High intensity training
1. List the following four schematic diagrams for reference.
2. Solution: (1)
Method 1: ∠ b = 90, the centerline EF, as shown in Figure 2- 1.
Method 2: AB = AC, midline (or high) AD, as shown in Figure 2-2.
⑵
AB = 2bc (or ∠ C = 90, ∠ A = 30), and the centerline EF, as shown in Figure 3.
⑶
Method 1: ∠ B = 90 and AB = 2bc, and the centerline EF, as shown in Figure 4- 1.
Method 2: AB = AC and ∠ AB=AC = 90, with median (or high) AD, as shown in Figure 4-2.
⑷
Method 1: Let ∠ b > ∠ c, take a point D on the side of BC, let ∠ gdb = ∠ b pass through AB to G, let's pass through the midpoint of AC, and let ∠ GD pass through BC to F, then EF is the shear line, as shown in Figure 5- 1.
Method 2: let ∠ b > ∠ c, take the midpoint D and E of AB and AC respectively, the crossing D and E are the vertical lines of BC, G and H are the vertical feet, the intercept HF = GB on HC, and EF is the shear line, as shown in Figure 5-2.
Method 3: Let ∠ b > ∠ c be high AD, intercept DG = DB on DC, connect AG, the midpoint of AC E is EF‖AG, and BC to F, then EF is the shear line, as shown in Figure 5-2.
Example 4
Solution: (1) y =,
...................................................., two points.
When x=60, the maximum value of y =1800; Four points.
(2) passing through the be⊥ad b point of E,CF⊥AD in F,
Let AB=CD=xcm and the trapezoidal area be Scm2, then BC = EF = (120-2x) cm.
AE=DF=x，BE=CF=x
，AD= 120-x，
∴S= x type (240-3x)
When x=40, the maximum value of s = 1200, 7 points.
Maximum value of s > maximum value of y 8 points.
(2) The scheme is correct, with 2 points and 4 points for * * *.
Scheme: ① Regular octagon and a half, ② Regular decagon and a half, ③ Semicircle, etc.