Question

P and Q are respectively the midpoints of sides AB and BC of a triangle ABC and R is the midpoint of AP,show that i) ar(PRQ) =½ ar(ARC) ii) ar (RQC) =⅜ ar(ABC) iii) ar(PBQ) = ar(ARC)

Answer 1

Hi,
Here is ur ans:-
Let ABC is a triangle. P and Q are the mid points of AB and BC respectively and R is the mid-point of AP.

Join PQ, QR, AQ, PC, RC as shown in the figure.

Now we know that median of a triangle divides it into two triangle of equal areas.

In triangle CAP, Cr is the mid point.

So Area(ΔCRA) = (1/2)* Area(ΔCAP) .......1

Again in triangle CAB, CP is the mid point.

So Area(ΔCAP) = (1/2)* Area(ΔCPB) ............2

from eqaution 1 and 2, we get

Area(ΔCAP) = (1/2)*Area(ΔCPB) ..............3

Again in triangle PBC, PQ is the mid point.

So (1/2)*Area(ΔCPB) = Area(ΔPBQ) ............4

From equation 3 and 4, we get

Area(ΔARC) = Area(ΔPBQ) ...............5



Now QP, And QR are the medians of triangle QAB and QAP respectively

So Area(ΔQAP) = Area(ΔQBP) ............6

and Area(ΔQAP) = 2*Area(ΔQRP) ............7

from equation 6 and 7, we get

Area(ΔPRQ) = (1/2)*Area(ΔPBQ) ............8

from equation 5 and 8, we get

Area(ΔPRQ) = (1/2)*Area(ΔARC)

Now CR is the median of triangle CAP

So Area(ΔARC) = (1/2)*Area(ΔCAP)

= (1/2)*(1/2)*Area(ΔABC)

= (1/4)*Area(ΔABC)

Again RQ is the median of triangle RBC

So Area(ΔRQC) = (1/2)*Area(ΔRBC)

= (1/2)*Area(ΔABC) - (1/2)*Area(ΔARC)

=(1/2)*Area(ΔABC) - (1/2)*(1/2)*(1/2)*Area(ΔABC)

=(1/2)*Area(ΔABC) - (1/8)*Area(ΔABC)

= (3/8)*Area(ΔABC)

So Area(ΔRQC) = = (3/8)*Area(ΔABC)
Hope it helps u

Answer 2

Answer:

Data: 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.

To prove: ar.(△RQC)=83ar.(△ABC)

Proof: P and Q are respectively the mid points of sides AB and BD

∴PQ∣∣AC and PQ=21AC

(∵ Mid-point theorem)

Marks S such that it is the mid point of AC and join QS

∴PQSA is a parallelogram. Its diagonals bisects quadrilateral into two congruent triangles.

ar.(△PAS)=ar.(△PQS)=ar.(△PQA)=ar.(△QAS)

Similarly, PSCQ,QSCT and PSQB are parallelogram.

∴ar.(△PSQ)=ar.(△CQS)

ar.(△QSC)=ar.(△CTQ)

ar.(△PSQ)=ar.(△BQP)

∴△PAS=△SQP=△PAQ=△SQA=△QSC=△CTQ=△QBP....(i)

and △ABC=△PBQ+△PAS+△PQS+△QSC

△ABC=△PBQ+△PBQ+△PBQ+△PBQ

△ABC=4×△PBQ

ar.(△ABC)=4×ar.(△PBQ)

ar(△PBQ)=

Step-by-step explanation:

HOPE IT HELPS