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

## NCERT Solutions Class 9 Maths Chapter 9 โ Access Free PDF

*** According to the Revised CBSE Syllabus 2020-21, this chapter has been removed.**

**NCERT Solutions for Class 9 Maths Chapter 9 Areas of Parallelograms and Triangles** is a valuable resource from an exam point of view. It will help students to score well in the final exams. In order to help students, we have provided detailed chapter-wise solutions to clear their doubts and understanding of the concepts. The **NCERT** Solutions provided here contain a detailed explanation of all the problems mentioned under Chapter 9 Areas of Parallelograms and Triangles. The NCERT Solutions are created as per the latest syllabus and exam pattern guidelines of the CBSE Board.

Our subject experts have designed these **NCERT Solutions** for the benefit of students which covers all the exercise questions from Class 9 Maths NCERT textbook. Students should solve questions from these NCERT Solutions for Class 9 which will help them to prepare well for their exams. Along with the NCERT solutions students should also go through other study materials like sample papers, previous year papers, important questions, etc.Chapter 9 Areas of Parallelograms and Triangles

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**List of Exercises in Class 9 Maths Chapter 9**

Exercise 9.1 Solutions

### Access Answers to NCERT Class 9 Maths Chapter 9 โ Areas of Parallelograms and Triangles

## Exercise 9.1 Page: 155

**1. Which of the following figures lie on the same base and in-between the same parallels? In such a case, write the common base and the two parallels.**

Solution:

(i) Trapezium ABCD and ฮPDC lie on the same DC and in-between the same parallel lines AB and DC.

(ii) Parallelogram PQRS and trapezium SMNR lie on the same base SR but not in-between the same parallel lines.

(iii) Parallelogram PQRS and ฮRTQ lie on the same base QR and in-between the same parallel lines QR and PS.

(iv) Parallelogram ABCD and ฮPQR do not lie on the same base but in-between the same parallel lines BC and AD.

(v) Quadrilateral ABQD and trapezium APCD lie on the same base AD and in-between the same parallel lines AD and BQ.

(vi) Parallelogram PQRS and parallelogram ABCD do not lie on the same base SR but in-between the same parallel lines SR and PQ.

## Exercise 9.2 Page: 159

**1. In Fig. 9.15, ABCD is a parallelogram, AE โฅ DC and CF โฅ AD. If AB = 16 cm, AE = 8 cm and CF = 10 cm, find AD.**

Solution:

Given,

AB = CD = 16 cm (Opposite sides of a parallelogram)

CF = 10 cm and AE = 8 cm

Now,

Area of parallelogram = Base ร Altitude

= CDรAE = ADรCF

โ 16ร8 = ADร10

โ AD = 128/10 cm

โ AD = 12.8 cm

**2. If E, F, G and H are respectively the mid-points of the sides of a parallelogram ABCD, show that ar (EFGH) = 1/2 ar(ABCD).**

Solution:

Given,

E, F, G and H are the mid-points of the sides of a parallelogram ABCD respectively.

To Prove,

ar (EFGH) = ยฝ ar(ABCD)

Construction,

H and F are joined.

Proof,

AD || BC and AD = BC (Opposite sides of a parallelogram)

โ ยฝ AD = ยฝ BC

Also,

AH || BF and and DH || CF

โ AH = BF and DH = CF (H and F are mid points)

โด, ABFH and HFCD are parallelograms.

Now,

We know that, ฮEFH and parallelogram ABFH, both lie on the same FH the common base and in-between the same parallel lines AB and HF.

โด area of EFH = ยฝ area of ABFH โ (i)

And, area of GHF = ยฝ area of HFCD โ (ii)

Adding (i) and (ii),

area of ฮEFH + area of ฮGHF = ยฝ area of ABFH + ยฝ area of HFCD

โ area of EFGH = area of ABFH

โด ar (EFGH) = ยฝ ar(ABCD)

**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).**

Solution:

ฮAPB and parallelogram ABCD lie on the same base AB and in-between same parallel AB and DC.

ar(ฮAPB) = ยฝ ar(parallelogram ABCD) โ (i)

Similarly,

ar(ฮBQC) = ยฝ ar(parallelogram ABCD) โ (ii)

From (i) and (ii), we have

ar(ฮAPB) = ar(ฮBQC)

**4. In Fig. 9.16, P is a point in the interior of a parallelogram ABCD. Show that**

**(i) ar(APB) + ar(PCD) = ยฝ ar(ABCD)**

**(ii) ar(APD) + ar(PBC) = ar(APB) + ar(PCD)**

**[Hint : Through P, draw a line parallel to AB.]**

Solution:

(i) A line GH is drawn parallel to AB passing through P.

