Please give me the best way for solving the problem of class 9^{th} ncert math of Areas of Parallelograms and Triangles chapter of math of class 9^{th} of exercise 9.4 of question no 7 (2) what is the tricky way 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:(ii) ar (RQC) = (3/8) ar (ABC)

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# 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:(ii) ar (RQC) = (3/8) ar (ABC) Q.7(2)

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PQ is the median of Î”BPC

ar (Î”PQC) = Â½ ar (Î”BPC)

= (Â½) Ã—(1/2 )ar (Î”ABC)

= Â¼ ar (Î”ABC) â€¦â€¦â€¦.(ix)

Also,

ar (Î”PRC) = Â½ ar (Î”APC) [From (iv)]

ar (Î”PRC) = (1/2)Ã—(1/2)ar ( ABC)

= Â¼ ar(Î”ABC) â€¦â€¦â€¦.(x)

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

ar (Î”PQC) + ar (Î”PRC) = (1/4)Ã—(1/4)ar (Î”ABC)

ar (quad. PQCR) = Â¼ ar (Î”ABC) â€¦â€¦â€¦.(xi)

Subtracting ar (Î”PRQ) from L.H.S and R.H.S,

ar (quad. PQCR)â€“ar (Î”PRQ) = Â½ ar (Î”ABC)â€“ar (Î”PRQ)

ar (Î”RQC) = Â½ ar (Î”ABC) â€“ Â½ ar (Î”ARC) [From result (i)]

ar (Î”ARC) = Â½ ar (Î”ABC) â€“(1/2)Ã—(1/2)ar (Î”APC)

ar (Î”RQC) = Â½ ar (Î”ABC) â€“(1/4)ar (Î”APC)

ar (Î”RQC) = Â½ ar (Î”ABC) â€“(1/4)Ã—(1/2)ar (Î”ABC) [ As, PC is median of Î”ABC]

ar (Î”RQC) = Â½ ar (Î”ABC)â€“(1/8)ar (Î”ABC)

ar (Î”RQC) = [(1/2)-(1/8)]ar (Î”ABC)

ar (Î”RQC) = (3/8)ar (Î”ABC)