NCERT Solutions
Class 9 Maths
Areas of Parallelograms and Triangles
Ex.9.4 Q.7
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) = 
2. ar (RQC) = 
3. ar (PBQ) = ar (ARC)
P and Q are respectively the mid-points of sides AB.
and BC of ΔABC and R is the mid-point of AP.
Join AQ and PC.
1. We have,
ar(ΔPQR) = 
[Since QR is a median and ΔAPQ and it divides the triangle into
two other triangles of equal area]
= 
× × ar(ΔABQ)
[Since QP is a median of ΔABQ]
= 
× ar(ΔABQ)
= × × ar(ΔABQ)
[Since AQ is a median of ΔABQ]
= 
× ar(ΔABQ) ……………... (1)
Again, ar(ΔARC) = × ar(ΔAPC)
[Since CR is a median of ΔAPC]
= × 
× ar(ΔABQ)
[Since CP is a median of ΔABC]
= × ar(ΔABQ) ……… (2)
From equation (1) and (2) we get
ar(ΔPQR) = × ar(ΔABC)
= × × ar(ΔABC)
= × ar(ΔARC)
2. We have,
ar(ΔRQC) = ar(ΔRQA) + ar(ΔAQC) - ar(ΔARC) …………... (3)
Now, ar(ΔRQA) = × ar(ΔPQA)
[Since RQ is a median of ΔPQA]
= × × ar(ΔAQB)
[Since PQ is a median of ΔAQB]
= × ar(ΔAQB)
= × × ar(ΔABC)
[Since AQ is a median of ΔABC]
= × ar(ΔABC) ………... (4)
ar(ΔAQC) = × ar(ΔABC) …… (5)
[Since AQ is a median of ΔABC]
ar(ΔARC) = × ar(ΔAPC)
[Since CR is a median of ΔAPC]
= × × ar(ΔABC)
[Since CP is a median of ΔABC]
= × ar(ΔABC) ……… (6)
From equation (3), (4), (5) and (6), we get
ar(ΔRQC) = × ar(ΔABC) + × ar(ΔABC) - × ar(ΔABC)
= 
× ar(ΔABC)
3. We have,
ar(ΔPBQ) = × ar(ΔABQ)
[Since PQ is a median of ΔABQ]
= × × ar(ΔABC)
[Since AQ is a median of ΔABC]
= × ar(ΔABC)
= ar(ΔARC)
[From equation 6]