मराठी

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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प्रश्न

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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उत्तर

(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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पाठ 14: Areas of Parallelograms and Triangles - Exercise 14.3 [पृष्ठ ४६]

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आर.डी. शर्मा Mathematics [English] Class 9
पाठ 14 Areas of Parallelograms and Triangles
Exercise 14.3 | Q 19 | पृष्ठ ४६

व्हिडिओ ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्‍न

In the given figure, ABC and ABD are two triangles on the same base AB. If line-segment CD is bisected by AB at O, show that ar (ABC) = ar (ABD).


ABCD is a trapezium with AB || DC. A line parallel to AC intersects AB at X and BC at Y. Prove that ar (ADX) = ar (ACY). 

[Hint: Join CX.]


Diagonals AC and BD of a quadrilateral ABCD intersect at O in such a way that ar (AOD) = ar (BOC). Prove that ABCD is a trapezium.


In the following figure, D and E are two points on BC such that BD = DE = EC. Show that ar (ABD) = ar (ADE) = ar (AEC).

Can you answer the question that you have left in the ’Introduction’ of this chapter, whether the field of Budhia has been actually divided into three parts of equal area?

[Remark: Note that by taking BD = DE = EC, the triangle ABC is divided into three triangles ABD, ADE and AEC of equal areas. In the same way, by dividing BC into n equal parts and joining the points of division so obtained to the opposite vertex of BC, you can divide ΔABC into n triangles of equal areas.]


Diagonals AC and BD of a quadrilateral ABCD intersect each other at P. Show that ar (APB) × ar (CPD) = ar (APD) × ar (BPC).

[Hint : From A and C, draw perpendiculars to BD.]


In the below fig. D and E are two points on BC such that BD = DE = EC. Show that ar
(ΔABD) = ar (ΔADE) = ar (ΔAEC).


In a ΔABC, if L and M are points on AB and AC respectively such that LM || BC. Prove
that:

(1) ar (ΔLCM ) = ar (ΔLBM )
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(3) ar (ΔABM) ar (ΔACL)
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The area of the parallelogram ABCD is 90 cm2 (see figure). Find ar (ΔBEF)


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