Answered

What value of \( x \) would make \( \overline{KM} \parallel \overline{JN} \)?

Complete the statements to solve for \( x \):

1. By the converse of the side-splitter theorem, if \( \frac{JK}{KL} = \frac{x-5}{x} \), then \( \overline{KM} \parallel \overline{JN} \).

2. Substitute the expressions into the proportion: [tex]\[ \frac{x-5}{x} = \frac{x-3}{x+4} \][/tex]

3. Cross-multiply: [tex]\[ (x-5)(x+4) = x(x-3) \][/tex]

4. Distribute: [tex]\[ x^2 + 4x - 5x - 20 = x^2 - 3x \][/tex]

5. Multiply and simplify: [tex]\[ x^2 - x - 20 = x^2 - 3x \][/tex]

6. Solve for \( x \): [tex]\[ -x - 20 = -3x \][/tex]
[tex]\[ 2x = 20 \][/tex]
[tex]\[ x = 10 \][/tex]



Answer :

GinnyAnswer
To determine the value of \( x \) that would make the segment \( \overline{KM} \) parallel to \( \overline{JN} \), we use the converse of the side-splitter theorem. According to this theorem, if \( \frac{JK}{KL} = \frac{JM}{MN} \), then \( \overline{KM} \parallel \overline{JN} \).

Given the proportion:
[tex]\[ \frac{x-5}{x} = \frac{x-3}{x+4} \][/tex]

we proceed as follows:

1. Set up the proportion:
By the converse of the side-splitter theorem, if:
[tex]\[ \frac{JK}{KL} = \frac{x-5}{x} \][/tex]

and:
[tex]\[ \frac{JM}{MN} = \frac{x-3}{x+4} \][/tex]

then \( \overline{KM} \parallel \overline{JN} \).

2. Cross-multiply:
[tex]\[ (x-5)(x+4) = x(x-3) \][/tex]

3. Distribute:
Expand both sides:
[tex]\[ x(x) + x(4) - 5(x) - 5(4) = x(x) + x(-3) \][/tex]
\( \Rightarrow x^2 + 4x - 5x - 20 = x^2 - 3x \)

4. Simplify:
Combine like terms:
[tex]\[ x^2 - x - 20 = x^2 - 3x \][/tex]

5. Solve for \( x \):
Subtract \( x^2 \) from both sides to eliminate \( x^2 \):
[tex]\[ -x - 20 = -3x \][/tex]

Add \( 3x \) to both sides:
[tex]\[ 2x - 20 = 0 \][/tex]

Add 20 to both sides:
[tex]\[ 2x = 20 \][/tex]

Divide by 2:
[tex]\[ x = 10 \][/tex]

Therefore, the value of \( x \) that will make \( \overline{KM} \parallel \overline{JN} \) is:
[tex]\[ x = 10 \][/tex]

[tex]\(\boxed{10}\)[/tex]

Other Questions