Question

1. if $overline{lk}congoverline{mk}, lk = 7x - 10, kn=x + 3, mn = 9x - 11$, and $kj = 28$, find $lj$.
2. if $t$ is the mid - point of $overline{su}$, find $x$.
3. if $g$ is the mid - point of $overline{fh}$, find $fg$.
4. if $r$ is the mid - point of $overline{qs}$, find $qs$.
5. if $b$ is the mid - point of $overline{ac}$, and $ac = 8x - 20$, find $bc$.
6. if $overline{ef}$ bisects $overline{cd}, cg = 5x - 1, gd = 7x - 13, ef = 6x - 4$, and $gf = 13$, find $eg$.
7. if $r$ is the mid - point of $overline{qs}, rs = 2x - 4, st = 4x - 1$, and $rt = 8x - 43$, find $qs$.

Explanation:

Step1: Identify relevant mid - point and congruence properties

Use the fact that if a point is a mid - point of a line segment, the two sub - segments are equal in length. If two line segments are congruent, their lengths are equal.

Step2: Solve for \(x\) in each case

For problem 9:

Since \(\overline{LK}\cong\overline{MK}\), we have \(LK = MK\). Given \(LK = 7x - 10\), \(KN=x + 3\), and \(MN=9x - 11\), then \(MK=MN - KN=(9x - 11)-(x + 3)=8x-14\). So, \(7x - 10=8x - 14\), which gives \(x = 4\). Then \(LK=7\times4 - 10 = 18\), and \(LJ=LK+KJ=18 + 28=46\).

For problem 10:

Since \(T\) is the mid - point of \(\overline{SU}\), then \(ST = TU\). So, \(8x+11 = 12x - 1\). Rearranging gives \(12=4x\), and \(x = 3\).

For problem 11:

Since \(G\) is the mid - point of \(\overline{FH}\), then \(FG=GH\). So, \(11x - 7=3x + 9\). Rearranging gives \(8x=16\), and \(x = 2\). Then \(FG=11\times2-7 = 15\).

For problem 12:

Since \(R\) is the mid - point of \(\overline{QS}\), then \(QR = RS\). So, \(5x-3=21 - x\). Rearranging gives \(6x=24\), and \(x = 4\). Then \(QS=(5x - 3)+(21 - x)=4x + 18=4\times4+18 = 34\).

For problem 13:

Since \(B\) is the mid - point of \(\overline{AC}\), then \(AB = BC\). Given \(AC = 8x - 20\) and \(AB = 3x - 1\), then \(BC=AC - AB=(8x - 20)-(3x - 1)=5x-19\). So, \(3x - 1=5x - 19\), which gives \(x = 9\). Then \(BC=3\times9 - 1=28\).

For problem 14:

Since \(\overline{EF}\) bisects \(\overline{CD}\), then \(CG = GD\). So, \(5x-1=7x - 13\), which gives \(x = 6\). Then \(EF=6x - 4=6\times6 - 4 = 32\), and \(EG=EF - GF=32 - 13=19\).

For problem 15:

Since \(R\) is the mid - point of \(\overline{QS}\), then \(QR = RS\). Also, \(RT=RS+ST\). Given \(RS = 2x - 4\), \(ST=4x - 1\), and \(RT=8x - 43\), then \(8x - 43=(2x - 4)+(4x - 1)\). Simplifying gives \(8x - 43=6x - 5\), which gives \(2x=38\) and \(x = 20\). Then \(QS = 2RS=2(2x - 4)=2\times(2\times20 - 4)=72\).

Answer:

1. \(LJ = 46\)
2. \(x = 3\)
3. \(FG = 15\)
4. \(QS = 34\)
5. \(BC = 28\)
6. \(EG = 19\)
7. \(QS = 72\)