Question

in the figure,sides AB,BC,CA of triangle ABC are produced upto points R,P,F respectively such that AB=BR,BC=CP,CA=AF. Prove that:A(triangle PFR)=7A(triangle ABC)​

Answer 1

We need to prove that A(triangle PFR) = 7A(triangle ABC)

• Now, according to the information provided and as illustrated in the diagram, we see that  

A(triangle PFR) = A(triangle ABC) + A(triangle BRP) + A(triangle CPF) + A(triangle AFR)

• We know that given a triangle with sides a , b and the angle between them C, area of triangle is A(triangle ABC) = (a*b*sinC)/2

• In triangle BRP, BR=AB=c, BP = BC+CP = 2a, angleRBP = 180-B. So, A(triangle BRP) = (BR*BP*sin(RBP))/2 = (c*2a*sin(180-B))/2 = (2acsinB)/2 = acsinB = 2A(triangle ABC), since sin(180-P)=sinP .

• Similarly, in triangle CPF, CP=BC=a, CF = CA+AF = 2b, anglePCF = 180-C. So, A(triangle CPF) = (CP*CF*sin(PCF))/2 = (a*2b*sin(180-C))/2 = (2absinC)/2 = absinC = 2A(triangle ABC) , since sin(180-P)=sinP .

• Similarly, in triangle AFR, AF=CA=b, AR = AB+BR = 2c, angleFAR = 180-A. So, A(triangle AFR) = (AF*AR*sin(FAR))/2 = (b*2c*sin(180-A))/2 = (2bcsinA)/2 = bcsinA = 2A(triangle ABC) , since sin(180-P)=sinP .
• Thus, we get,
• A(triangle PFR) = A(triangle ABC) + A(triangle BRP) + A(triangle CPF) + A(triangle AFR)
• A(trinagle PFR) =  A(triangle ABC) + 2A(triangle ABC) + 2A(triangle ABC) + 2A(triangle ABC) = 7A(triangle ABC)

Hence, proved.

Answer 2

Area of Δ PFR = 7 * Area of ΔABC

Step-by-step explanation:

Area of Triangle ABC =

(1/2) AB * BC * Sin ∠B  = (1/2) BC * CA * Sin ∠C = (1/2) CA * AB * Sin ∠A

Area of Δ PFR =

Area of ΔABC + Area of ΔAFR + Area of ΔBPR + Area of ΔCFP

Area of ΔAFR = (1/2) AF * AR * Sin(180 - A)  = (1/2) AC * 2(AB) * SinA

= (1/2) * 2 * AB * AC * SinA

= 2 (1/2) CA * AB * Sin ∠A

= 2 * area of ΔABC

Similarly  Area of ΔBPR =  Area of ΔCFP =  2 * area of ΔABC

Area of Δ PFR = Area of ΔABC + 2 * area of ΔABC  +  2 * area of ΔABC +  2 * area of ΔABC

=> Area of Δ PFR = 7 * Area of ΔABC

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