NCERT solution Class 9 areas of parallelograms and triangles

NCERT Solutions

Class 9 Maths

Areas of Parallelograms and Triangles

NCERT Questions

Ex.9.1 Q.1
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Ex.9.2 Q.3

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

Show that ar (APB) = ar (BQC).

Given: A parallelogram ABCD. P and Q are any points on DC and AD respectively.

To prove: ar (APB) = ar (BQC)

Construction: Draw PS || AD and QR || AB.

Proof: In parallelogram ABRQ, BQ is the diagonal.

So, area of ∆BQR = area of .............(1)

In parallelogram CDQR, CQ is a diagonal.

So, area of ∆RQC = area of ............(2)

Adding equation (1) and (2), we get

area of ∆BQR + area of ∆RQC = (area of ABRQ + area of CDQR) ÷ 2

=> area of ∆BQC = area of .........(3)

Again, in parallelogram DPSA, AP is a diagonal.

So, area of ∆ASP = area of ..........(4)

In parallelogram BCPS, PB is a diagonal.

Now, area of ∆BPS = area of ………. (5)

Adding equation (4) and (5), we get

area of ∆ASP + area of ∆BPS = (area of DPSA + area of BCPS) ÷ 2

=> area of ∆APB = (area of ABCD) ……... (6)

From equation (3) and (6), we have

area of ∆APB = area of ∆BQC.

Hence, Proved.

Video solution:

Areas of Parallelograms and Triangles Class 9 Maths NCERT Chapter 9 NCERT Solutions

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