Q. 175.0( 1 Vote )

# P and Q are respectively the mid- points of sides AB and BC of 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)

Answer :

**Proof:**

P and Q are respectively the mid- points of sides AB and BC of Δ ABC and R is the mid- point of AP.

QR is a median of Δ APQ and it divides the Δ into two other triangles of equal area.

ar (PQR) = 1/2 ar (APQ)

[Since QP is a median of Δ ABQ]

= 1/2 ×1/2 ar (ABQ)

=1/4 ar (ABQ) = 1/4 × 1/2 ar (ABC)

[Since AQ is a median of Δ ABC]

= 1/8 ar (ABC) ... (1)

[Since CR is a median of Δ APC]

ar(ARC) = 1/2 ar (APC)

[Since CP is a median of Δ ABC]

= 1/2 × 1/2 ar (ABC)

= 1/4 ar (ABC) ... (2)

From (1) and (2) , we get,

ar (PQR) = 1/8 ar (ABC) = 1/2 × 1/4 ar (ABC)

= 1/2 ar (ARC).

**(ii)** We have,

ar (RQC) = ar(RQA) + ar (AQC) – ar (ARC) ... (3)

[Since RQ is a median of Δ PQA]

So, ar (Δ RQA) = 1/2 ar (PQA)

[Since PQ is a median of Δ AQB]

= 1/2 × 1/2 ar (AQB)

= 1/4 ar (AQB)

[Since AQ is a median of Δ ABC]

= 1/4 × 1/2 ar(ABC)

= 1/8 ar (ABC) ... (4)

[Since AQ is a median of Δ ABC]

ar (AQC) = 1/2 ar (ABC) ... (5)

[Since CR is a median of Δ APC]

ar (Δ ARC) = 1/2 ar (APC)

[Since CP is a median of Δ ABC]

= 1/2 × 1/2 ar(ABC)

= 1/4 ar (ABC) ... (6)

From (3), (4) , (5) and (6) we have,

ar (RQC) = 1/8 ar (ABC) + 1/2 ar (ABC) – 1/4 ar (ABC)

=(1/8 + 1/2 − 1/4 ) ar (ABC)

= 3/8 ar (ABC)

**(iii)** We have,

[Since PQ is a median of Δ ABQ]

ar (PBQ) = 1/2 ar (ABQ)

[Since AQ is a median of Δ ABC]

= 1/2 × 1/2 ar(ABC)

= 1/4 ar (ABC)

= ar(ARC) [Using (6)]

ar (PBQ) = ar (ARC)

Hence Proved

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