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
Ex.9.2 Q.1
Ex.9.2 Q.2
Ex.9.2 Q.3
Ex.9.2 Q.4
Ex.9.2 Q.5
Ex.9.2 Q.6
Ex.9.3 Q.1
Ex.9.3 Q.2
Ex.9.3 Q.3
Ex.9.3 Q.4
Ex.9.3 Q.5
Ex.9.3 Q.6
Ex.9.3 Q.7
Ex.9.3 Q.8
Ex.9.3 Q.9
Ex.9.3 Q.10
Ex.9.3 Q.11
Ex.9.3 Q.12
Ex.9.3 Q.13
Ex.9.3 Q.14
Ex.9.3 Q.15
Ex.9.3 Q.16
Ex.9.4 Q.1
Ex.9.4 Q.2
Ex.9.4 Q.3
Ex.9.4 Q.4
Ex.9.4 Q.5
Ex.9.4 Q.6
Ex.9.4 Q.7
Ex.9.4 Q.8

Ex.9.2 Q.5

In the figure, PQRS and ABRS are parallelograms and X is

any point on side BR. Show that

(i) ar (PQRS) = ar (ABRS)

(ii) ar (AXS) = ar

Given: PQRS and ABRS are parallelograms and X is any point on side BR.

To prove:

(i) ar (PQRS) = ar (ABRS)

(ii) ar (AXS) =

(i) In ∆ASP and BRQ, we have,

∠SPA = ∠RQB

[Corresponding angles] ... (1)

∠PAS = ∠QBR

[Corresponding angles] ... (2)

So, ∠PSA = ∠QRB

[Angle sum property of a triangle] ... (3)

Also, PS = QR

[Opposite sides of the parallelogram PQRS] ... (4)

So, ∆ASP ≅ ∆BRQ

[ASA axiom, using (1), (3) and (4)]

Therefore, area of ∆PSA = area of ∆QRB

[Congruent figures have equal areas] ... (5)

Now, ar (PQRS) = ar (PSA) + ar (ASRQ)

= ar (QRB) + ar (ASRQ)

= ar (ABRS)

So, ar (PQRS) = ar (ABRS), Proved.

(ii) Now, ∆AXS and ||gm ABRS are on the same base AS

and between same parallels AS and BR

So, area of ∆AXS = area of

=> area of ∆AXS = area of

[Since ar (PQRS) = ar (ABRS]

=> ar of (AXS) = area of

Video solution:

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

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