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)
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:(ii) ar (RQC) = (3/8) ar (ABC) Q.7(2)
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PQ is the median of 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)