P and Q are the mid-points of sides AB and BC of triangle ABC respectively and R is the mid-point of AP, then ar (â–³PRQ) is equal to
P and Q are the mid-points of sides AB and BC of triangle ABC respectively and R is the mid-point of AP, then ar (â–³PRQ) is equal to
The correct option is D ar (△PRQ) = 12 ar (△ARC)
Take a point S on AC such that S is the midpoint of AC.
Extend PQ to T such that PQ = QT.
Join TC, QS, PS, and AQ.
In ABC, P and Q are the mid-points of AB and AC respectively. Hence, by using the mid-point theorem, we obtain:
PQ∥AC and PQ=12AC
PQ∥AS and PQ=AS (As S is the mid-point of AC)
Therefore, PQSA is a parallelogram. We know that diagonals of a parallelogram bisect it into two triangles of equal areas.
ar(△PAS)= ar(△QSP)= ar(△PAQ)= ar(△SQA)
Similarly, it can also be proved that quadrilaterals PSCQ, QSCT, and PSQB are also parallelograms and therefore,
ar(△PSQ)=ar(△CQS) (For parallelogram PSCQ)
ar(△QSC)=ar(△CTQ) (For parallelogram QSCT)
ar(△PSQ)=ar(△QBP) (For parallelogram PSQB)
Thus,
ar(△PAS)=ar(△SQP)=ar(△PAQ)=ar(△SQA)=ar(△QSC)=ar(△CTQ)=ar(△QBP) ----(1)
Also, ar(△ABC)=ar(△PBQ)+ar(△PAS)+ar(△PQS)+ar(△QSC)
Now, ar(△ABC)=ar(△PBQ)+ar(△PBQ)+ar(△PBQ)+ar(△PBQ)
= 4ar(△PBQ)
ar(△PBQ)=14ar(△ABC) -----(2)
Now, join point P to C.
In △PAQ,QR is the median.
ar (△PRQ) = 12 ar (△PAQ)
= 12x14 ar (△ABC)
= 18 ar (△ABC) -----(3)
In △ABC, P and Q are the mid-points of AB and BC respectively. Hence, by using mid-point theorem, we obtain
PQ = 12AC
AC = 2PQ, AC = PT
Also, PQ ∥ AC, PT ∥ AC
Hence, PACT is a parallelogram.
ar(PACT)=ar(PACQ)+ar(△QTC)
= ar(PACQ) + ar (△PBQ) [Using equation (1)]
ar(PACT)=ar(△ABC) ------(4)
ar (△ARC) = 12 ar (△PAC) (CR is median of △PAC)
= 12x12ar(PACT) (PC is the diagonal of parallelogram PACT)
= 14ar(PACT) = 14 ar (△ABC)
12 ar (△ARC) = 18 ar (△ABC)
12 ar (△ARC) = ar (△PRQ) [Using equation (3)]
Take a point S on AC such that S is the midpoint of AC.
Extend PQ to T such that PQ = QT.
Join TC, QS, PS, and AQ.
In ABC, P and Q are the mid-points of AB and AC respectively. Hence, by using the mid-point theorem, we obtain:
PQ∥AC and PQ=12AC
PQ∥AS and PQ=AS (As S is the mid-point of AC)
Therefore, PQSA is a parallelogram. We know that diagonals of a parallelogram bisect it into two triangles of equal areas.
ar(△PAS)= ar(△QSP)= ar(△PAQ)= ar(△SQA)
Similarly, it can also be proved that quadrilaterals PSCQ, QSCT, and PSQB are also parallelograms and therefore,
ar(△PSQ)=ar(△CQS) (For parallelogram PSCQ)
ar(△QSC)=ar(△CTQ) (For parallelogram QSCT)
ar(△PSQ)=ar(△QBP) (For parallelogram PSQB)
Thus,
ar(△PAS)=ar(△SQP)=ar(△PAQ)=ar(△SQA)=ar(△QSC)=ar(△CTQ)=ar(△QBP) ----(1)
Also, ar(△ABC)=ar(△PBQ)+ar(△PAS)+ar(△PQS)+ar(△QSC)
Now, ar(△ABC)=ar(△PBQ)+ar(△PBQ)+ar(△PBQ)+ar(△PBQ)
= 4ar(△PBQ)
ar(△PBQ)=14ar(△ABC) -----(2)
Now, join point P to C.
In △PAQ,QR is the median.
ar (△PRQ) = 12 ar (△PAQ)
= 12x14 ar (△ABC)
= 18 ar (△ABC) -----(3)
In △ABC, P and Q are the mid-points of AB and BC respectively. Hence, by using mid-point theorem, we obtain
PQ = 12AC
AC = 2PQ, AC = PT
Also, PQ ∥ AC, PT ∥ AC
Hence, PACT is a parallelogram.
ar(PACT)=ar(PACQ)+ar(△QTC)
= ar(PACQ) + ar (△PBQ) [Using equation (1)]
ar(PACT)=ar(△ABC) ------(4)
ar (△ARC) = 12 ar (△PAC) (CR is median of △PAC)
= 12x12ar(PACT) (PC is the diagonal of parallelogram PACT)
= 14ar(PACT) = 14 ar (△ABC)
12 ar (△ARC) = 18 ar (△ABC)
12 ar (△ARC) = ar (△PRQ) [Using equation (3)]
