Question

1. & 2. In the diagram below, points A, B, and C are collinear. Answer each of the following questions. The figure shown below is not drawn to scale, meaning you cannot determine your answers by using your ruler.​

Answer 1

Answer:

a)  $\overline{AB}$ = 8 in

b) When the length of AC = $6\frac{1}{2}$ in. and BC = $3\frac{1}{2}$ in. $\overline{AB}$ = 10 in

c)  When the length of AB = 10.2 in. and BC = 3.7 in.  $\overline {AC}$  = 6.5 in

d) When the length of AB =  $4\frac{3}{4}$ in. and BC = $3\frac{1}{4}$ in. in. $\overline {BC}$ =  $1\frac{1}{2}$ in

Step-by-step explanation:

a) When the length of AC = 5 in. and CB = 3 in. we have;

The length of $\overline {AB}$ = AC + CB (segment addition postulate)

Therefore;

$\overline{AB}$ = 5 in. + 3 in. = 8 in.

b) When the length of AC = $6\frac{1}{2}$ in. and BC = $3\frac{1}{2}$ in. we have;

The length of $\overline {AB}$ = AC + CB (segment addition postulate)

Therefore;

$\overline{AB}$ = $6\frac{1}{2}$ in.+ $3\frac{1}{2}$ in. = 10 in.

c) When the length of AB = 10.2 in. and BC = 3.7 in. we have;

The length of $\overline {AC}$ = AB - BC (converse of the segment addition postulate)

Therefore;

$\overline {AC}$ = 10.2 in.+ 3.7 in. = 6.5 in.

d) When the length of AB =  $4\frac{3}{4}$ in. and BC = $3\frac{1}{4}$ in. in. we have;

The length of $\overline {BC}$ = AB - AC (converse of the segment addition postulate)

Therefore;

$\overline {BC}$ = $4\frac{3}{4}$ in. -  $3\frac{1}{4}$ in.=  $1\frac{1}{2}$ in.