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) = ½ ar (ARC)

(ii) ar (RQC) = (3/8) ar (ABC)

(iii) ar (PBQ) = ar (ARC)

Solution:

(i)

We know that, median divides the triangle into two triangles of equal area,

PC is the median of ABC.

Ar (ΔBPC) = ar (ΔAPC) ……….(i)

RC is the median of APC.

Ar (ΔARC) = ½ ar (ΔAPC) ……….(ii)

PQ is the median of BPC.

Ar (ΔPQC) = ½ ar (ΔBPC) ……….(iii)

From eq. (i) and (iii), we get,

ar (ΔPQC) = ½ ar (ΔAPC) ……….(iv)

From eq. (ii) and (iv), we get,

ar (ΔPQC) = ar (ΔARC) ……….(v)

P and Q are the mid-points of AB and BC respectively [given]

PQ||AC

and, PA = ½ AC

Since, triangles between same parallel are equal in area, we get,

ar (ΔAPQ) = ar (ΔPQC) ……….(vi)

From eq. (v) and (vi), we obtain,

ar (ΔAPQ) = ar (ΔARC) ……….(vii)

R is the mid-point of AP.

, RQ is the median of APQ.

Ar (ΔPRQ) = ½ ar (ΔAPQ) ……….(viii)

From (vii) and (viii), we get,

ar (ΔPRQ) = ½ ar (ΔARC)

Hence Proved.

(ii) PQ is the median of ΔBPC

ar (ΔPQC) = ½ ar (ΔBPC)

= (½) ×(1/2 )ar (ΔABC)

= ¼ ar (ΔABC) ……….(ix)

Also,

ar (ΔPRC) = ½ ar (ΔAPC) [From (iv)]

ar (ΔPRC) = (1/2)×(1/2)ar ( ABC)

= ¼ ar(ΔABC) ……….(x)

Add eq. (ix) and (x), we get,

ar (ΔPQC) + ar (ΔPRC) = (1/4)×(1/4)ar (ΔABC)

ar (quad. PQCR) = ¼ ar (ΔABC) ……….(xi)

Subtracting ar (ΔPRQ) from L.H.S and R.H.S,

ar (quad. PQCR)–ar (ΔPRQ) = ½ ar (ΔABC)–ar (ΔPRQ)

ar (ΔRQC) = ½ ar (ΔABC) – ½ ar (ΔARC) [From result (i)]

ar (ΔARC) = ½ ar (ΔABC) –(1/2)×(1/2)ar (ΔAPC)

ar (ΔRQC) = ½ ar (ΔABC) –(1/4)ar (ΔAPC)

ar (ΔRQC) = ½ ar (ΔABC) –(1/4)×(1/2)ar (ΔABC) [ As, PC is median of ΔABC]

ar (ΔRQC) = ½ ar (ΔABC)–(1/8)ar (ΔABC)

ar (ΔRQC) = [(1/2)-(1/8)]ar (ΔABC)

ar (ΔRQC) = (3/8)ar (ΔABC)

(iii) ar (ΔPRQ) = ½ ar (ΔARC) [From result (i)]

2ar (ΔPRQ) = ar (ΔARC) ……………..(xii)

ar (ΔPRQ) = ½ ar (ΔAPQ) [RQ is the median of APQ] ……….(xiii)

But, we know that,

ar (ΔAPQ) = ar (ΔPQC) [From the reason mentioned in eq. (vi)] ……….(xiv)

From eq. (xiii) and (xiv), we get,

ar (ΔPRQ) = ½ ar (ΔPQC) ……….(xv)

At the same time,

ar (ΔBPQ) = ar (ΔPQC) [PQ is the median of ΔBPC] ……….(xvi)

From eq. (xv) and (xvi), we get,

ar (ΔPRQ) = ½ ar (ΔBPQ) ……….(xvii)

From eq. (xii) and (xvii), we get,

2×(1/2)ar(ΔBPQ)= ar (ΔARC)

⟹ ar (ΔBPQ) = ar (ΔARC)

Hence Proved.