Question

1. if $overline{ef}$ bisects $overline{cd}$, $cg = 5x - 1$, $gd = 7x - 13$, $ef = 6x - 4$, and $gf = 13$, find $eg$. 15. if $r$ is the mid - point of $overline{qs}$, $rs = 2x - 4$, $st = 4x - 1$, and $rt = 8x - 43$, find $qs$.

Explanation:

14.

Step1: Use the property of a bisector

Since $\overline{EF}$ bisects $\overline{CD}$, then $CG = GD$. So we set up the equation $5x - 1=7x - 13$.
\[

$$\begin{align*} 5x-1&=7x - 13\\ 13 - 1&=7x-5x\\ 12&=2x\\ x& = 6 \end{align*}$$

\]

Step2: Find the length of $EF$

Substitute $x = 6$ into the expression for $EF$. $EF=6x - 4=6\times6 - 4=36 - 4=32$.

Step3: Find the length of $EG$

We know that $EF=EG + GF$. So $EG=EF - GF$. Since $EF = 32$ and $GF = 13$, then $EG=32-13 = 19$.

Step1: Use the mid - point property

Since $R$ is the mid - point of $\overline{QS}$, then $QR=RS$. Also, $RT=RS + ST$. Substitute the given expressions: $8x-43=(2x - 4)+(4x - 1)$.
\[

$$\begin{align*} 8x-43&=2x-4 + 4x-1\\ 8x-43&=6x-5\\ 8x-6x&=- 5 + 43\\ 2x&=38\\ x&=19 \end{align*}$$

\]

Step2: Find the length of $RS$

Substitute $x = 19$ into the expression for $RS$. $RS=2x - 4=2\times19-4=38 - 4 = 34$.

Step3: Find the length of $QS$

Since $R$ is the mid - point of $\overline{QS}$, then $QS = 2RS$. So $QS=2\times34=68$.

Answer:

$19$

15.