Question 870741
NOTE: I am answering as fast as possible, so check my calculation, just in case I make a mistake.
 
The surface of that prism includes 2 hexagons that I'll call the bases, and 6 rectangular faces that I will call the sides.
The sides seem to be rectangles measuring 12 cm by 4 cm,
so each one of those faces has an area of (12 cm)(4 cm) = 48 square centimeters.
The hexagonal bases can be divided into 6 equilateral triangles with sides measuring 4 cm. (The dividing lines connect each vertex to the center, like spokes in a wheel).
The area of each triangle can be calculated as (1/2)(base)(height).
The height of those triangles is the distance from the middle of each side of the hexagon to the center of the hexagon, and it is called the "apothem" of the polygon
So, the area of the hexagon can be calculated as 6(1/2)(side)(apothem)=(1/2)(6(side))(apothem).
Since 6(side)=perimeter, the formula is written as (1/2)(perimeter)(apothem by those who love formulas and like the word "apothem".
That is how my 6th grade teacher taught it to me, and we were supposed to get the ratio of apothem to side from a table.
We would not have needed a table if we had been told about the Pythagorean theorem.
The height (the length of the altitude segment) of one of those triangles in your problem measures {{{2sqrt(3)}}} cm.
The altitude divides each equilateral triangle into 2 right triangles and  we can apply the Pythagorean theoren to find the height, {{{x}}} .
{{{drawing(300,300,-1,0.2,-0.55,0.55,
green(triangle(-0.866,0,0,-0.5,0,0.5)),
triangle(-0.866,0,0,0.5,0,0),
rectangle(0,0,-0.05,0.05),
locate(-0.03,-0.2,2),locate(-0.433,-0.25,4),
locate(-0.03,0.3,2),locate(-0.433,0.25,4),
locate(-0.4,0.05,x)
)}}} {{{x=sqrt(4^2-2^2)=sqrt(16-4)=sqrt(12)=sqrt(4*3)=sqrt(4)*sqrt(3)=2sqrt(3)}}}
So the surface area of each base is
{{{6*(1/2)*4*(2sqrt(3))=24sqrt(3)}}} square centimeters
and since the area of each of the 6 rectangular sides was {{{48}}} square centimeters, the total surface area (in square centimeters) is
{{{2(24sqrt(3))+6*48=highlight(288+48sqrt(3))}}}
The approximate value is {{{highlight371.14)}}}
 
(b)Water flows through a cylindrical pipe of radius 0.74 cm. It fills a 12 litre bucket in 4 minutes. 
 
(i) Calculate the speed of the water through the pipe in centimetres per minute.
The rate the water is flowing is {{{12L/"4 minutes"=3}}}{{{L/min}}} .
Since 1L is 1 cubic decimeter, a cube 10cm to a side,
1L is (10cm)(10cm)(10cm) = 1000 cubic centimeters.
So in one minute 3,000 cubic centimeters of water come out of the pipe.
Before it started coming out, those 3,000 cubic centimeters of water coming out of the pipe each minute were in the form of
a cylinder of radius 0.74cm, and height {{{h}}} cm.
The water at the back of that {{{3000}}}{{{cm^3}}} cylinder that just came out traveled a distance {{{h}}} in one minute.
That {{{h}}} is the speed of the water through the pipe in centimetres per minute.
The volume of a cylinder of radius {{{r}}} and height {{{h}}} is {{{pi*r^2*h}}}
So {{{pi*0.74^2*h=3000}}}
Rounding, I get {{{pi*0.74^2=1.7195}}} , so
{1.7195h=3000}}}--->{{{h=3000/1.7195}}}-->{{{highlight(h=1745)}}}
 
(ii) When the 12 litre bucket is emptied into a circular pool, the water level rises by 5 millimetres. Calculate the radius of the pool correct to the nearest centimetre.
That {{{12L=12000}}}{{{cm^3}}} is the volume of water added.
Once it was added to the pool, it formed a cylinder of radius {{{r}}} cm and height {{{5mm=0.5cm}}} at the top of the pool.
The volume of that cylinder (in {{{cm^3}}} ) is
{{{pi*r^2*0.5=12000}}}-->{{{r^2=12000/(pi*0.5)}}}
The approximate value is {{{r^2=7639}}} ,
so the approximate value for {{{r}}} in cm is
{{{r=sqrt(7639)=approximately}}}{{{highlight(87.4)}}}