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.9.4 Q.4

In Fig. 9.32, ABCD is a parallelogram and BC is produced to a point Q such that AD = CQ. If AQ intersect DC at P, show that ar (BPC) = ar (DPQ).

[Hint: Join AC.]

It is given that ABCD is a parallelogram.

AD || BC and AB || DC

[Opposite sides of a parallelogram are parallel to each other]

Join point A to point C.

Consider ∆APC and ∆BPC,

∆APC and ∆BPC are lying on the same base PC and between the same parallels PC and AB.

Therefore,

Area (∆APC) = Area (∆BPC) ..........(1)

In quadrilateral ACDQ, it is given that

AD = CQ

Since ABCD is a parallelogram,

AD || BC (Opposite sides of a parallelogram are parallel)

CQ is a line segment which is obtained when line segment BC is produced.

So, AD || CQ

We have,

AC = DQ and AC || DQ

Hence, ACQD is a parallelogram.

Consider ∆DCQ and ∆ACQ

These are on the same base CQ and between the same parallels CQ and AD.

Therefore,

Area (∆DCQ) = Area (∆ACQ)

=> Area (∆DCQ) − Area (∆PQC) = Area (∆ACQ) − Area (∆PQC)

=> Area (∆DPQ) = Area (∆APC) ... (2)

From equations (1) and (2), we obtain

Area (∆BPC) = Area (∆DPQ)

Video solution:

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

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