Question

In triangle ABC,side BC is extended to P such that CP=BC,side CA is extended to F such that AF=AC and the side AB is extended to R such that BR =AB.Prove that area of triangle PFR = 7×area of triangle ABC​

Answer 1

Answer:

Area of Δ PFR = 7 Area of Δ ABC

Step-by-step explanation:

Let say BC = a , AC= b & AB = c

CP = a  , AF = b  BR = c

BP = BC + CP = 2a

CF = AC + AF = 2b

AR = AB + BR = 2c

Area of Δ ABC =  (1/2)acSinB  = (1/2)abSinC  = (1/2)bcSinA

now Area of Δ BPR

= (1/2)2acSin(180 - ∠B)  = acSinB  = 2 (1/2)acSinB = 2 Area of Δ ABC

Similarly

Area of Δ CPF = 2 Area of Δ ABC

Area of Δ ARF = 2 Area of Δ ABC

Area of Δ PFR = Area of Δ ABC + Area of Δ BPR +  Area of Δ CPF +  Area of Δ ARF

=> Area of Δ PFR =  Area of Δ ABC + 2 Area of Δ ABC + 2 Area of Δ ABC + 2 Area of Δ ABC

=> Area of Δ PFR = 7 Area of Δ ABC

QED

Proved

Answer 2

In the figure sides AB,BC,CA of △ABC are produced upto points R,F,P respectively such that AB=BR,BC=CP,CA=AF

Prove that : A(△PER)=7A(△ABC)

Given that Sides AB,BC and CA of △ABC are produced up to points R,P,F respectively such that AB=BR,BC=CP, and CA=AF

Construction: Join PA,FB and RC

Let Area(△ABC)=a

We know that median of a triangle divides it into triangles of equal area.

In △PAB,AC is the median

⇒Area(△ABC)=Area(△APC)=a

In △PCF,PA is the median

⇒Area(△APC)=Area(△PFA)=a

In △BCF,BA is the median

⇒Area(△ABC)=Area(△ABF)=a

In △RAF,FB is the median

⇒Area(△ABF)=Area(△RBF)=a

In △ACR,BC is the median

⇒Area(△ABC)=Area(△RBC)=a

In △PRB,RC is the median

⇒Area(△RBC)=Area(△RPC)=a

Thus,Area(△ABC)=Area(△APC)=Area(△PFA)=Area(△ABF)=Area(△RBF)=Area(△RBC)=Area(△RPC)

Hence Area(△PRF)=a+a+a+a+a+a+a=7a=7(Area△ABC)