As shown in Figure 1, it is an additional problem in mathematics for the fifth grade of primary school: after the two squares of one large and one small are stacked, the area of the remaining part of the small square!
Figure 1
Pei Xiao Mathematics Problem 17: One large and one small square are stacked together, one vertex of the small square coincides with the center of the large square, the side length of the large and small squares is 6cm and 4cm respectively, find the area of the shadowed part.
This problem is actually a "wrong question" and needs to be solved in three types of situations, but two of the "serious over-level" need to be solved by high school knowledge, and 99% of people will miss two types of over-level situations!
First, the root cause of the "error": 3√2>4.
From "the side length of the small square is 4, the side length of the large square is 6", it is easy to know that 3√2>4, which is the root cause of the "wrong question". "3√2>4" will cause points D and E on the small square to fall into the inside of the large square, D and E are shown in Figure 2.
Second, how to correct the error? Remember that the sides of the large and small squares are a and b, and they need to meet a√2≤2b.
For such question types for elementary school students, generally only consider point D on the outside or side of the large square, otherwise it is seriously overclass, and high school knowledge needs to be used for classification and solving.
For convenience, remember that the sides of the large and small squares are lengths a and b, and in order to ensure that point D is outside or on the side of the large square, it is necessary to meet a√2≤2b.
Specific to question 17, if the value of the side length of the small square remains unchanged to 4, the value of the side length of the large square must meet the requirements of "4≤a≤4√2". If the value of the side length of the large square remains unchanged to 6, the value of the side length of the small square must meet the requirements of "6≥b≥3√2".
For example, if the side length of a large square is 6, and the side length of a small square is 5, you can avoid point D falling into the inside of the large square.
For another example, if the side length of the small square is 4, and the side length of the large square is 5, the point D can also be prevented from falling into the inside of the large square.
Third, the analysis of wrong questions: For primary school students, question 17 is actually "wrong questions"!
For illustrative purposes,
In Figure 1, the vertex of the large square is C, the midpoint of its upper side is A, the vertex of the small square is E, the intersection point on the sides of the two squares is recorded as B, and the side connecting OC and the side of the small square intersects at point D, as shown in Figure 2.
Figure II
Note: In Figure 2, (1) points O, A and C are fixed and do not change with the intersection or coincidence positions of the two squares; (2) Point B will vary with the intersection position of the two squares, point D moves on OC and point E moves on OB.
From Figure 2, question 17 needs to consider three situations:
Case 1) Point D is on the outside or side of a large square, and point E is on the outside of a large square.
Case 2) Point D is inside the large square, and point E is on the outside or side of the large square.
Case 3) Points D and E are both inside the large square.
Among them, only case 1) can be solved with elementary school knowledge. Therefore, question 17 is actually a "wrong question", there is no doubt about it!
4. Scenario 1) is not over-the-top solution
In this case, OD≥OC=3√2 and OB<OE=4. It can be understood as "from the side of the small square OB and OA coincide as the initial case, and the small square rotates around the point O on both sides of the OA <45°-arctan√2/4."
Solution 1:
Since point D is on the outside of the large square or coincides with point C, point E must be on the outside of the large square.
Figure III
Noting that OB is vertical OA', and the reverse extension OB and OB' intersect on the sides of the large square, then the large square is divided into four regions by two straight lines perpendicular to each other and passing through the center. That is, S shadow = 16-36÷4 = 7cm²
Solution 2:
Rotate △OAB around point O 90° counterclockwise to △OA'B', then S△OAB=S△OA'B', that is, S quadrilateral OBCB'=S square OACA' = 1/4S large square = 9cm², that is, S shadow = 16-9=7cm².
5. Scenario 2) description and super-level analysis
In this case, OD<OC=3√2 and OB≤OE=4, that is, 45°-arctan√2/4<∠AOB≤arctan√7/3.
When OE=OB is that B coincides with E, AB=√7 and BC=3-√7
Figure IV
The small square is actually OBGB', and D falls inside the large square, the inside of the OB, and is written as D'. THE SHADED FIGURE IS △B'GH, GB'=OB=4, where H is the intersection point of BG and the left side of the large square.
It is noted that △BCH is similar to △OAB, and its corresponding side length ratio is (3-√7)/3.
It is also known that S△OAB=3√7/2, so S△BCH=3√7/2×(16-6√7)/9=(8√7-21)/3.
Therefore, S shadow = S small square - S quadrilateral OBCB' + S △BCH
=7+(8√7-21)/3=8√7/3＞7。
When the OE is greater than the OB, it can be solved similarly.
6. Scenario 3) description and super-class analysis
Figure V
In this case, OD<OC=3√2 and OB>OE=4, that is, ∠AOB>arctan√7/3. The position relationship between small squares and large squares, as shown in Figure 5. Small squares are OEMNs. The intersections of EM, MN, and the left side of the large square are denoted as P and Q respectively, and the shade is △PMQ. Note that △PMQ is similar to △B'NQ.
At this time, the area of the shadow part changes with the change of E and N landing points.
In the special case E and N fall on the diagonal of the large square, that is, the point E is on OB and OE=4.
Only for this special case, the solution is as follows:
At this time, the shadowed part △PMQ is an isosceles right triangle.
OP=OQ,OM=4√2, OM vertical PQ, write the intersection point of OM and PQ as K, then OK=3, MK=4√2-3. Thus, the hypotenuse length of △PMQ is 8√2-6, so the S shadow △PMQ=(164-96√2)/4=41-24√2.
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