【Haichuan Teahouse】Brain-teaser Series: Is this image clever or over the top?
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A teacher at a key primary school gave the students a homework question for them to think about. But that night, in the parents’ group for students, they had a heated discussion about this question. Some parents believe that although this question is difficult, it is very clever ； Some parents believe that this question is beyond the curriculum scope and cannot be solved using elementary school knowledge at all ； Some parents also believe that this question is wrong and has no solution at all. It left the teacher in an awkward situation for a while. Dear sailors, what do you think? Feel free to leave comments and share your views. It is given that ABCD is a square, OF is perpendicular to OC, the length of OF is 4, and the length of OC is 6. Find the area of the shaded region.
Is this a primary school question? . . .
It’s for key primary schools; it seems way above the level required
This post was last edited by Yang Qing on 2021-7-10 at 12:48. With the extensions of OF and CD intersecting at G, the Pythagorean theorem and similar triangles can be used to determine that FG=4, DG=3.2, and FD=2.4… Thus, this problem has no solution if it is a square, but it can be solved if it is a rectangle. ;P
It doesn’t seem quite right either; let me think about it again Why can’t I delete the image from the post edited on this forum? ?
This post was last edited by Yang Qing on 2021-7-10 12:59. If B’C=DC=6.8, then the endpoint B of BF is not at a corner of the square~
Therefore, the original question should be: Given that ABCD is a rectangle, OF is perpendicular to OC, with OF having a length of 4 and OC having a length of 6, find the area of the shaded region.
The graphical method is the basic approach for solving these problems
Because, given the two length conditions of this problem and the assumption of a \"square\", it is not possible to draw a diagram. I suddenly thought of extending BF to point G; I’m not sure if that’s correct, but it seems like that’s the way it should be. Let’s see how others would solve this.
Draw a perpendicular from point F to line BC; this perpendicular intersects BC at H and EC at R. Triangles FDC are congruent to triangle FHC, and triangles FBH are congruent to triangle BCE. (In right triangles, if one angle is equal and the adjacent sides are equal, then the triangles are congruent – this is beyond the scope of the curriculum.) The area of the shaded region equals FHC + BOE, which is also equal to FRC + RHC + BOE. Since RHC + BOE + quadrilateral BORH equals FRO + BORH, the area of the shaded region is 4*6/2
This post was last edited by Yang Qing on 2021-7-13 at 12:20. The main issue is that the premise of the question is incorrect; it isn’t a square that meets the conditions given, but rather a rectangle. The area of the shadow can be calculated as 15.75375:Q