3 Ways of Solving The Red Shaded Area With Limited Data

🔴 3 Ways of Solving The RED Shaded Area With Limited Data?
In this video, we tackle a fascinating geometry problem where you're challenged to find the area of the red-shaded region-using only limited information! This problem is perfect for anyone who loves solving puzzles, working through tricky math challenges, or sharpening their geometry skills.
Watch as we break down the steps to approach the problem, explore different methods, and find the solution. Whether you're a student, math enthusiast, or just someone curious about how this works, this video is for you!
🔍 Topics Covered:
- Geometry Problem Solving
- Area Calculation Techniques
- Thinking Critically with Limited Data
💡 Can you solve it? Let's find out together!
Don't forget to Like, Subscribe, and Share if you enjoy the video and want to see more math challenges like this one. Hit the notification bell to stay updated with our latest content!
Join the Chanel 👉 / @thephantomofthemath
-------------------------------------------------------------------
📧 Contact Me:
✉️ [e-mail]
------------------------------------------------------------------
Happy learning and see you in the video!
#Geometry #MathChallenge #AreaProblem #PuzzleSolver #ShadedArea

[@ludosmets2018]

Point A can also be slid down to point O. The area of the small quarter circle then becomes 0 and the diameter of the semicircle is then the same as the radius of the large quarter circle. The area of the large quarter circle minus the area of the semicircle is 4π - 2π = 2π.

[@jacquespictet5363]

Nice. I would emphasize the fact that the area is independent of r. But can we know it before making some calculations ?

[@devondevon4366]

Answer 2pi
Let the radius of the quarter circle to the LEFT =n , Hence, the area = n^2 pi /4
Hence, the radius of the quarter to the RIGHT = 4-n , Hence, the area = ( n^2 + 16- 8n pi) /4
Hence, the diameter of the semi-circle on TOP = (4-n-n) or 4- 2n)
Hence its radius is 2-n, Hence, the area of the semi-circle on Top = (n^2+4- 4n + 4pi)/2
Since the other two have a denominator of 4, Let's change this also to 4 by multiplying the numerator
and denominator by 2 (2n^2 + 8-8n + 8pi)/4.
Since the area of the shaded region
is the Large quarter circle - the Large semi-circle + the small quarter circle, then
( n*2+ 16 -8n pi) /4 - ( 2n^2+8 -8n + 8pi)/4 + n^2 pi/4
pi/4 ( n^2 + 16-8n - 2n^2 -8 + 8n - 8pi + n^2 factor out pi/4
pi/4 ( 2n^2- 2n^2 -8n+ 8n+ 16-8)
pi/4 ( 0 + 16-8)
pi/4 ( 8)
8/4 pi
2 pi

[@burakozc3079]

calculated in 4 seconds.

[@santiagoarosam430]

Cómo no se da el valor de los radios del cuarto de círculo pequeño y del semicírculo, podemos desplazar verticalmente el punto de tangencia entre ambos hasta que el radio del semicírculo se anule y la zona roja se transforme en un semicírculo de diámetro 4 ud→ r=4/2=2→ Área roja =2²π/2=2π ud².
Gracias y saludos.

[@marioalb9726]

Limit condition: r₁=R=2cm. ; r₂=0
A = ½πR² = 2π cm² (Solved √ )
With these modifications, the calculation is easier, and original conditions are meeting
Other limit condition:
r₁=0 , r₂=½R , R=4cm
A = ¼πR² - ½πr₂²
A = ¼π4² - ½π2²
A = 4π - 2π = 2π cm² ( Solved √ )

[@jaspervantwout5583]

1/2π in 2 seconds

[@devondevon4366]

Answer 2pi

[@jmpeax3596]

Background music is super annoying