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In a δAbc, P and Q Are Respectively the Mid-points of Ab and Bc and R is the Mid-point of Ap. Prove that : (1) Ar (δ Pbq) = Ar (δ Arc) (2) Ar (δ Prq) =`1/2`Ar (δ Arc) (3

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Question

In a ΔABC, P and Q are respectively the mid-points of AB and BC and R is the mid-point
of AP. Prove that :

(1) ar (Δ PBQ) = ar (Δ ARC)

(2) ar (Δ PRQ) =`1/2`ar (Δ ARC)

(3) ar (Δ RQC) =`3/8` ar (Δ ABC) .

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Solution

(1)   We know that each median of a Δle  divides it into two triangles of equal area
        Since, OR is a median of  ΔCAP

       ∴ ar (ΔCRA) = `1/2` ar (ΔCAP)       ....... (1) 

        Also, CPis a median of ΔCAB

       ∴ ar  ( ΔCAP) ar  (ΔCPB)            ....... (2) 
        From (1) and (2) we get

       ∴ area (Δ ARC ) = `1/2 ar (CPB)` ....... (3)

         PQ is the median of  ΔPBC

        ∴ area( Δ CPB) =  2area (Δ PBQ)    ......... (4)

     From (3) and (4) we get

   ∴ area (Δ ARC) = area (PBQ)    .......  (5)

(2)     Since QP and QR medians of s QAB and QAP                respectively.

       ∴ ar (ΔQAP) = area (ΔPBQ)      ............ (6)

        And area  (ΔQAP)  =  2ar (QRP)  ......... (7)

        From (6) and (7) we have

        Area (ΔPRQ) = `1/2` ar (ΔPBQ)    ......... (8)

         From (5)  and (8)  we get 

        Area (ΔPRQ) = `1/2` area (ΔARC)

(3)   Since, ∠R is a median of  ΔCAP

       ∴ area (ΔARC) = `1/2` ar (ΔCAP) 

        `= 1/2 xx1/2[ ar (ABC)]` 

         = `1/4` area (ABC)

Since RQ is a median of  ΔRBC

        ∴ ar (ΔRQC) =`1/2` ar (Δ RBC)

         = `1/2`[ ar (ΔABC)- ar (ARC) ]

         = `1/2`[ar (ΔABC) - `1/4`(Δ ABC )]

          = `3/8`(Δ ABC)

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Chapter 14: Areas of Parallelograms and Triangles - Exercise 14.3 [Page 46]

APPEARS IN

R.D. Sharma Mathematics [English] Class 9
Chapter 14 Areas of Parallelograms and Triangles
Exercise 14.3 | Q 19 | Page 46

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