h = √27.25

Area of the rectangular base
= 3 × 8 ft²
= 24 ft²

Total area of the two isos. Δs with base 3 ft
= 2 × [(1/2) × 3 × √41] ft²
= 3√41 ft²

Total area of the two isos. Δs with base 8 ft
= 2 × [(1/2) × 8 × √27.25]
= 8√27.25 ft²

Surface area of the prism
= (24 + 3√41 + 8√27.25) ft²
= 85.0 ft² (to 3 sig. figs.)

okay so surface area is relatively easy as soon as you know how to do it. you must find the area of all of the faces (assuming it's a rectangular prism you would do 3x5, 5x8, 3x8) as soon as you have done that, you have the area of half of the faces, so simply multiply by 2 to get the other half. (3x5x2, 8x5x2, 3x8x2) when you have done that, you must add the sums of 3x5x2, 8x5x2, 3x8x2 and you will get your answer.

h = 5, w = 3, l = 8 (in feet)

surface area = 2x sum of area of each pair of dimensions

2[ 5x3 + 5x8 + 3x8 ]
2[ 15 + 40 + 24 ]
= 2(79) = 158 ft²