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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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Please give me the best way for solving the problem of class 9th ncert math of Areas of Parallelograms and Triangles chapter of math of class 9th 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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1. 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)

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