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(Δ PBQ) = ar(Δ ARC) (ii) ar(Δ PQR) = ar(Δ ARC) (iii) ar(Δ RQC) = ar(Δ ABC)

Question

In a &#x394; 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(&#x394; PBQ) = ar(&#x394; ARC) (ii) ar(&#x394; PQR) = ar(&#x394; ARC) (iii) ar(&#x394; RQC) = ar(&#x394; ABC)

Philoid · Accepted 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) = 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) = Area (ΔCAP)

= * Area (ΔABC) (Therefore, CP is a median of Δ ABC)

= Area (ΔABC) (x)

Since,

RQ is a median of Δ RBC.

Therefore,

Area (ΔRQC) = Area (ΔRBC)

=[Area (ΔABC) – Area (ΔARC)]

= [Area (ΔABC) - Area ()]

= Area (Δ ABC)

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(Δ PBQ) = ar(Δ ARC)(ii) ar(Δ PQR) = ar(Δ ARC)(iii) ar(Δ RQC) = ar(Δ ABC)

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(Δ PBQ) = ar(Δ ARC)

(ii) ar(Δ PQR) = ar(Δ ARC)

(iii) ar(Δ RQC) = ar(Δ ABC)