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Question

# In the figure sides AB,BC,CA of △ABC are produced upto points R,F,P respectively such that AB=BR,BC=CP,CA=AFProve that : A(△PER)=7A(△ABC)

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Solution

## 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)=aIn △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)=aIn △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)

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