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प्रश्न
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) = 1/2 ar(ARC)
(ii) ar(RQC) = 3/8 ar(ABC)
(iii) ar(PBQ) = ar(ARC)
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उत्तर
Take a point S on AC such that S is the mid-point of AC.
Extend PQ to T such that PQ = QT.
Join TC, QS, PS, and AQ.
In ΔABC, P and Q are the mid-points of AB and BC respectively. Hence, by using mid-point theorem, we obtain
PQ || AC and PQ = 1/2AC
⇒ PQ || AS and PQ = AS (As S is the mid-point of AC)
∴ PQSA is a parallelogram. We know that diagonals of a parallelogram bisect it into equal areas of triangles.
∴ ar (ΔPAS) = ar (ΔSQP) = ar (ΔPAQ) = ar (ΔSQA)
Similarly, it can also be proved that quadrilaterals PSCQ, QSCT, and PSQB are also parallelograms and therefore,
ar (ΔPSQ) = ar (ΔCQS) (For parallelogram PSCQ)
ar (ΔQSC) = ar (ΔCTQ) (For parallelogram QSCT)
ar (ΔPSQ) = ar (ΔQBP) (For parallelogram PSQB)
Thus,
ar (ΔPAS) = ar (ΔSQP) = ar (ΔPAQ) = ar (ΔSQA) = ar (ΔQSC) = ar (ΔCTQ) = ar (ΔQBP) ... (1)
Also, ar (ΔABC) = ar (ΔPBQ) + ar (ΔPAS) + ar (ΔPQS) + ar (ΔQSC)
ar (ΔABC) = ar (ΔPBQ) + ar (ΔPBQ) + ar (ΔPBQ) + ar (ΔPBQ)
= 4 ar (ΔPBQ)
⇒ ar (ΔPBQ) = 1/4 ar(ΔABC) ... (2)
(i)Join point P to C.
In ΔPAQ, QR is the median.
`therefore ar(trianglePRQ)=1/2ar(trianglePAQ)=1/2xx1/4ar(triangleABC)=1/8ar(triangleABC)" .........(3)"`
In ΔABC, P and Q are the mid-points of AB and BC respectively. Hence, by using mid-point theorem, we obtain
PQ = 1/2AC
AC = 2PQ ⇒ AC = PT
Also, PQ || AC ⇒ PT || AC
Hence, PACT is a parallelogram.
ar (PACT) = ar (PACQ) + ar (ΔQTC)
= ar (PACQ) + ar (ΔPBQ [Using equation (1)]
∴ ar (PACT) = ar (ΔABC) ... (4)
`ar(triangleARC) = 1/2ar(trianglePAC)" (CR is the median of ΔPAC)"`
`= 1/2xx1/2ar(PACT)" (PC is the diagonal parallelogram PACT)"`
`= 1/4 ar(PACT) = 1/4 ar(triangleABC)`
`rArr 1/2ar(triangleARC) = 1/8ar(triangleABC)`
`rArr1/2ar(triangleARC)=ar(trianglePRQ)" [Using equation (3)] .......(5)"`
(ii)
ar(PACT) = ar(ΔPRQ) + ar(ΔARC) + ar(ΔQTC) + ar(ΔRQC)
Putting the values from equations (1), (2), (3), (4), and (5), we obtain
`ar(ΔABC) = 1/8ar(ΔABC) + 1/4ar(ΔABC) + 1/4ar(ΔABC) + ar(ΔRQC)`
`ar(ΔABC) = 5/8ar(ΔABC) + ar(ΔRQC)`
`ar(ΔRQC) = (1-5/8)ar(ΔABC)`
`ar(ΔRQC) = 3/8ar(ΔABC)`
(iii) In parallelogram PACT,
`ar(ΔARC) = 1/2ar(ΔPAC)" (CR is the median of ΔPAC)"`
`= 1/2xx1/2ar(PACT)" (PC is the diagonal of parallelogram PACT)"`
`= 1/4ar(PACT)`
`= 1/4ar(ABC)`
= ar(ΔPBQ)
संबंधित प्रश्न
A villager Itwaari has a plot of land of the shape of a quadrilateral. The Gram Panchayat of the village decided to take over some portion of his plot from one of the corners to construct a Health Centre. Itwaari agrees to the above proposal with the condition that he should be given equal amount of land in lieu of his land adjoining his plot so as to form a triangular plot. Explain how this proposal will be implemented.
In the following figure, ABC and BDE are two equilateral triangles such that D is the mid-point of BC. If AE intersects BC at F, show that
(i) ar (BDE) = 1/4 ar (ABC)
(ii) ar (BDE) = 1/2 ar (BAE)
(iii) ar (ABC) = 2 ar (BEC)
(iv) ar (BFE) = ar (AFD)
(v) ar (BFE) = 2 ar (FED)
(vi) ar (FED) = 1/8 ar (AFC)
[Hint : Join EC and AD. Show that BE || AC and DE || AB, etc.]
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 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) .
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 )
(2) ar (ΔLBC) = ar (ΔMBC)
(3) ar (ΔABM) ar (ΔACL)
(4) ar (ΔLOB) ar (ΔMOC)
PQRS is a parallelogram whose area is 180 cm2 and A is any point on the diagonal QS. The area of ∆ASR = 90 cm2.
O is any point on the diagonal PR of a parallelogram PQRS (Figure). Prove that ar (PSO) = ar (PQO).
A point E is taken on the side BC of a parallelogram ABCD. AE and DC are produced to meet at F. Prove that ar (ADF) = ar (ABFC)
The medians BE and CF of a triangle ABC intersect at G. Prove that the area of ∆GBC = area of the quadrilateral AFGE.
