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In a Δ ABC, P and Q are respectively the mid-point of AB and BC and R is the mid-point of AP. Prove that : (i) ar(ΔPBQ) = ar(ΔARC) (ii) ar(ΔPQR) = \(\frac{1}{2}\)ar(ΔARC) (iii) ar(ΔRQC) = \(\frac{3}{8}\)ar(ΔABC) |
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Answer» (i) We know that, Each median of a triangle divides it into two triangles of equal area. Since, CR is a median of ΔCAP Therefore, Area (ΔCRA) = Area (ΔCAP) ...(i) Also, CP is a median of ΔCAB Therefore, Area (ΔCAP) = Area (ΔCPB) ...(ii) From (i) and (ii), we get Therefore, Area (ΔARC) = \(\frac{1}{2}\)Area (ΔCPB) ...(iii) PQ is a median of ΔPBC Therefore, Area (ΔCPB) = 2 Area (ΔPQB) ...(iv) From (iii) and (iv), we get Area (ΔARC) = Area (ΔPBQ) ...(v) (ii) Since, QP and QR medians of ΔQAB and QAP respectively. Area (ΔQAP) = Area (ΔQBP) ...(vi) And, Area (ΔQAP) = 2 Area (ΔQRP) ...(vii) From (vi) and (vii), we get Area (ΔPRQ) = Area (ΔPBQ)...(viii) From (v) and (viii), we get Area (ΔPRQ) = Area (ΔARC) ...(ix) (iii) Since, CR is a median of ΔCAP Therefore, Area (ΔARC) = \(\frac{1}{2}\)Area (ΔCAP) = \(\frac{1}{2}\)x \(\frac{1}{2}\) Area (ΔABC) (Therefore, CP is a median of Δ ABC) =\(\frac{1}{4}\) Area (ΔABC) ...(x) Since, RQ is a median of Δ RBC. Therefore, Area (ΔRQC) = \(\frac{1}{2}\)Area (ΔRBC) = \(\frac{1}{2}\)[Area (ΔABC) – Area (ΔARC)] = \(\frac{1}{2}\)[Area (ΔABC) - Area (ΔABC)] = \(\frac{3}{4}\)Area (ΔABC) |
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