
Area and Pythagoras
Hey! I couldn't find a section here that was dedicated to this topic, so i posted here:
I'm having trouble.
In my Textbook it has: Use (pie sign) = 3.14
1. (a) Draw the net of this cylinder/[it's a normal cylinder only it's on it's side and the length is 10 cm, and the width is 8 cm.
(b) What shapes have you drawn?
(c)What is the radius of each circle?
(d) What is the area of each circle?
(e) what are the dimentions of the rectangle?
(f) What is the area of the rectangle?
(g) What is the total surface area
And well in my other textbook it has the answers like:
a)a rectangle with 2 circles, one on each side,
b)1 rectangle 2 circles
c)4 cm
d)50.3 cm2
e)25.1 cm x 10 cm =
f)251.3 cm2
g)351.9 cm2
I don’t see how they got (d) and from then on!
And for workings they have: A = Pie R 2 (s is squared)
= pie x 4 x 4
= 50.265
= 50.3 cm2
C = pie d
= pie x 8
= 25.132
= 25.1 cm

Since the width is $\displaystyle 8 \ \text{cm} $, the radius of each circle is $\displaystyle 4 \ \text{cm} $. So the area of a circle is $\displaystyle A = \pi r^2 $ or $\displaystyle A = \pi (4)^2 = 16 \pi \approx 50.3 \ \text{cm}^2 $.
The net of a cylinder has a rectangle with its width= circumference of cylinder and length= height of cylinder because it is rolled up to make the cylinder.
So we know that the length = $\displaystyle 10 \ \text{cm} $ and the width = $\displaystyle C = 2 \pi r = 2 \pi (4) = 8 \pi \approx 25.1 \ \text{cm} $
Then the area of the rectangle is $\displaystyle A = \ \text{length} \times \ \text{width} = 25.1 \times 10 = 251 \ \text{cm}^2 $
The surface area is $\displaystyle 2 \pi \cdot 16 + 251 \approx 351.9 \ \text{cm}^2 $ or the area all the individual shapes. So its the area of 2 circles + the area of 1 rectangle.

"So we know that the length = http://www.mathhelpforum.com/mathhe...5c7e65a11.gif and the width = http://www.mathhelpforum.com/mathhe...bbf8d0d01.gif"
actually it's the other way around! 10 is the width and 25.1 is the length~