In a parallelogram,

AB || GH (by construction) โ (i)

โด,

AD || BC โ AG || BH โ (ii)

From equations (i) and (ii),

ABHG is a parallelogram.

Now,

ฮAPB and parallelogram ABHG are lying on the same base AB and in-between the same parallel lines AB and GH.

โด ar(ฮAPB) = ยฝ ar(ABHG) โ (iii)

also,

ฮPCD and parallelogram CDGH are lying on the same base CD and in-between the same parallel lines CD and GH.

โด ar(ฮPCD) = ยฝ ar(CDGH) โ (iv)

Adding equations (iii) and (iv),

ar(ฮAPB) + ar(ฮPCD) = ยฝ [ar(ABHG)+ar(CDGH)]

โ ar(APB)+ ar(PCD) = ยฝ ar(ABCD)

(ii) A line EF is drawn parallel to AD passing through P.

In the parallelogram,

AD || EF (by construction) โ (i)

โด,

AB || CD โ AE || DF โ (ii)

From equations (i) and (ii),

AEDF is a parallelogram.

Now,

ฮAPD and parallelogram AEFD are lying on the same base AD and in-between the same parallel lines AD and EF.

โดar(ฮAPD) = ยฝ ar(AEFD) โ (iii)

also,

ฮPBC and parallelogram BCFE are lying on the same base BC and in-between the same parallel lines BC and EF.

โดar(ฮPBC) = ยฝ ar(BCFE) โ (iv)

Adding equations (iii) and (iv),

ar(ฮAPD)+ ar(ฮPBC) = ยฝ {ar(AEFD)+ar(BCFE)}

โar(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) = ยฝ ar (PQRS) **

Solution:

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

โด ar(PQRS) = ar(ABRS) โ (i)

(ii) ฮAXS and parallelogram ABRS are lying on the same base AS and in-between the same parallel lines AS and BR.

โด ar(ฮAXS) = ยฝ ar(ABRS) โ (ii)

From (i) and (ii), we find that,

ar(ฮAXS) = ยฝ ar(PQRS)

**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?**

Solution:

The field is divided into three parts each in triangular shape.

Let, ฮPSA, ฮPAQ and ฮQAR be the triangles.

Area of (ฮPSA + ฮPAQ + ฮQAR) = Area of PQRS โ (i)

Area of ฮPAQ = ยฝ area of PQRS โ (ii)

Here, the triangle and parallelogram are on the same base and in-between the same parallel lines.

From (i) and (ii),

Area of ฮPSA +Area of ฮQAR = ยฝ area of PQRS โ (iii)

From (ii) and (iii), we can conclude that,

The farmer must sow wheat or pulses in ฮPAQ or either in both ฮPSA and ฮQAR.

## Exercise 9.3 Page: 162

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

Solution:

Given,

AD is median of ฮABC. โด, it will divide ฮABC into two triangles of equal area.

โดar(ABD) = ar(ACD) โ (i)

also,

ED is the median of ฮABC.

โดar(EBD) = ar(ECD) โ (ii)

Subtracting (ii) from (i),

ar(ABD) โ ar(EBD) = ar(ACD) โ ar(ECD)

โar(ABE) = ar(ACE)

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

Solution:

ar(BED) = (1/2)รBDรDE

Since, E is the mid-point of AD,

AE = DE

Since, AD is the median on side BC of triangle ABC,

BD = DC

,

DE = (1/2) AD โ (i)

BD = (1/2)BC โ (ii)

From (i) and (ii), we get,

ar(BED) = (1/2)ร(1/2)BC ร (1/2)AD

โ ar(BED) = (1/2)ร(1/2)ar(ABC)

โ ar(BED) = ยผ ar(ABC)

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

Solution:

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

In ฮABC, BO is the median.

