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 (RQC) = \(\frac{3}{8}\) ar (ABC)
(iii) ar (PBQ) = ar (ARC)
(i) figure.png
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) = 1/2 ar (ΔAPC) ……….(ii)
PQ is the median of BPC.
Ar (ΔPQC) = 1/2 ar (ΔBPC) ……….(iii)
From eq. (i) and (iii), we get,
ar (ΔPQC) = 1/2 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 = 1/2 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) = 1/2 ar (ΔAPQ) ……….(viii)
From (vii) and (viii), we get,
ar (ΔPRQ) = 1/2 ar (ΔARC)
Hence Proved.
(ii) PQ is the median of ΔBPC
ar (ΔPQC) = 1/2 ar (ΔBPC)
= (1/2) × (1/2 )ar (ΔABC)
= ¼ ar (ΔABC) ……….(ix)
Also,
ar (ΔPRC) = 1/2 ar (ΔAPC) [From (iv)]
ar (ΔPRC) = (1/2) × (1/2)ar ( ABC)
= 1/4 ar(ΔABC) ……….(x)
Add eq. (ix) and (x), we get,
ar (ΔPQC) + ar (ΔPRC) = (1/4) × (1/4)ar (ΔABC)
ar (quad. PQCR) = 1/4 ar (ΔABC) ……….(xi)
Subtracting ar (ΔPRQ) from L.H.S and R.H.S,
ar (quad. PQCR)–ar (ΔPRQ) = 1/2 ar (ΔABC) – ar (ΔPRQ)
ar (ΔRQC) = 1/2 ar (ΔABC) – 1/2 ar (ΔARC) [From result (i)]
ar (ΔARC) = 1/2 ar (ΔABC) – (1/2) × (1/2)ar (ΔAPC)
ar (ΔRQC) = 1/2 ar (ΔABC) –(1/4)ar (ΔAPC)
ar (ΔRQC) = 1/2 ar (ΔABC) – (1/4) × (1/2)ar (ΔABC) [ As, PC is median of ΔABC]
ar (ΔRQC) = 1/2 ar (ΔABC) – (1/8)ar (ΔABC)
ar (ΔRQC) = [(1/2) - (1/8)]ar (ΔABC)
ar (ΔRQC) = (3/8) ar (ΔABC)
(iii) ar (ΔPRQ) = 1/2 ar (ΔARC) [From result (i)]
2ar (ΔPRQ) = ar (ΔARC) ……………..(xii)
ar (ΔPRQ) = 1/2 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) = 1/2 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) = 1/2 ar (ΔBPQ) ……….(xvii)
From eq. (xii) and (xvii), we get,
2 × (1/2)ar(ΔBPQ)= ar (ΔARC)
⟹ ar (ΔBPQ) = ar (ΔARC)
Hence Proved.