SOLUTION: Mr. Javier designed an arch made of bent iron for the top of a school’s main entrance. The 12 segments between the two concentric semicircles are each 0.8 meter long. Suppose

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Question 997069: Mr. Javier designed an arch made of
bent iron for the top of a school’s main
entrance. The 12 segments between
the two concentric semicircles are
each 0.8 meter long. Suppose the
diameter of the inner semicircle is 4
meters. What is the total length of the
bent iron used to make this arch?
please help me to solve this problem, thatk you
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
It may look like this:
or maybe the 12 straight segments are placed between the two semicircles
in a different arrangement.
There are 12 straight pieces measuring 0.8meters each, for a total of
12%280.8meters%29=9.6meters .
There are also 2 half-circle arcs.
The inside arc has a radius of 4meters%2F2=2+meters .
The outside arc has a radius of 2meters%2B0.8meters=2.8meters .
Since the circumference of a circle of radius R can be calculated as 2pi%2AR ,
the length of a semicircle of radius R can be calculated as
pi%2AR .
So, the length needed for the inside semicircle is
pi%2A%282meters%29=about6.28meters ,
and the length needed for the outside semicircle is
pi%2A%282.8meters%29=about8.80meters .
The total length of the bent iron used to make the arch is about
9.6meters%2B6.28meters%2B8.80meters=highlight%2824.68meters%29 .