Yesterday i was doing the question from class 9th ncert book of math of Areas of Parallelograms and Triangles chapter of exercise 9.4 What is the easiest way for solving it because i was not able to do this question please help me for solving this question 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:(iii) ar (PBQ) = ar (ARC)
AnilSinghBoraGuru
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:(iii) ar (PBQ) = ar (ARC) Q.7(3)
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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.