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) = \(\frac{1}{2} \) ar(ARC)
(ii) ar(PBQ) = ar(ARC)
(iii) ar(RQC) = \(\frac{3}{8} \) ar(ABC)

We have a \(\triangle \)ABC such that P is the mid-point of AB and Q is the mid-point of BC.
Also, R is the mid-point of AP. Let us join AQ, RQ, PC and PC.


(i) In \(\triangle \) APQ , R is the mid point of AP(Given )
\(\therefore \) RQ is the median of \(\triangle \) APQ
\(\Rightarrow \) ar(PRQ) = \(\frac{1}{2} \) ar (APQ) ..... (1 )

In triangle ABQ ,P is the mid point of AB
\(\therefore \) QP is the median of \(\triangle \) ABQ
\(\Rightarrow \) ar(APQ) = \(\frac{1}{2} \) ar (ABQ) ..... (2 )

from (1) and (2) we get
\(\Rightarrow \) ar(PRQ) = \(\frac{1}{2} × \frac{1}{2} \) ar (ABQ)
\(\Rightarrow \) ar(PRQ) = \(\frac{1}{4} \) ar (ABQ) = \(\frac{1}{4} × \frac{1}{2} \) ar (ABC)
[ \(\because \) AQ is the Median of \(\triangle \) ABC ]
\(\therefore \) ar(PRQ) = \(\frac{1}{8} \) × ar (ABC).....(3)

Now ar(ARC) = \(\frac{1}{2} \) ×ar(APC)
[ \(\because \) CR is a Median of \(\triangle APC \) ]

\(\Rightarrow \) ar(ARC) = \(\frac{1}{2} \frac{1}{2} \) × ar(ABC)
[ \(\because \) CP is a Median of \(\triangle ABC \) ]
\(\therefore \) ar(ARC) = \(\frac{1}{4} \) × ar(ABC) .....(4)

Now from (3) we get
ar(PRQ) = \(\frac{1}{8} \) (ABC)
ar(PRQ) = \(\frac{1}{2} \frac{1}{4} \) ar(ABC)
ar(PRQ) = \(\frac{1}{2} \) (ARC)


(iii) In triangle RBC , RQ is a Median
\(\therefore \) ar(RQC) = ar(RBQ)
\(\Rightarrow \) ar(RQC) = ar(PRQ) + ar(BPQ)
\(\Rightarrow \) ar(RQC) = \(\frac{1}{8} \)× ar(ABC) + ar(BPQ)
\(\Rightarrow \) ar(RQC) = \(\frac{1}{8} \) × ar(ABC) + ar(BPQ)
\(\Rightarrow \) ar(RQC) = \(\frac{1}{8} × ar(ABC) + \frac{1}{2} \) ar(BPQ)
[\(\because \) PQ is the Median of BPC ]
\(\Rightarrow \) ar(RQC) = \(\frac{1}{8} × ar(ABC) + \frac{1}{2} \frac{1}{2} \) ar(ABC)
[\(\because \) CP is the Median of ABC ]
\(\Rightarrow \) ar(RQC) = ( \(\frac{1}{4} + \frac{1}{8} \) ar(ABC)

Thus , ar(RQC) = ( \(\frac{3}{8} \) ar(ABC)


(ii) QP is the Median of \(\triangle \) ABQ
\(\therefore \) ar\(\triangle \) PBQ = \(\frac{1}{2} \) ar(ABQ)
= \(\frac{1}{2}× \frac{1}{2} \) × ar(ABC)
= \(\frac{1}{4} \) × ar(ABC) = ar(ARC )

From equation (4)
Thus ar(PBQ) = ar(ARC)