โดar(AOB) = ar(BOC) โ (i)

also,

In ฮBCD, CO is the median.

โดar(BOC) = ar(COD) โ (ii)

In ฮACD, OD is the median.

โดar(AOD) = ar(COD) โ (iii)

In ฮABD, AO is the median.

โดar(AOD) = ar(AOB) โ (iv)

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

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

Hence, we get, 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).**

Solution:

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

โดar(AOC) = ar(AOD) โ (i)

also,

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

โดar(BOC) = ar(BOD) โ (ii)

Adding (i) and (ii),

We get,

ar(AOC)+ar(BOC) = ar(AOD)+ar(BOD)

โar(ABC) = ar(ABD)

**5. D, E and F are respectively the mid-points of the sides BC, CA and AB of a ฮABC.Show that**

**(i) BDEF is a parallelogram. **

**(ii) ar(DEF) = ยผ ar(ABC)**

**(iii) ar (BDEF) = ยฝ ar(ABC)**

Solution:

(i) In ฮABC,

EF || BC and EF = ยฝ BC (by mid point theorem)

also,

BD = ยฝ BC (D is the mid point)

So, BD = EF

also,

BF and DE are parallel and equal to each other.

โด, the pair opposite sides are equal in length and parallel to each other.

โด BDEF is a parallelogram.

(ii) Proceeding from the result of (i),

BDEF, DCEF, AFDE are parallelograms.

Diagonal of a parallelogram divides it into two triangles of equal area.

โดar(ฮBFD) = ar(ฮDEF) (For parallelogram BDEF) โ (i)

also,

ar(ฮAFE) = ar(ฮDEF) (For parallelogram DCEF) โ (ii)

ar(ฮCDE) = ar(ฮDEF) (For parallelogram AFDE) โ (iii)

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

ar(ฮBFD) = ar(ฮAFE) = ar(ฮCDE) = ar(ฮDEF)

โ ar(ฮBFD) +ar(ฮAFE) +ar(ฮCDE) +ar(ฮDEF) = ar(ฮABC)

โ 4 ar(ฮDEF) = ar(ฮABC)

โ ar(DEF) = ยผ ar(ABC)

(iii) Area (parallelogram BDEF) = ar(ฮDEF) +ar(ฮBDE)

โ ar(parallelogram BDEF) = ar(ฮDEF) +ar(ฮDEF)

โ ar(parallelogram BDEF) = 2ร ar(ฮDEF)

โ ar(parallelogram BDEF) = 2ร ยผ ar(ฮABC)

โ ar(parallelogram BDEF) = ยฝ ar(ฮABC)

**6. In Fig. 9.25, diagonals AC and BD of quadrilateral ABCD intersect at O such that OB = OD.If AB = CD, then show that:**

**(i) ar (DOC) = ar (AOB)**

**(ii) ar (DCB) = ar (ACB)**

**(iii) DA || CB or ABCD is a parallelogram.[Hint : From D and B, draw perpendiculars to AC.]**

Solution:

Given,

OB = OD and AB = CD

Construction,

DE โฅ AC and BF โฅ AC are drawn.

Proof:

(i) In ฮDOE and ฮBOF,

โ DEO = โ BFO (Perpendiculars)

โ DOE = โ BOF (Vertically opposite angles)

OD = OB (Given)

โด, ฮDOE โ ฮBOF by AAS congruence condition.

โด, DE = BF (By CPCT) โ (i)

also, ar(ฮDOE) = ar(ฮBOF) (Congruent triangles) โ (ii)

Now,

In ฮDEC and ฮBFA,

โ DEC = โ BFA (Perpendiculars)

CD = AB (Given)

DE = BF (From i)

โด, ฮDEC โ ฮBFA by RHS congruence condition.

