# NCERT Class 9 Solutions: Areas of Parallelograms and Triangles (Chapter 9) Exercise 9.2 – Part 2 (For CBSE, ICSE, IAS, NET, NRA 2022)

Glide to success with Doorsteptutor material for CBSE : fully solved questions with step-by-step explanation- practice your way to success.

Q-3 P and Q are any two points lying on the sides DC and AD respectively of a parallelogram ABCD. Show that

Solution:

Given,

P and Q are any two points lying on the sides DC and AD respectively of a parallelogram ABCD.

and parallelogram ABCD are on the same base AB and between the same parallel lines AB and DC. Therefore, (Parallelogram ABCD) … equation (1)

- Similarly, and parallelogram ABCD are on the same base BC and these are between the same parallel lines AD and BC
- Therefore, (Parallelogram ABCD) … equation (2)
- From (1) and (2) , we have

Q-4 In the figure is a point in the interior of a parallelogram ABCD. Show that

Solution (i)

A line GH is drawn parallel to AB passing through P. In a parallelogram, (By construction) … equation (1)

Thus,

- and
- … equation (2)

From equations (1) and (2) ,

- is a parallelogram.

Now,

In and parallelogram ABHG are lying on the same base AB and between the same parallel lines AB and GH. Therefore, … equation (3)

Also,

In and parallelogram CDGH are lying on the same base CD and between the same parallel lines CD and GH.

… equation (4)

Adding equations (3) and (4) ,

- … equation (5)

Solution (ii)

A line EF is drawn parallel to AD passing through P.

In a parallelogram, (by construction) … equation (6)

Thus, … equation (7)

From equations (6) and (7) ,

- AEFD is a parallelogram.

Now,

In ΔAPD and parallelogram AEFD are lying on the same base AD and between the same parallel lines AD and EF, … equation (8)

Also, in and parallelogram BCFE are lying on the same base BC and between the same parallel lines BC and EF … equation (9)

Adding equations (8) and (9) ,

- … equation (10)

From equation (5) and (10)