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 (1) 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: (i) ar (PRQ) = ½ ar (ARC)

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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: (i) ar (PRQ) = ½ ar (ARC) Q.7(1)

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We know that, median divides the triangle into two triangles of equal area,

PC is the median of ABC.

Ar (ΔBPC) = ar (ΔAPC) ……….(i)

RC is the median of APC.

Ar (ΔARC) = ½ ar (ΔAPC) ……….(ii)

PQ is the median of BPC.

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 the mid-points of AB and BC respectively [given]

PQ||AC

and, PA = ½ AC

Since, triangles between 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 of AP.

, RQ is the median of APQ.

Ar (ΔPRQ) = ½ ar (ΔAPQ) ……….(viii)

From (vii) and (viii), we get,

ar (ΔPRQ) = ½ ar (ΔARC)

Hence Proved.