【Haichuan Teahouse】Too many lines are also a source of distraction
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It’s okay, let’s all use our brains
It’s a bit tricky; it seems that definite integrals could be used to solve it. Unfortunately, I already returned it to the teacher. Maybe geometry could help, as there are properties related to squares inscribed in or circumscribed around circles that might lead to a solution. But I can’t remember those properties anymore. Homework and exams for kids these days are indeed quite difficult.
Let the radius of the larger circle be R, the radius of the smaller circle be r, the side length of the larger square be X, and the side length of the smaller square be x. Since the area of the shaded region is 25, we have X2 – x2 = 100. Moreover, X2 = 2R2 and x2 = 2r2; therefore, R2 – r2 = 50. The area of the annulus is π(R2 – r2) = 3.14 * 50 = 157
This post was last edited by Yang Qing on 2021-6-1 at 13:59. The solution can indeed be found; by utilizing the properties of the inscribed circle in a square or the Pythagorean theorem, one can determine the values as stated in the question. By solving the relevant equations, it is found that the radius of the outer circle is 10, while the radius of the inner circle is 5 times the square root of 2. With these two radii known, the area of the annulus is calculated to be 50π – 50 multiplied by pi
The system of equations states that: the area of one-quarter of the circle minus the area of the two white regions equals the area of the shaded region (the value of pi can be canceled out during the calculation)
Well, it’s relatively simpler; it can be calculated based on the difference in the areas of the triangles~
The outer circle radius corresponds to the longer side of the trapezoid, the inner circle radius corresponds to the shorter side of the trapezoid, and the height of the trapezoid forms a 45-degree angle with the radii of the inner and outer circles.
The simplest approach – no need to calculate the actual radius
This post was last edited by wang*neng on 2021-6-2 13:01: 50π
It seems like we don’t need such advanced concepts as definite integrals