Question

#4.) in the diagram below, $overline{cf}=x + 21$, $overline{ce}=18 + x$, $overline{df}=2x + 13$, and $overline{de}=10$. what is the length of $overline{cd}$?
a.) 8
b.) 10
c.) 13
d.) 18
#5.) in the diagram below $overline{ac}$ bisects $overline{de}$ at $b$. if $overline{bc}=(5x - 11)$in., $overline{db}=(4x + 4)$in., $overline{ab}=(x + 17)$in., and $overline{be}=(8x - 12)$in. find the length of $overline{ac}$.
(a) 24 in.
(b) 48 in.
(c) 30 in.
(d) 76 in.

Explanation:

Step1: Use segment - addition postulate for #4

Since \(CF=CE + EF\) and \(CF = CD+DF\), and \(CE=18 + x\), \(DF = 2x+13\), \(DE = 10\), \(CF=x + 21\). Also, \(CF=CD + DF\) and \(CE=CD + DE\). So, \((18 + x)=CD + 10\), then \(CD=(18 + x)-10=8 + x\). And \(x + 21=CD+(2x + 13)\). Substitute \(CD = 8 + x\) into \(x + 21=CD+(2x + 13)\):
\[x + 21=(8 + x)+(2x + 13)\]
\[x + 21=8 + x+2x + 13\]
\[x + 21=2x+21\]
\[x=0\]
Since \(CD = 8 + x\), when \(x = 0\), \(CD = 8\).

Step2: Use the property of bisector for #5

Since \(\overline{AC}\) bisects \(\overline{DE}\) at \(B\), then \(DB=BE\).
Set up the equation \(4x + 4=8x-12\).
Subtract \(4x\) from both sides: \(4=4x-12\).
Add 12 to both sides: \(4x=16\), so \(x = 4\).
\(AB=x + 17=4 + 17=21\) and \(BC=5x-11=5\times4-11=20 - 11 = 9\).
\(AC=AB + BC=21+9=30\)

Answer:

a. 8
C. 30 in.