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/Class-9 : get questions, notes, tests, video lectures and more- for all subjects of CBSE/Class-9.

Paralleligram ABCD is Equal to Area of Parallelogram ABEF Al …

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

Solution:

Parallelogram ABCD

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

Parallelogram ABCD

Solution (i)

Parallelogram ABCD with Line GH

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)

Developed by: