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)

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)=Â½QSx

and area(QPR)=Â½QS(x/2)=area(QBP)/2 = ?/4x2 = ?/8   ...............(2)

Now, area(QBR)= area(QBP)+area(PQR) = ?/4 +?/8 = 3?/8  ...........(3)
and
also, area(QRC)=Â½TRy
and area(QBR)=Â½TRy
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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