Question

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
(1) ar (PRQ) = 1/2ar (ARC)
(m) ar (RQC) - 3/8ar (ABC)
(iii) ar (PBQ) = ar (ARC)​

Answer 1

Answer:

Step-by-step explanation:

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(ΔR= (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)