Question

topic 2: segment addition postulate

1. if $df = 42$, find $de$.

$7x + 1$
$4x - 3$
$d$
$e$
$f$
10.

Explanation:

Step1: Apply Segment Addition Postulate

According to the Segment Addition Postulate, \(DE + EF=DF\). Given \(DE = 7x + 1\), \(EF=4x - 3\), and \(DF = 42\), we substitute these into the equation: \((7x + 1)+(4x - 3)=42\).

Step2: Simplify the left - hand side of the equation

Combine like terms: \(7x+4x+1 - 3=42\), which simplifies to \(11x-2 = 42\).

Step3: Solve for \(x\)

Add 2 to both sides of the equation: \(11x-2 + 2=42 + 2\), so \(11x=44\). Then divide both sides by 11: \(x=\frac{44}{11}=4\).

Step4: Find the length of \(DE\)

Substitute \(x = 4\) into the expression for \(DE\): \(DE=7x + 1\). So \(DE=7\times4+1=28 + 1=29\).

Answer:

\(29\)