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Sagot :
Answer:
[tex]\fbox{Perimeter of ∆ABC = 54 units}[/tex]
Step-by-step explanation:
Given:
➳ AC || RP
➳ AC = 18 units
➳ AR = 14 units
➳ RB = 7 units
➳ CP = 10 units
To find:
perimeter of triangle ABC =?
Solution:
in ∆ABC and ∆RBP
since we know AC || RP,
then ∠ACP = ∠RPB
and ∠CAB = ∠PRB (Property of parallel lines)
and ∠B is a common angle in both traingles,
hence ∆ABC ~ ∆RBP from AAA property!
now,
using the properties of similar triangle,
[tex] \frac{AR }{RB} = \frac{CP}{PB} \\ \\ \frac{14}{7} = \frac{10}{PB} \\ \\ X = \frac{10}{2} \\ \\ \fbox{PB = 5} \\ \\ \\ [/tex]
To find perimeter of ∆ABC, add all the components.
Perimeter of ∆ABC = AR + RB + BP +PC + CA
Perimeter of ∆ABC= 14 + 7 +5 +10 + 18 units
[tex]\fbox{Perimeter of ∆ABC = 54 units}[/tex]
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[tex]\qquad\qquad\huge\underline{{\sf Answer}}♨[/tex]
In the given figure, PR is parallel to CA, therefore two angles of Triangle ABC is equal to two corresponding angles of Triangle RBP.
that is
[tex]\qquad \sf \dashrightarrow \:\angle BCA = \angle BPR[/tex]
[tex]\qquad \sf \dashrightarrow \:\angle BAC= \angle BRP [/tex]
[ They form pairs of corresponding angles ]
So, we can infer that :
[tex]\qquad \sf \dashrightarrow \:\triangle BAC \sim \triangle BRP [/tex]
The Triangles are similar, by AA criteria ~
let's assume measure of BP = x
Now, let's use its result :
[tex] \qquad \sf \dashrightarrow \:\dfrac{BA}{BR}= \dfrac{BC}{BP}[/tex]
[tex] \qquad \sf \dashrightarrow \:\dfrac{14 + 7}{7}= \dfrac{x + 10 }{x}[/tex]
[tex] \qquad \sf \dashrightarrow \:\dfrac{21}{7}= \dfrac{x + 10 }{x}[/tex]
[tex] \qquad \sf \dashrightarrow \:3= \dfrac{x + 10 }{x}[/tex]
[tex] \qquad \sf \dashrightarrow \:3x={x + 10 }{}[/tex]
[tex] \qquad \sf \dashrightarrow \:3x - {x = 10 }{}[/tex]
[tex]\qquad \sf \dashrightarrow \:2x = 10[/tex]
[tex]\qquad \sf \dashrightarrow \:x = 5 \: \: units[/tex]
So, the required Sides of Triangle ABC are :
- AB = 14 + 7 = 21 units
- AC = 18 units
- BC = 10 + 5 = 15 units
Therefore, Perimeter = AB + BC + AC :
[tex]\qquad \sf \dashrightarrow \:21 + 18 + 15[/tex]
[tex]\qquad \sf \dashrightarrow \:54 \: \: units[/tex]
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