Question
Fri November 18, 2011 By:

P and Q are respectively the mid points of sides AB and BC of a triangle ABC and R is the mid point of AP. Show that: i)ar(PRQ)=1/2 ar(ARC) ii)ar(RQC)-3/8 ar(ABC) iii)ar(PBQ)=ar(ARC)

Expert Reply
Fri November 18, 2011
 
See in the above figure,
 
Let us suppose that QS and RT are the perpendiculars drawn on the sides AB and BC respectively.
 
Now, Let us say area(ABC) = ?
 
Now, Q and P are midpoints of side BC and AB respectively.
Hence, area(QBP)=area(ABC)/4 =  ?/4  ...........(1) by mid-point theorem.
 
Now, area(QBP)=½QSx 
 
and area(QPR)=½QS(x/2)=area(QBP)/2 = ?/4x2 = ?/8   ...............(2)
 
Now, area(QBR)= area(QBP)+area(PQR) = ?/4 +?/8 = 3?/8  ...........(3)
and
also, area(QRC)=½TRy
and area(QBR)=½TRy
so, area(QRC)=area(QBR) = 3?/8  ...........(4)
 
So area(ARC)=area(ABC)-area(QRC)-area(QRB)=?-3?/8-3?/8=?/4 .............(5)
 
Hence from eqn2 and 5
we get area(PRQ)=area(ARC)/2   .............Proved
 
From eqn 4 we have area(RQC)=3area(ABC)/8   ......Proved
 
From eqn 1 and 5 we find that
area(PBQ)=area(ARC) ... Proved
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