A fifth-grade competition question was previously released:Two right-angled triangles share right-angled vertices and right-angled edges, and find the difference in their areasIt was extremely difficult, the whole class was wiped out, and the top students were not spared!
Question 448: As shown in the figure,
The hypotenuse lengths of the right-angled triangle ABC and FCE are 10 and 6, respectively, AF=CE, and the area of the blue-shaded quadrilateral ACEF is found.
1. Extracurricular analysis suitable for middle school students: Pythagorean theorem + unknown metaalgebraic operation!
Let ab=x,bc=y,af=ce=z, which is obtained from the Pythagorean theorem, x +y =100, (x-z) +y-z) =36, i.exz+yz-z²=32
S quadrilateral acef=s△abc-s△aef=1/2xy-1/2(x-z)(y-z)=1/2xz+1/2yz-1/2z²=32÷2
2. Suitable for the fifth grade: puzzles!
Entry point or topic: af=ce, the difference between the two right-angled sides is equal.
Assemble 4 triangles identical to ABC into 1 large square with AC as the side: the hook (i.e., the short right-angled side) and the strand (i.e., the long right-angled side) are alternately aligned. A side length of 10 large squares ABGH can be obtained, and there is a small square EFIJ inside, the side length of which is equal to BC-AB. See Figure 2.
If gn=hm=af=ce is intercepted on gj and hi respectively, and ef, fm, mn, nc are connected, then efmn is a square with a side length of 6. As shown in Figure 3.
S quadrilateral acef=(s square acgh-s square efmn) 4 = (100-36) 4
Friends have good ideas or methods, welcome to leave a message to share!