Area of Sector Removed

[jaegers]

Sounds so dumb, but this isnt making any sense. I know the area of the circle is 30.190in^2

 

[jaegers]

I just cant figure out where to include the 400?

 

[Dr.Peterson]

I presume the sector they are removing is the one with the 400 inch arc! They ought to say so.

The area of a sector is proportional to the arc length. What is the arc length (circumference) of the entire circle?

But that makes it clear that the problem is impossible, since the circumference is far less than 400 inches (assuming that the units for both diameter and arc are in inches, which should have been indicated!). It looks like a misprint. Or could 400 be a measure of angle using some units I've never heard of? Does the context of the question say anything about that?

 

[pka]

Agree, the problem should have stated more information.
Have a look HERE.

 

[jonah2.0]

Beer soaked conjecture follows.

View attachment 21245

Sounds so dumb, but this isnt making any sense. I know the area of the circle is 30.190in^2

I just cant figure out where to include the 400?

I thought 400 might be 400 mils; so Area*400/6400. No cigar.
Three more beer bottles and after washing some dishes, it occurred to me that maybe 400 may have been 40° all along. If it really was 40°, then it was a sloppy rounding in addition to a sloppy labeling.

 

[HallsofIvy]

The problem is that the information given makes no sense! We are told that the diameter or the circle is 6.2 so its circumference must be \(\displaystyle 6.2\pi\) which is approximately 19.5, much smaller that "400". If that 400 is in different units than the diameter, we need to be told what those units are!

 

[Deleted member 4993]

I think most reasonable interpretation is what Sir Jonah gave us above (#6):

400 \(\displaystyle \to \ \ \) 40^o

 

[HallsofIvy]

I think most reasonable interpretation is what Sir Jonah gave us above (#6):

400 \(\displaystyle \to \ \ \) 40^o

Very good! and 40/360= 1/9 so the area of the sector is 1/9 the area of the entire disk and the remaining area is 8/9 the area of the entire disk.