Question

checkpoint 2. use the line segment addition postulate to solve for a missing segment.
c-level: for each problem, find the length of the missing line segment.

1. diagram: line with points, segments labeled
2. diagram: a---b---c, ab length x, bc length 3, ac length 10
3. diagram: l---m---n, lm length 2, mn length 7, ln length 7? (text: 7 below)
4. diagram: f---e---d, fe length x, ed length 4, fd length 8

b-level: solve for \x\ in the following problems using the segment addition postulate.

1. diagram: l---k---j, lk length 2x-4, kj length -2+x, lj length 15
2. diagram: j---i---h, ji length x, ih length 2x-13, jh length 17
3. diagram: f---e---d, fe length 11, ed length 2x-9, fd length 3x-3
4. diagram: b---c---d, bc length x+2, cd length x+11, bd length 13

Explanation:

B-level Problem 1:

Step1: Apply Segment Addition Postulate

The segment addition postulate states that if \( K \) is between \( L \) and \( J \), then \( LK + KJ = LJ \). So we have \((2x - 4)+(-2 + x)=15\).

Step2: Combine Like Terms

Combine the \( x \)-terms and the constant terms: \(2x+x-4 - 2=15\), which simplifies to \(3x-6 = 15\).

Step3: Solve for \( x \)

Add 6 to both sides: \(3x=15 + 6=21\). Then divide both sides by 3: \(x=\frac{21}{3}=7\).

Step1: Apply Segment Addition Postulate

If \( I \) is between \( J \) and \( H \), then \( JI+IH = JH \). So \(x+(2x - 13)=17\).

Step2: Combine Like Terms

Combine the \( x \)-terms: \(3x-13 = 17\).

Step3: Solve for \( x \)

Add 13 to both sides: \(3x=17 + 13=30\). Then divide by 3: \(x=\frac{30}{3}=10\).

Step1: Apply Segment Addition Postulate

If \( E \) is between \( F \) and \( D \), then \( FE+ED = FD \). So \(11+(2x - 9)=3x - 3\).

Step2: Simplify Left Side

Simplify \(11+2x - 9\) to \(2x + 2\). So we have \(2x+2=3x - 3\).

Step3: Solve for \( x \)

Subtract \(2x\) from both sides: \(2=x - 3\). Then add 3 to both sides: \(x=2 + 3=5\).

Answer:

\( x = 7 \)

B-level Problem 2: