# NCERT Solutions for Class 9 Maths Chapter 9 – Areas of Parallelograms and Triangles

Here we provide NCERT Solutions for Class 9 Maths Chapter 9 – Areas of Parallelograms and Triangles for English medium students, Which will very helpful for every student in their exams. Students can download the latest NCERT Class 9 Maths solution Chapter 9 pdf. Now you will get step by step solution to each question. In a parallelogram,
AD || EF (by construction) — (i)
Thus,
AB || CD ⇒ AE || DF — (ii)
From equations (i) and (ii),
AEDF 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.
∴ ar(ΔAPD) = 1/2 ar(AEFD) — (iii)
also,
In ΔPBC and parallelogram BCFE are lying on the same base BC and between the same parallel lines BC and EF.
∴ ar(ΔPBC) = 1/2 ar(BCFE) — (iv)
Adding equations (iii) and (iv),
ar(ΔAPD) + ar(ΔPBC) = 1/2 {ar(AEFD) + ar(BCFE)}
⇒ ar(APD) + ar(PBC) = ar(APB) + ar(PCD)

5. In Fig. 9.17, PQRS and ABRS are parallelograms and X is any point on side BR. Show that
(i) ar (PQRS) = ar (ABRS)
(ii) ar (AXS) = 1/2 ar (PQRS)

(i) Parallelogram PQRS and ABRS lie on the same base SR and between the same parallel lines SR and PB.

∴ ar(PQRS) = ar(ABRS) — (i)
(ii) In ΔAXS and parallelogram ABRS are lying on the same base AS and between the same parallel lines AS and BR.
∴ ar(ΔAXS) = 1/2 ar(ABRS) — (ii)
From (i) and (ii),
ar(ΔAXS) = 1/2 ar(PQRS)

Page No: 106

6. A farmer was having a field in the form of a parallelogram PQRS. She took any point A on RS and joined it to points P and Q. In how many parts the fields is divided? What are the shapes of these parts? The farmer wants to sow wheat and pulses in equal portions of the field separately. How should she do it?

The field is divided into three parts. The three parts are in the shape of triangle. ΔPSA, ΔPAQ and ΔQAR.
Area of ΔPSA + ΔPAQ + ΔQAR = Area of PQRS — (i)
Area of ΔPAQ = 1/2 area of PQRS — (ii)
Triangle and parallelogram on the same base and between the same parallel lines.
From (i) and (ii),
Area of ΔPSA + Area of ΔQAR = 1/2 area of PQRS — (iii)
Clearly from (ii) and (iii),
Farmer must sow wheat or pulses in ΔPAQ or either in both ΔPSA and ΔQAR.

Page No: 162

Exercise 9.3

1. In Fig.9.23, E is any point on median AD of a ΔABC. Show that ar (ABE) = ar(ACE).

Given,

AD is median of ΔABC. Thus, it will divide ΔABC into two triangles of equal area.

∴ ar(ABD) = ar(ACD) — (i)

also,

ED is the median of ΔABC.

∴ ar(EBD) = ar(ECD) — (ii)

Subtracting (ii) from (i),

ar(ABD) – ar(EBD) = ar(ACD) – ar(ECD)

⇒ ar(ABE) = ar(ACE)

2. In a triangle ABC, E is the mid-point of median AD. Show that ar(BED) = 1/4 ar(ABC).

ar(BED) = (1/2) × BD × DE
As E is the mid-point of AD,

Thus, AE = DE
As AD is the median on side BC of triangle ABC,
Thus, BD = DC
Therefore,

DE = (1/2)AD — (i)
BD = (1/2)BC — (ii)

From (i) and (ii),

ar(BED) = (1/2) × (1/2) BC × (1/2)AD ⇒ ar(BED) = (1/2) × (1/2) ar(ABC)

⇒ ar(BED) = 1/4 ar(ABC)

3. Show that the diagonals of a parallelogram divide it into four triangles of equal area.

O is the mid point of AC and BD. (diagonals of bisect each other)

In ΔABC, BO is the median.

∴ ar(AOB) = ar(BOC) — (i)

also,

In ΔBCD, CO is the median.

∴ ar(BOC) = ar(COD) — (ii)

In ΔACD, OD is the median.

∴ ar(AOD) = ar(COD) — (iii)

In ΔABD, AO is the median.

∴ ar(AOD) = ar(AOB) — (iv)

From equations (i), (ii), (iii) and (iv),

ar(BOC) = ar(COD) = ar(AOD) = ar(AOB)

So, the diagonals of a parallelogram divide it into four triangles of equal area.

4. In Fig. 9.24, ABC and ABD are two triangles on the same base AB. If line- segment CD is bisected by AB at O, show that:

ar(ABC) = ar(ABD).

