What's the total surface area of a covered box with a length of 4 ft, a width of 3 ft, and a height of 6 ft? Question 17 options: A) 108 sq ft B) 38 sq ft C) 84 sq ft D) 96 sq ft

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A. 108 ft².

Step-by-step explanation:

Find the area of each face using the dimensions 4, 3 and 6. Remember, there are 6 total faces with 3 sets of congruent faces. The faces are rectangular, so the formula A = l × w will be used to solve for the area of each:

2(4 ·3) = 2(12) = 24 ft²

2(4· 6) = 2(24) = 48 ft²

2(3 · 6) = 2(18) = 36ft².

Add up the areas of the faces:

24 + 48 + 36 = 108 ft².

A) 108 sq ft

Step-by-step explanation:

SA = PH + 2LW

P = 2(L + W)

SA = 2(L + W)H + 2LW

SA = 2(4 ft + 3 ft)(6 ft) + 2(4 ft)(3 ft)

SA = 108 sq ft

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!

560

hope this : )