O is any point on the diagonal PR of a parallelogram PQRS (Fig. 9.16). Prove that ar (PSO) = ar (PQO).
Solution:
Given, PQRS is a parallelogram
PR is the diagonal
O is any point on the diagonal PR
We have to prove that ar(PSO) = ar(PQO)
Join SQ which intersects PR at B.
We know that the diagonals of the parallelogram bisect each other.
So, B is the midpoint of SQ
We know that the median of a triangle divides it into two triangles of equal areas.
PB is a median of triangle QPS
ar(BPQ) = ar(BPS) -------------------- (1)
OB is a median of triangle OSQ
ar(OBQ) = ar(OBS) ---------------------- (2)
Adding (1) and (2),
ar(BPQ) + ar(OBQ) = ar(BPS) + ar(OBS)
From the figure,
BPQ + BOQ = PQO
ar(POQ) = ar(BPS) + ar(OBS)
From the figure,
BPS + OBS = PSO
Therefore, ar(PQO) = ar(PSO)
✦ Try This: D, E and F are respectively the mid-points of the sides BC, CA, and AB of a ABC. Show that ar(BDEF) = 1/2 ar(ABC)
☛ Also Check: NCERT Solutions for Class 9 Maths Chapter 9
NCERT Exemplar Class 9 Maths Exercise 9.3 Problem 6
O is any point on the diagonal PR of a parallelogram PQRS (Fig. 9.16). Prove that ar (PSO) = ar (PQO)
Summary:
O is any point on the diagonal PR of a parallelogram PQRS (Fig. 9.16). It is proven that ar (PSO) = ar (PQO)
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