Answered

Discover the answers to your questions at Westonci.ca, where experts share their knowledge and insights with you. Experience the ease of finding precise answers to your questions from a knowledgeable community of experts. Get quick and reliable solutions to your questions from a community of experienced experts on our platform.

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]

Thanks for joining brainly community!

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]

Thank you for visiting our platform. We hope you found the answers you were looking for. Come back anytime you need more information. We appreciate your visit. Our platform is always here to offer accurate and reliable answers. Return anytime. Westonci.ca is committed to providing accurate answers. Come back soon for more trustworthy information.