P and Q are the mid-points of sides AB and BC of nbsp-triangle ABC respectively nbsp-and R is the mid-point of AP- then nbsp-ar -PRQ- is equal to ar -PRQ- - 2-3 ar -ARC-ar -PRQ- - 4-3 ar -ARC-ar -PRQ- - 3-8 ar -ARC-ar -PRQ- - 1-2 ar -ARC-

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Standard IX

Mathematics

The Mid-Point Theorem

P and Q are t...

Question

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 ( $△$ PRQ) is equal to

ar ( $△$ PRQ) = $\frac{2}{3}$ ar ( $△$ ARC)

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ar ( $△$ PRQ) = $\frac{4}{3}$ ar ( $△$ ARC)

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ar ( $△$ PRQ) = $\frac{3}{8}$ ar ( $△$ ARC)

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ar ( $△$ PRQ) = $\frac{1}{2}$ ar ( $△$ ARC)

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Solution

The correct option is D
ar ( $△$ PRQ) = $\frac{1}{2}$ 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:

$P Q ∥ A C a n d P Q = \frac{1}{2} A C$

$P Q ∥ A S a n d P Q = A S$ (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.

$a r (PAS) = a r (QSP) = a r (PAQ) = a r (SQA)$

Similarly, it can also be proved that quadrilaterals PSCQ, QSCT, and PSQB are also parallelograms and therefore,

$a r (△ PSQ) = a r (△ CQS)$ (For parallelogram PSCQ)
$a r (△ QSC) = a r (△ CTQ)$ (For parallelogram QSCT)
$a r (△ PSQ) = a r (△ QBP)$ (For parallelogram PSQB)

Thus,

$a r (△ PAS) = a r (△ SQP) = a r (△ PAQ) = a r (△ SQA) = a r (△ QSC) = a r (△ CTQ) = a r (△ QBP)$ ----(1)

Also, $a r (△ ABC) = a r (△ PBQ) + a r (△ PAS) + a r (△ PQS) + a r (△ QSC)$

Now, $a r (△ ABC) = a r (△ PBQ) + a r (△ PBQ) + a r (△ PBQ) + a r (△ PBQ)$

= $4 a r (△ PBQ)$

$a r (△ PBQ) = \frac{1}{4} a r (△ ABC)$ -----(2)

Now, join point P to C.

In $△ P A Q, Q R$ is the median.

ar $(△ PRQ)$ = $\frac{1}{2}$ ar $(△ P A Q)$
= $\frac{1}{2}$ x $\frac{1}{4}$ ar $(△ A B C)$
= $\frac{1}{8}$ 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 = $\frac{1}{2}$ AC

AC = 2PQ, AC = PT

Also, PQ $∥$ AC, PT $∥$ AC

Hence, PACT is a parallelogram.

$a r (P A C T) = a r (P A C Q) + a r (△ Q T C)$

= $a r (P A C Q)$ + ar $(△ PBQ)$ [Using equation (1)]

$a r (P A C T) = a r (△ ABC)$ ------(4)

ar $(△ ARC)$ = $\frac{1}{2}$ ar $(△ PAC)$ (CR is median of $△ P A C)$

= $\frac{1}{2}$ x $\frac{1}{2} a r (P A C T)$ (PC is the diagonal of parallelogram PACT)

= $\frac{1}{4} a r (P A C T)$ = $\frac{1}{4}$ ar $(△ ABC)$

$\frac{1}{2}$ ar $(△ ARC)$ = $\frac{1}{8}$ ar $(△ ABC)$

$\frac{1}{2}$ ar $(△ ARC)$ = ar $(△ PRQ)$ [Using equation (3)]

Suggest Corrections

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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 (△PRQ) 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 ( $△$ PRQ) is equal to