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 #1
avatar+1484 
-1

Identify Similar Triangles:

 

Since ABCD is a parallelogram, lines AD and BC are parallel. When line DE intersects line BC at point F, it creates two transversal lines (AD and BC).

 

Because of this, corresponding angles on alternate sides of DE are congruent (alternate interior angles). Therefore, triangles EAD and EBF are similar by Angle-Angle Similarity (AA).

 

Ratio of Areas:

 

Since triangles EAD and EBF are similar, the ratio of their areas is equal to the square of the ratio of their corresponding side lengths.

 

e are given that the area of triangle EBF is 4 and the area of triangle EAD is 9. Let the ratio between a corresponding side length in EBF and the corresponding side length in EAD be k. Therefore:

 

Area(EBF) / Area(EAD) = k^2

 

4 / 9 = k^2

 

k = ± (2/3) (We can take the positive or negative value of k since it represents a ratio of side lengths, which can be positive or negative depending on direction.)

 

Relating Side Lengths in Similar Triangles:

 

Since k represents the ratio between corresponding side lengths in the similar triangles, we can express the lengths of sides BE and AE in terms of k:

BE = k * AE (BE corresponds to a side in EBF and AE corresponds to a side in EAD)

 

Area of Triangle EBF:

 

We are given that the area of triangle EBF is 4. The area of a triangle can be calculated as:

 

Area = (base * height) / 2

 

Since BE is the base of triangle EBF and DF is the height with respect to that base (DF is perpendicular to BE), we can set up an equation:

 

4 = (BE * DF) / 2

 

8 = BE * DF (multiply both sides by 2)

 

Relating Heights in Similar Triangles:

 

Since triangles EAD and EBF are similar, the ratio between corresponding heights is also equal to k:

 

DF = k * AH (DF corresponds to the height in EBF and AH corresponds to the height in EAD with respect to bases BE and AE, respectively)

 

Substituting and Solving for AE:

 

We can substitute the expression for DF from step 5 into the equation for the area of triangle EBF from step 4:

 

8 = BE * (k * AH)

 

8 = (k * AE) * (k * AH) (substitute BE with k * AE from step 3)

 

8 = k^2 * AE * AH

 

Since k ≠ 0 (otherwise, the triangles wouldn't be similar), we can divide both sides by k^2 * AH:

 

AE = 8 / (k^2 * AH)

 

Area of Triangle EAD:

 

We are given that the area of triangle EAD is 9. Using the same formula for the area of a triangle as before:

 

9 = (AE * AH) / 2

 

18 = AE * AH (multiply both sides by 2)

 

Substitute and Solve for AH:

 

We can substitute the expression for AE from step 6 into the equation for the area of triangle EAD from step 7:

 

18 = (8 / (k^2 * AH)) * AH

 

18 = 8 / k^2

 

k^2 = 8 / 18 = 4 / 9

 

Since we found k to be ± (2/3) in step 2, in this case, k = 2/3 (the positive value makes sense since it represents a ratio of side lengths where the corresponding side in EBF is shorter than the corresponding side in EAD).

 

Area of Parallelogram ABCD:

 

The area of a parallelogram is equal to the base times the height. Since AD is parallel to BC, the height of parallelogram ABCD is the same as the height (AH) of triangle EAD that we just solved for.

 

The base of parallelogram ABCD is the same length as side AE of triangle EAD that we also solved for. Therefore:

 

Area(ABCD) = AE * AH = 8 * 9/4 = 18.

19 abr 2024
 #1
avatar+1484 
0

Let the roots of Babette's monic quadratic be r and s. Since the leading coefficient is 1, the quadratic can be written as:

 

p(x)=(x−r)(x−s)=x2−(r+s)x+rs.

 

We are given that squaring the roots gives another monic quadratic with those squares as its roots. In other words:

 

(x−r2)(x−s2)=x2−(r2+s2)x+r2s2

 

Expanding the left side gives:

 

x2−(r2+2rs+s2)x+r2s2

 

Equating the coefficients of the corresponding terms in both quadratics, we get the system of equations:

 

r+s=r2+s2

 

r2s2=rs (notice this simplifies to r2s2−rs=0)

 

From the first equation, we can rearrange to get r2+s2−r−s=0. Factoring this gives (r−1)(s−1)=0. This means either r=1 or s=1.

 

We can consider two cases:

 

Case 1: r=1

 

Substituting r=1 into the second equation gives s2−s=0, which factors as s(s−1)=0. This means either s=0 or s=1. However, if s=0, then the leading coefficient of the quadratic wouldn't be 1, so we reject this case.

 

Therefore, in this case, s=1 and the quadratic is simply x2−2x+1=(x−1)2. There is only 1​ quadratic possible in this case.

 

Case 2: s=1

 

Substituting s=1 into the second equation gives r2−r=0, which factors as r(r−1)=0. This means either r=0 or r=1. However, if r=0, then the leading coefficient of the quadratic wouldn't be 1, so we reject this case.

 

Therefore, in this case, r=1 and the quadratic is again simply x2−2x+1=(x−1)2. There is only 1​ quadratic possible in this case.

 

Since both cases lead to only one possible quadratic, Babette could have been thinking of at most 2​ different monic quadratics.

19 abr 2024