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Find the perimeter of ΔABC and how you got it.

Find The Perimeter Of ΔABC And How You Got It class=

Sagot :

Ankit

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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DᴀʀᴋPᴀʀᴀᴅᴏx

[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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