โด, ar(ฮDEC) = ar(ฮBFA) (Congruent triangles) โ (iii)

Adding (ii) and (iii),

ar(ฮDOE) + ar(ฮDEC) = ar(ฮBOF) + ar(ฮBFA)

โ ar (DOC) = ar (AOB)

(ii) ar(ฮDOC) = ar(ฮAOB)

Adding ar(ฮOCB) in LHS and RHS, we get,

โar(ฮDOC) + ar(ฮOCB) = ar(ฮAOB) + ar(ฮOCB)

โ ar(ฮDCB) = ar(ฮACB)

(iii) When two triangles have same base and equal areas, the triangles will be in between the same parallel lines

ar(ฮDCB) = ar(ฮACB)

DA || BC โ (iv)

For quadrilateral ABCD, one pair of opposite sides are equal (AB = CD) and other pair of opposite sides are parallel.

โด, ABCD is parallelogram.

**7. D and E are points on sides AB and AC respectively of ฮABC such that ar(DBC) = ar(EBC). Prove that DE || BC.**

Solution:

ฮDBC and ฮEBC are on the same base BC and also having equal areas.

โด, they will lie between the same parallel lines.

โด, DE || BC.

**8. XY is a line parallel to side BC of a triangle ABC. If BE || AC and CF || AB meet XY at E and F respectively, show thatar(ฮABE) = ar(ฮACF)**

Solution:

Given,

XY || BC, BE || AC and CF || AB

To show,

ar(ฮABE) = ar(ฮAC)

Proof:

BCYE is a || gm as ฮABE and ||gm BCYE are on the same base BE and between the same parallel lines BE and AC.

โด,ar(ABE) = ยฝ ar(BCYE) โฆ (1)

Now,

CF || AB and XY || BC

โ CF || AB and XF || BC

โ BCFX is a || gm

As ฮACF and || gm BCFX are on the same base CF and in-between the same parallel AB and FC .

โด,ar (ฮACF)= ยฝ ar (BCFX) โฆ (2)

But,

||gm BCFX and || gm BCYE are on the same base BC and between the same parallels BC and EF.

โด,ar (BCFX) = ar(BCYE) โฆ (3)

From (1) , (2) and (3) , we get

ar (ฮABE) = ar(ฮACF)

โ ar(BEYC) = ar(BXFC)

As the parallelograms are on the same base BC and in-between the same parallels EF and BCโ(iii)

Also,

โณAEB and ||gm BEYC are on the same base BE and in-between the same parallels BE and AC.

โ ar(โณAEB) = ยฝ ar(BEYC) โ (iv)

Similarly,

โณACF and || gm BXFC on the same base CF and between the same parallels CF and AB.

โ ar(โณ ACF) = ยฝ ar(BXFC) โ (v)

From (iii), (iv) and (v),

ar(โณABE) = ar(โณACF)

**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 thatar(ABCD) = ar(PBQR).[Hint : Join AC and PQ. Now compare ar(ACQ) and ar(APQ).]**

Solution:

AC and PQ are joined.

Ar(โณACQ) = ar(โณAPQ) (On the same base AQ and between the same parallel lines AQ and CP)

โ ar(โณACQ)-ar(โณABQ) = ar(โณAPQ)-ar(โณABQ)

โ ar(โณABC) = ar(โณQBP) โ (i)

AC and QP are diagonals ABCD and PBQR.

โด,ar(ABC) = ยฝ ar(ABCD) โ (ii)

ar(QBP) = ยฝ ar(PBQR) โ (iii)

From (ii) and (ii),

ยฝ ar(ABCD) = ยฝ ar(PBQR)

โ 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).**

Solution:

โณDAC and โณDBC lie on the same base DC and between the same parallels AB and CD.

Ar(โณDAC) = ar(โณDBC)

โ ar(โณDAC) โ ar(โณDOC) = ar(โณDBC) โ ar(โณDOC)

โ ar(โณAOD) = ar(โณBOC)

**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(โณACB) = ar(โณACF)**

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

Solution:

- โณACB and โณACF lie on the same base AC and between the same parallels AC and BF.