In ΔABC,

AO is the median. (CD is bisected by AB at O)

∴ ar(AOC) = ar(AOD) — (i)

also,

In ΔBCD,

BO is the median. (CD is bisected by AB at O)

∴ ar(BOC) = ar(BOD) — (ii)
Adding (i) and (ii) we get,
ar(AOC) + ar(BOC) = ar(AOD) + ar(BOD)
⇒ ar(ABC) = ar(ABD)  From (iii), (iv) and (v),
ar(△AEB) = ar(△ACF)

9. The side AB of a parallelogram ABCD is produced to any point P. A line through A and parallel to CP meets CB produced at Q and then parallelogram PBQR is completed (see Fig. 9.26). Show that
ar(ABCD) = ar(PBQR).
[Hint : Join AC and PQ. Now compare ar(ACQ) and ar(APQ).]

AC and PQ are joined.
ar(△ACQ) = ar(△APQ) (On the same base AQ and between the same parallel lines AQ and CP)
⇒ ar(△ACQ) – ar(△ABQ) = ar(△APQ) – ar(△ABQ)
⇒ ar(△ABC) = ar(△QBP) — (i)
AC and QP are diagonals ABCD and PBQR. Thus,
ar(ABC) = 1/2 ar(ABCD) — (ii)
ar(QBP) = 1/2 ar(PBQR) — (iii)
From (ii) and (ii),
1/2 ar(ABCD) = 1/2 ar(PBQR)
⇒ ar(ABCD) = ar(PBQR)

10. Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at O. Prove that ar (AOD) = ar (BOC).

△DAC and △DBC lie on the same base DC and between the same parallels AB and CD.
∴  ar(△DAC) = ar(△DBC)
⇒ ar(△DAC) − ar(△DOC) = ar(△DBC) − ar(△DOC)
⇒ ar(△AOD) = ar(△BOC)

11. In Fig. 9.27, ABCDE is a pentagon. A line through B parallel to AC meets DC produced at F.
Show that
(i) ar(ACB) = ar(ACF)
(ii) ar(AEDF) = ar(ABCDE)

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

∴ ar(△ACB) = ar(△ ACF)

(ii) ar(△ACB) = ar(△ACF)

⇒ ar(△ACB) + ar(△ACDE) = ar(△ACF) + ar(△ACDE)

⇒ ar(ABCDE) = ar(△AEDF)

Page No: 164

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.

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

Construction,

Diagonal BD is joined. AE is drawn parallel BD. BE is joined which intersected AD at O. △BCE is the shape of the original field and △AOB is the area for constructing health centre. Also, △DEO land joined to the plot.
To prove:
ar(△DEO) = ar(△AOB)
Proof:
△DEB and △DAB lie on the same base BD and between the same parallel lines BD and AE.
ar(△DEB) = ar(△DAB)
⇒ ar(△DEB) – ar△DOB) = ar(△DAB) – ar(△DOB)

⇒ ar(△DEO) = ar(△AOB)

13. ABCD is a trapezium with AB || DC. A line parallel to AC intersects AB at X and BC at Y. Prove that ar (ADX) = ar (ACY).

[Hint : Join CX.]

Given,

ABCD is a trapezium with AB || DC.

XY || AC

Construction,

CX is joined.

To Prove,

Proof:

ar(△ADX) = ar(△AXC) — (i) (On the same base AX and between the same parallels AB and CD)
also,

ar(△ AXC)=ar(△ ACY) — (ii) (On the same base AC and between the same parallels XY and AC.)

From (i) and (ii),

14. In Fig.9.28, AP || BQ || CR. Prove that
ar(AQC) = ar(PBR).

Given,

AP || BQ || CR

To Prove,

ar(AQC) = ar(PBR)

Proof:

ar(△AQB) = ar(△PBQ) — (i) (On the same base BQ and between the same parallels AP and BQ.)

also,

ar(△BQC) = ar(△BQR) — (ii) (On the same base BQ and between the same parallels BQ and CR.)

Adding (i) and (ii),
ar(△AQB) + ar(△BQC) = ar(△PBQ) + ar(△BQR)

⇒ ar(△ AQC) = ar(△ PBR)

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

Given,
ar(△AOD) = ar(△BOC)

To Prove,

ABCD is a trapezium.

Proof:

ar(△AOD) = ar(△BOC)

⇒ ar(△AOD) + ar(△AOB) = ar(△BOC) + ar(△AOB)

⇒ ar(△ADB) = ar(△ACB)

Areas of △ADB and △ACB are equal. Therefore, they must lying between the same parallel lines.
Thus, AB ∥  CD

Therefore, ABCD is a trapezium.

16. In Fig.9.29, ar(DRC) = ar(DPC) and ar(BDP) = ar(ARC). Show that both the quadrilaterals ABCD and DCPR are trapeziums.

Given,

ar(DRC) = ar(DPC) and ar(BDP) = ar(ARC)

To Prove,

ABCD and DCPR are trapeziums.

Proof:

ar(△BDP) = ar(△ARC)

⇒ ar(△BDP) – ar(△DPC) = ar(△DRC)

⇒ ar(△BDC) = ar(△ADC)

ar(△BDC) = ar(△ADC). Therefore, they must lying between the same parallel lines.

Thus, AB ∥ CD

Therefore, ABCD is a trapezium.

also,

ar(DRC) = ar(DPC). Therefore, they must lying between the same parallel lines.

Thus, DC ∥ PR

Therefore, DCPR is a trapezium.

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