Question

1. if ( angle acongangle m), (la = 2x - 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: Use mid - point property

If a point is a mid - point of a line segment, the two sub - segments are equal. For example, in problem 10, since $T$ is the mid - point of $SU$, we have $8x + 11=12x - 1$.
Solve for $x$:
$11 + 1=12x-8x$
$4x = 12$
$x = 3$

Step2: Find the required length

We can then use the value of $x$ to find the required lengths in each problem.

Answer:

1. $x = 3$
2. First, set $11x-7 = 3x + 9$, $11x-3x=9 + 7$, $8x=16$, $x = 2$, $FG=11x-7=11\times2-7 = 15$
3. Set $5x-3=21 - x$, $5x+x=21 + 3$, $6x=24$, $x = 4$, $QS=(5x-3)+(21 - x)=4x + 18=4\times4+18=34$
4. Since $B$ is the mid - point of $AC$ and $AC = 8x-20$, $BC=\frac{AC}{2}=4x-10$. Also, $3x-1=4x - 10$, $10 - 1=4x-3x$, $x = 9$, $BC=4\times9-10=26$
5. Since $EF$ bisects $CD$, $CG = GD$, $5x-1=7x-13$, $13 - 1=7x-5x$, $2x=12$, $x = 6$, $EF=6x-4=6\times6-4 = 32$, $EG=EF - GF=32-13 = 19$
6. Since $RS+ST=RT$, $(2x - 4)+(4x-1)=8x-43$, $6x-5=8x-43$, $43 - 5=8x-6x$, $2x=38$, $x = 19$, $QS = 2RS=2(2x-4)=2(2\times19-4)=68$