This may be an old problem - from Quora

K

khansaheb

Full Member
Joined
Apr 6, 2023
Messages
887
1686596994050.png
What's the area of the total shaded region?

Given:
  • AOB is a quarter cycle
  • OA || DC
  • DC = 12 (any unit - cm or in or mile or whatever you want)

    This problem could be solved without knowing "almost" any geometry or algebra.
 
khansaheb said:
View attachment 35971
What's the area of the total shaded region?

Given:
  • AOB is a quarter cycle
  • OA || DC
  • DC = 12 (any unit - cm or in or mile or whatever you want)

    This problem could be solved without knowing "almost" any geometry or algebra.
Click to expand...
Using very little geometry, together with the assumption that there is an answer (a fixed number not dependent on the facts we don't know), it's easy to give an answer: Just pick a very simple size for the white circle ...

I've learned, though, not to trust that there actually is an answer to every question; sometimes there just isn't enough data, or the question was stated incorrectly. So I don't like that approach, as elegant as it is.

There is another easy way to get an answer with no written work, which uses the Pythagorean Theorem applied to triangle OCD. To use it directly, I have to use the fact that the area of a circle is the same as that of a quadrant of a circle whose radius is the diameter of the given circle. Does that count as using geometry?
 
Circle in quadrant.png​

Pursuing @Dr.Peterson's "hints" my solution was...

Consider the complete (purple) circle that AOB is a quadrant of; its area would be πR² therefore the area of the quadrant (AOB) is ¼πR².

The area of the smaller, white circle is πr², therefore, the Shaded Area = ¼πR² - πr².

But the length OD is 2r, therefore, by Pythagoras:-

R² - 4r² = 12² = 144​

If we multiply both sides by π, we get:-

πR² - 4πr² = 144π​

and dividing throughout by 4 gives:-

¼πR² - πr² = 36π​

but, as shown above, ¼πR² - πr² is the Shaded Area

Therefore, the Shaded Area = 36π cm² or in² or square poronkusemaa ? or whatever else you prefer.

QED ?​
 
My quick way was to shrink the white circle down to nothing, so that the radius of the large circle is 12 units; then the shaded area is 1/4 of pi times 12 squared, which is ... 36 pi square units.

My visual way was to reshape it as an annulus (in which form it is definitely an old problem):

1686613635615.png​

So the shaded area is 1/4 pi R^2 - 1/4 pi r^2 = 1/4 pi 12^2 = 36 pi square units.

(I'm not using Latex because the whole point is to minimize the work.)
 
Altered the GIF slightly but it was too late to replace in the post so here's better one...

shrinking the circle.gif​
 
You must log in or register to reply here.
Top