Question

84.) in the diagram below, $cf = x + 21$, $ce = 18 + x$, $df = 2x + 13$, and $de = 10$. what is the length of $cd$?
a.) 8
b.) 10
c.) 13
d.) 18

85.) in the diagram below $ac$ bisects $de$ at $b$. if $bc=(5x - 11)$ in., $db=(4x + 4)$ in., $ab=(x + 17)$ in., and $be=(8x - 12)$ in. find the length of $ac$
(a) 24 in.
(b) 48 in.
(c) 30 in.
(d) 76 in.

Explanation:

Step1: Use segment - addition postulate

Since $CF=CE + EF$, and $CF=x + 21$, $CE=18 + x$, $EF = DF+DE$, $DF = 2x+13$, $DE = 10$. But we can also use the fact that if we consider the segments on the line, we know that $CF - CE=EF$ and $EF=DF + DE$. So $(x + 21)-(18 + x)=(2x + 13)+10$. Simplifying the left - hand side: $(x + 21)-(18 + x)=x + 21-18 - x=3$. Then the equation becomes $3=2x+13 + 10$.

Step2: Solve for $x$

First, combine like terms on the right - hand side: $3=2x+23$. Subtract 23 from both sides: $3-23=2x$, so $- 20=2x$. Divide both sides by 2: $x=-10$.

Step3: Find the length of $CD$

We know that $CD=CF - DF$. Substitute $CF=x + 21$ and $DF = 2x+13$ into the formula. $CF=-10 + 21 = 11$ and $DF=2\times(-10)+13=-20 + 13=-7$. But we can also use another way. Since $CE=18+x$ and $DE = 10$, and $CD=CE - DE$. Substitute $x=-10$ into $CE$, we get $CE=18-10 = 8$. Then $CD=CE - DE=8 - 0=8$.

for second problem:

Step1: Use the property of segment bisector

Since $AC$ bisects $DE$ at $B$, then $DB=BE$. So $4x + 4=8x-12$.

Step2: Solve for $x$

Subtract $4x$ from both sides: $4=4x-12$. Add 12 to both sides: $4 + 12=4x$, so $16=4x$. Divide both sides by 4: $x = 4$.

Step3: Find the length of $AC$

$AC=AB + BC$. Substitute $x = 4$ into the expressions for $AB$ and $BC$. $AB=x + 17=4+17 = 21$ and $BC=5x-11=5\times4-11=20 - 11 = 9$. Then $AC=21+9=30$.

Answer:

a. 8