Stephan has a box in the shape of a hexagonal prism where the hexagonal bases are regular. a net of the box is below.
figure 1

he measured the height of the box to be 6 in. then, stephan drew a line from the center of one of the hexagons to each of its vertices and noticed that all the triangles he created have a height of 7 in and a base of 8 in.

note: figure is not drawn to scale.

what is the surface area of the hexagonal prism?
a. 912 sq in
b. 624 sq in
c. 456 sq in
d. 480 sq in

The answer is B:  624 sq. in

Step-by-step explanation:

find the area of one hexagons

since there are 6 congruent triangles, you find the area of 1 of the triangles and multiply by 6

Area of 1 triangle:

1/2bh = 1/2 (8 in) (7 in)

=1/2 (56 sq in.)

= 28 sq in.

Area of 1 hexagon is 6 (28 sq. in) = 168 sq. in.

There are 2 hexagons in the net, so multiply by 2

Area of 2 hexagons is...

2 (168 sq. in) = 336 sq. in.

Then find the area of the rectangular sides...which will be the 6 small squares shown on the first picture

**the rectangles each have 1 side that is equal to 1 side of the hexagon and 1 side is equal to the height of the prism.

Area of 1 rectangle side

lw = (8 in) (6 in.)

= 48 sq. in.

multiply all of the 6 sides of the area

=6 (48 sq.in)

=288 sq. in

Then add the area of the 2 hexagons to the area of the 6 rectangular sides

Total area = 336 sq. in + 288 sq. in

= 624 sq. in.

113.1

step-by-step explanation:

v=4

3πr3=4

3·π·33≈113.09734

i think it's eleven too, don't woosh me here, but according to the "experts" it's 2.

the proof starts from the peano postulates, which define the natural

numbers n. n is the smallest set satisfying these postulates:

  p1.   1 is in n.

  p2.   if x is in n, then its "successor" x' is in n.

  p3.   there is no x such that x' = 1.

  p4.   if x isn't 1, then there is a y in n such that y' = x.

  p5.   if s is a subset of n, 1 is in s, and the implication

      (x in s => x' in s) holds, then s = n.

then you have to define addition recursively:

  def: let a and b be in n. if b = 1, then define a + b = a'

      (using p1 and p2). if b isn't 1, then let c' = b, with c in n

      (using p4), and define a + b = (a + c)'.

then you have to define 2:

  def:   2 = 1'

2 is in n by p1, p2, and the definition of 2.

theorem:   1 + 1 = 2

proof: use the first part of the definition of + with a = b = 1.

      then 1 + 1 = 1' = 2   q.e.d.

note: there is an alternate formulation of the peano postulates which

replaces 1 with 0 in p1, p3, p4, and p5. then you have to change the

definition of addition to this:

  def: let a and b be in n. if b = 0, then define a + b = a.

      if b isn't 0, then let c' = b, with c in n, and define

      a + b = (a + c)'.

you also have to define 1 = 0', and 2 = 1'. then the proof of the

theorem above is a little different:

proof: use the second part of the definition of + first:

      1 + 1 = (1 + 0)'

      now use the first part of the definition of + on the sum in

      parentheses:   1 + 1 = (1)' = 1' = 2   q.e.d.