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The eighth grade last semester mathematics monthly examination questions.
2 1, a: The rectangle BCDE in Figure 4 is a golden rectangle. ?

Proof: If MN=2, we can know that AC= 1 and BC=2 from the folding process. AD=AB。

In the triangle ABC, AB = √ (AC 2+BC 2) = √ (1+22) = √ 5?

So CD=AD-AC=(√5)- 1?

And BC=2?

So the aspect ratio of rectangular BCDE is CD/BC=[(√5)- 1]/2?

So the rectangle BCDE is a golden rectangle. ?

22, see figure

24、

(1), the quadrilateral ABCD is prismatic.

Prove:

For convenience, I marked some ∠ 1 = ∠ ADB, ∠2=∠DBC, ∠3=∠DAC, ∠4=∠ACB, ∞.

Because the quadrilateral ABMN is a parallelogram,

So a ‖BM

So: ∠ 1=∠2, ∠3=∠4.

And because BD and AC share ∠ABC and ∠BAD respectively.

So: ∠5=∠2, ∠7=∠3.

So: ∠ 1=∠5, ∠4=∠7.

So: AB=BC=AD

So: the quadrilateral ABCD is prismatic. A quadrilateral with four equilateral sides is prismatic.

(2) Solution: A (4,4 √ 3), C (8,0), D (12,4 √ 3).

Let: the inverse proportional function passing through point d be y = k/x.

Then: replace the coordinates of d (12,4 √ 3) with y=k/x to get the equation? 4√3=k/ 12。 ? Solution: k=48(√3)

Therefore, the inverse proportional function of crossing point D is y = 48 (√ 3)/x.

(3)

Solution: Let the equation of straight line MN be y=kx+b, M(m, 0), N(m+4, 4√3).

Then the equation: mk+b=o, (m+4)k+b=4√3.

Solution: k = √ 3, b = (-√ 3) m.

So the equation of line MN is: y = (√ 3) x-(√ 3) m.

The coordinates of the intersection of the straight line MN and the image whose inverse proportional function is y=48(√3)/x are:

Solve the equation y=(√3)x-(√3)m, y = 48 (√ 3)/x, and you get

y =[48(√3)]/{[m √(m^2+ 192)]/2}

According to the meaning, y=(4√3)/2=2√3.

So [48 (√ 3)]/{[m √ (m 2+192)]/2} = 2 √ 3.

M=22 of the solution

That is, when the length of BM is 22, point P is the midpoint of MN.