Geometry: What's the Area of the Red Rectangle?

Welcome back to our geometry challenge series! In today's video, we dive into an intriguing geometric puzzle. Given that the areas A1 and A2 are equal, can you determine what's the area of the red rectangle? 🟥📏
📊 Video Highlights:
- Understand the geometric relationships in the given figure.
- Step-by-step breakdown of the solution process.
- Key mathematical concepts and formulas used.
🧠 Challenge Yourself:
This puzzle is perfect for sharpening your problem-solving skills and enhancing your understanding of geometry. Pause the video at any point to take a closer look and try solving it on your own before we reveal the solution!
Don't forget to:
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Have questions or suggestions? Drop them in the comments below. I love hearing from you!
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Happy learning and see you in the video!

[@JLvatron]

Great puzzle! I solved it in my head!

[@johngutwirth7706]

Elegant indeed ! THX

[@andrec.2935]

Probleminha de uma beleza infinita!

[@cvb-bm5dg]

Well Done! Especially the graphics.

[@ThePhantomoftheMath]

@@cvb-bm5dg thank you friend!

[@quigonkenny]

We are given that the left blue region (call it S) and the right blue region (call it Q) are the same area. S is part of a semicircle, and Q is part of a quarter circle, and the remaining part (call it C) is the same part for both, so the area of Q+C is the same as the area of S+C. We know from the given length of the left side of the rectangle as 1 that the radius of the quarter circle is also 1, as the right side of the rectangle is that radius. If the radius of the semicircle is r, then r can be found by equating the formulas for the area of a quarter circle and the area of a semicircle and plugging in the respective radii: πR²/4 = πr²/2 π(1)²/4 = πr²/2 πr²/2 = π/4 r² = (π/4)(2/π) = 1/2 r = 1/√2 The width of the red rectangle is equal to the difference between the diameter of the semicircle and the radius of the quarter circle: w = 2r - R = 2/√2 - 1 = √2 - 1 Area = lw = 1(√2-1) = √2 - 1 sq units

[@ThePhantomoftheMath]

Nice! Good job!

[@jacquespictet5363]

Nice. It could be worth mentioning [1] the general formulae r=R/√2 and w=R(√2-1); [2] that, as r and w are linear functions of R, the shape of the figure won't change with R; [3] that, as π is out of the equation, it is possible to treat it as rectangles instead of circles.

[@amirhusseinmontazeran6685]

Thank you. Is so attractive for me.

[@ThePhantomoftheMath]

Glad you like it! Cheers!

[@vcvartak7111]

Fantastic

[@d.dpatel5010]

Excellant

[@mddmotivation]

❤❤

[@ThePhantomoftheMath]

❤️❤️

[@leonznidarsic]

What is the area of ​​the shaded fields? You didn't calculate that, you only calculated the area of ​​the red rectangle?!

[@ThePhantomoftheMath]

Hi friend. The challenge was to find the area of the red rectangle, and not the area of shaded fields!

[@alain_b_]

❤🧡💛💚💙💜🤎🖤🤍