Perimeter of triangle ABC: P=?
P=AB+BC+AC
AB=AR+RB
AR=14; RB=7
AB=14+7→AB=21
BC=PB+PC
PC=10; PB=?
BC=PB+10
AC=18
If AC is parallel to RP,and they are cut by the secant AB, the correspondent angles must be congruents, then angle CAB must be congruent with angle PRB.
The triangles ABC and RBP are similars, because they have to congruent angles:
Angle CAB is congruent with angle PRB
Angle ABC is common
Then theirs sides must be proportionals:
PB/RB=CB/AB
RB=7
BC=PB+10
AB=21
Replacing the known values:
PB/7=(PB+10)/21
Solving for PB. Multiplying both sides of the equation by 21:
21(PB/7)=21[(PB+10)/21]
3PB=PB+10
Subtracting PB both sides of the equation:
3PB-PB=PB+10-PB
2PB=10
Dividing both sides of the equation by 2:
(2PB)/2=(10)/2
PB=5
BC=PB+10
BC=5+10
BC=15
Then the perimeter is of triangle ABC is:
P=AB+BC+AC
P=21+15+18
P=54
Answer: The perimeter of triangle ABC is 54 units