A = ½(¼πc²) = ⅛π8² = 8π mm² (Solved √ ) Extremely easy !!!! Just need to apply the formula of the circular ring area, respect to the chord, previusly moving the yellow semicircle to the center of the blue semicircle (Which doesn't modify the required area)
A = ½(¼πd²) = ⅛π8² = 8π mm² (Solved √ ) Diameters of both semicircles are not given, we can modify these, and the original conditions are still being fulfilled. If we modify the not given diameter of yellow semicircle to zero, then the chord "8" becomes the diameter "d" of that blue semicircle, and this is the required area.
You can slide the center of the small semicircle untill it lies on the center of the big semicircle, then applying Pythagorean theorem R^2-r^2=4^2=16. And being semicircles the area is 16/2pi There was also a logical approach: you can shrink the small semicircle untill its radius =0 then the chord of 8 is the diameter and area semicircle is 16/2pi again 😊
A = ½(¼πc²) = ⅛π8² = 8π mm²
(Solved √ )
Extremely easy !!!!
Just need to apply the formula of the circular ring area, respect to the chord, previusly moving the yellow semicircle to the center of the blue semicircle (Which doesn't modify the required area)
I obtained the same answer by the same method. It took me about a minute to solve it.
A = ½(¼πd²) = ⅛π8² = 8π mm²
(Solved √ )
Diameters of both semicircles are not given, we can modify these, and the original conditions are still being fulfilled.
If we modify the not given diameter of yellow semicircle to zero, then the chord "8" becomes the diameter "d" of that blue semicircle, and this is the required area.
Personally I think this is the best, I find it really interesting, thanks for sharing
You can slide the center of the small semicircle untill it lies on the center of the big semicircle, then applying Pythagorean theorem R^2-r^2=4^2=16. And being semicircles the area is 16/2pi
There was also a logical approach: you can shrink the small semicircle untill its radius =0 then the chord of 8 is the diameter and area semicircle is 16/2pi again 😊
Wow these are two perfect methods, but the best one for me is the second, it just makes the Question interesting, thanks 👍