Question -
Answer -
(i)
We know that, mediandivides the triangle into two triangles of equal area,
PC is the median ofABC.
Ar (ΔBPC) = ar (ΔAPC)……….(i)
RC is the median ofAPC.
Ar (ΔARC) = ½ ar(ΔAPC) ……….(ii)
PQ is the median ofBPC.
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 themid-points of AB and BC respectively [given]
PQ||AC
and, PA = ½ AC
Since, trianglesbetween 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 ofAP.
, RQ is the median ofAPQ.
Ar (ΔPRQ) = ½ ar(ΔAPQ) ……….(viii)
From (vii) and (viii),we get,
ar (ΔPRQ) = ½ ar (ΔARC)
Hence Proved.
(ii) PQ is the medianof 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.