Question

Find the perimeter of ΔABC and how you got it.

Answer 1

$\qquad\qquad\huge\underline{{\sf Answer}}♨$

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

$\qquad \sf \dashrightarrow \:\angle BCA = \angle BPR$

$\qquad \sf \dashrightarrow \:\angle BAC= \angle BRP$

[ They form pairs of corresponding angles ]

So, we can infer that :

$\qquad \sf \dashrightarrow \:\triangle BAC \sim \triangle BRP$

The Triangles are similar, by AA criteria ~

let's assume measure of BP = x

Now, let's use its result :

$\qquad \sf \dashrightarrow \:\dfrac{BA}{BR}= \dfrac{BC}{BP}$

$\qquad \sf \dashrightarrow \:\dfrac{14 + 7}{7}= \dfrac{x + 10 }{x}$

$\qquad \sf \dashrightarrow \:\dfrac{21}{7}= \dfrac{x + 10 }{x}$

$\qquad \sf \dashrightarrow \:3= \dfrac{x + 10 }{x}$

$\qquad \sf \dashrightarrow \:3x={x + 10 }{}$

$\qquad \sf \dashrightarrow \:3x - {x = 10 }{}$

$\qquad \sf \dashrightarrow \:2x = 10$

$\qquad \sf \dashrightarrow \:x = 5 \: \: units$

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 :

$\qquad \sf \dashrightarrow \:21 + 18 + 15$

$\qquad \sf \dashrightarrow \:54 \: \: units$

Answer 2

Answer:

$\fbox{Perimeter of ∆ABC = 54 units}$

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,

$\frac{AR }{RB} = \frac{CP}{PB} \\ \\ \frac{14}{7} = \frac{10}{PB} \\ \\ X = \frac{10}{2} \\ \\ \fbox{PB = 5}$

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

$\fbox{Perimeter of ∆ABC = 54 units}$

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