SOLUTION: Figure 12.36 shows a cross-section of a sheet of corrugated iron.
{{{drawing(400,1400/9,-18,18,-7,7,
green(line(-10sqrt(3),0,10sqrt(3),0),
line(-10sqrt(3),0,-5sqrt(3),-5),
l
Algebra ->
Trigonometry-basics
-> SOLUTION: Figure 12.36 shows a cross-section of a sheet of corrugated iron.
{{{drawing(400,1400/9,-18,18,-7,7,
green(line(-10sqrt(3),0,10sqrt(3),0),
line(-10sqrt(3),0,-5sqrt(3),-5),
l
Log On
Question 1210270: Figure 12.36 shows a cross-section of a sheet of corrugated iron.
The sheet is a series of arcs of radius 10 cm., each arc subtending 120o at its centre.
If there are 14 such arcs in one sheet, how wide would the sheet be if flattened out? Found 2 solutions by greenestamps, Edwin McCravy:Answer by greenestamps(13203) (Show Source):
I agree with greenestamps that the way this problem was typed, it made no sense.
Here's how it was typed:
-------------
"Figure 12.32 shows, a, cross-sectional of a sheetis a series of ark of
radius 10 cm each acts of stained 120° at eating time if there are 14 such ax in
one sheet how wide would a sheet be if flatted out"
-------------
But, believe it or not, I googled and found this very problem online. Here's the
way it was stated online:
Figure 12.36 shows a cross-section of a sheet of corrugated iron.
The sheet is a series of arcs of radius 10 cm., each arc subtending 120o at its centre.
If there are 14 such arcs in one sheet, how wide would the sheet be if flattened out?
I only drew two of the 14 arcs above. So I'll find the length of one arc and
multiply it by 14.
The length of an arc is of a circumference. So the length
of one arc is
Multiplying by 14:
, approximately 293.2 cm.
Edwin
