# Class 9 NCERT Solutions- Chapter 9 Areas of Parallelograms And Triangles – Exercise 9.3 | Set 2

Last Updated : 05 Apr, 2021

### Show that

(i) ar(Î”ACB) = ar(Î”ACF)

(ii) ar(AEDF) = ar(ABCDE)

Solution:

(i) Î”ACB and Î”ACF lie on the same base AC and between the same parallels AC and BF.

Therefore,

ar(Î”ACB) = ar(Î”ACF)

(ii) ar(Î”ACB) = ar(Î”ACF)

=> ar(Î”ACB)+ar(Î”ACDE) = ar(Î”ACF)+ar(Î”ACDE)

=> ar(ABCDE) = ar(AEDF)

### Question 12. A villager Itwaari has a plot of land of the shape of a quadrilateral. The Gram Panchayat of the village decided to take over some portion of his plot from one of the corners to construct a Health Centre. Itwaari agrees to the above proposal with the condition that he should be given equal amount of land in lieu of his land adjoining his plot so as to form a triangular plot. Explain how this proposal will be implemented.

Solution:

Let us assume ABCD be the plot of the land of the shape of a quadrilateral.

Construction:

Join the diagonal BD.

Draw AE parallel to BD.

Join BE, that intersected AD at O.

We will get,

Î”BCE is the shape of Itwari’s original field

Î”AOB is the area for construction of the Health Centre.

Î”DEO is the land joined to the plot.

To prove:

ar(Î”DEO) = ar(Î”AOB)

Proof:

Î”DEB and Î”DAB lie on the same base BD, in-between the same parallels BD and AE.

Therefore,

Ar(Î”DEB) = ar(Î”DAB)

=> ar(Î”DEB) – ar(Î”DOB) = ar(Î”DAB) – ar(Î”DOB)

=> ar(Î”DEO) = ar(Î”AOB)

### [Hint: Join CX.]

Solution:

Given:

ABCD is a trapezium with,

AB || DC.

XY || AC

To Construct,

Join CX

Prove:

Proof:

ar(Î”ADX) = ar(Î”AXC)         -(equation 1)(Since they are on the same base AX and

in-between the same parallels AB and CD)

also,

ar(Î”AXC)=ar(Î”ACY)          -(equation-2)(Since they are on the same base AC and

in-between the same parallels XY and AC)

From equation (1) and (2),

### Question 14. In Figure AP || BQ || CR. Prove that ar(Î”AQC) = ar(Î”PBR).

Solution:

Given:

AP || BQ || CR

Prove:

ar(AQC) = ar(PBR)

Proof:

ar(Î”AQB) = ar(Î”PBQ)          -(equation 1)(Since they are on the same base BQ and

between the same parallels AP and BQ)

also,

ar(Î”BQC) = ar(Î”BQR)          (equation 2)(Since they are on the same base BQ and

between the same parallels BQ and CR)

ar(Î”AQB)+ar(Î”BQC) = ar(Î”PBQ)+ar(Î”BQR)

=> ar(Î”AQC) = ar(Î”PBR)

### Question 15. Diagonals AC and BD of a quadrilateral ABCD intersect at O in such a way that ar(â–³AOD) = ar(â–³BOC). Prove that ABCD is a trapezium.

Solution:

Given:

ar(Î”AOD) = ar(Î”BOC)

Prove:

ABCD is a trapezium.

Proof:

ar(Î”AOD) = ar(Î”BOC)

=> ar(Î”AOD) + ar(Î”AOB) = ar(Î”BOC)+ar(Î”AOB)

Areas of Î”ADB and Î”ACB are equal.

Therefore,

They should be lying between the same parallel lines.

Therefore,

AB âˆ¥  CD

Hence, ABCD is a trapezium.

### Question 16. In Figure, ar(DRC) = ar(DPC) and ar(BDP) = ar(ARC). Show that both the quadrilaterals ABCD and DCPR are trapeziums.

Solution:

Given:

ar(Î”DRC) = ar(Î”DPC)

ar(Î”BDP) = ar(Î”ARC)

Prove:

ABCD and DCPR are trapeziums.

Proof:

ar(Î”BDP) = ar(Î”ARC)

â‡’ ar(Î”BDP) â€“ ar(Î”DPC) = ar(Î”DRC)

Therefore,

ar(Î”BDC) and ar(Î”ADC) are lying in-between the same parallel lines.

Hence, AB âˆ¥ CD

ABCD is a trapezium.

Similarly,

ar(Î”DRC) = ar(Î”DPC).

Therefore,

ar(Î”DRC) and ar(Î”DPC) are lying in-between the same parallel lines.

Hence, DC âˆ¥ PR

Thus, DCPR is a trapezium.

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