Calculate the Area of the Red shaded Region | Geometry Olympiad Problems

Calculate the Area of the Red shaded Region | Geometry Olympiad Problems

#olympiad_geometry
#rmo_geometry_questions
#geometry_olympiad_problem,
#premath
#mindyourdecision

[KipIngram]

Just presuming that there must be a well-defined answer implies that the answer is independent of the scale. If you made the large circle bigger, you'd have to make the small circle bigger too, to push the line up such that its length remained 16. So, take the limiting case where the small circle has zero radius - then the length 16 becomes the diameter of the large half circle, and the entire area of that half circle becomes the red shaded area. So - diameter = 16 implies radius = 8, which implies a circle area of pi*8^2 = 64*pi, and half of that is 32*pi. So that must be the answer.

PROVING that the scale didn't change the answer would require more work, similar to what's shown here in the video. But... given that an answer is expected for this problem, it MUST be the case that it's scale independent.

[santiagoarosam]

La superficie buscada es la mitad de la corona circular delimitada por dos circunferencias con radios iguales a los de los semicírculos de la figura. ---> Área sombreada =(16/2)²π/2=32π.
Gracias y saludos.

[bobbybannerjee]

Well done.
Can you tell me what software (app) is that by which you can write ✍️ electronically on the screen with your stylus pen?

[botfeeder]

Nice problem but too easy to be Olympiad. A=(1/2)*pi*(R^2--r^2). By pythagorean, r^2+8^2=R^2. Thus R^2-r^2=64, hence A=32*pi