โดar(โณACB) = ar(โณ ACF)

- ar(โณACB) = ar(โณACF)

โ ar(โณACB)+ar(โณACDE) = ar(โณACF)+ar(โณACDE)

โ ar(ABCDE) = ar(โณAEDF)

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

To Construct,

Join the diagonal BD.

Draw AE parallel to BD.

Join BE, that intersected AD at O.

We get,

โณBCE is the shape of the original field

โณAOB is the area for constructing health centre.

โณDEO is the land joined to the plot.

To prove:

ar(โณDEO) = ar(โณAOB)

Proof:

โณDEB and โณDAB lie on the same base BD, in-between the same parallels BD and AE.

Ar(โณDEB) = ar(โณDAB)

โar(โณDEB) โ arโณDOB) = ar(โณDAB) โ ar(โณDOB)

โ ar(โณDEO) = ar(โณAOB)

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

**[Hint : Join CX.]**

Solution:

Given,

ABCD is a trapezium with AB || DC.

XY || AC

Construction,

Join CX

To Prove,

ar(ADX) = ar(ACY)

Proof:

ar(โณADX) = ar(โณAXC) โ (i) (Since they are on the same base AX and in-between the same parallels AB and CD)

also,

ar(โณ AXC)=ar(โณ ACY) โ (ii) (Since they are on the same base AC and in-between the same parallels XY and AC.)

(i) and (ii),

ar(โณADX) = ar(โณACY)

**14. In Fig.9.28, AP || BQ || CR. Prove that ar(โณAQC) = ar(โณPBR).**

Solution:

Given,

AP || BQ || CR

To Prove,

ar(AQC) = ar(PBR)

Proof:

ar(โณAQB) = ar(โณPBQ) โ (i) (Since they are on the same base BQ and between the same parallels AP and BQ.)

also,

ar(โณBQC) = ar(โณBQR) โ (ii) (Since they are on the same base BQ and between the same parallels BQ and CR.)

Adding (i) and (ii),

ar(โณAQB)+ar(โณBQC) = ar(โณPBQ)+ar(โณBQR)

โ ar(โณ AQC) = ar(โณ PBR)

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

Solution:

Given,

ar(โณAOD) = ar(โณBOC)

To Prove,

ABCD is a trapezium.

Proof:

ar(โณAOD) = ar(โณBOC)

โ ar(โณAOD) + ar(โณAOB) = ar(โณBOC)+ar(โณAOB)

โ ar(โณADB) = ar(โณACB)

Areas of โณADB and โณACB are equal. โด, they must lying between the same parallel lines.

โด, AB โฅ CD

โด, 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.**

Solution:

Given,

ar(โณDRC) = ar(โณDPC)

ar(โณBDP) = ar(โณARC)

To Prove,

ABCD and DCPR are trapeziums.

Proof:

ar(โณBDP) = ar(โณARC)

โ ar(โณBDP) โ ar(โณDPC) = ar(โณDRC)

โ ar(โณBDC) = ar(โณADC)

ar(โณBDC) = ar(โณADC).

โด, ar(โณBDC) and ar(โณADC) are lying in-between the same parallel lines.

โด, AB โฅ CD

ABCD is a trapezium.

Similarly,

ar(โณDRC) = ar(โณDPC).

โด, ar(โณDRC) andar(โณDPC) are lying in-between the same parallel lines.

โด, DC โฅ PR

โด, DCPR is a trapezium.

## Exercise 9.4(Optional)* Page: 164

**1. Parallelogram ABCD and rectangle ABEF are on the same base AB and have equal areas. Show that the perimeter of the parallelogram is greater than that of the rectangle.**

Solution:

Given,

|| gm ABCD and a rectangle ABEF have the same base AB and equal areas.

To prove,

Perimeter of || gm ABCD is greater than the perimeter of rectangle ABEF.

Proof,

We know that, the opposite sides of a|| gm and rectangle are equal.

, AB = DC [As ABCD is a || gm]

and, AB = EF [As ABEF is a rectangle]

, DC = EF โฆ (i)

Adding AB on both sides, we get,

โAB + DC = AB + EF โฆ (ii)

We know that, the perpendicular segment is the shortest of all the segments that can be drawn to a given line from a point not lying on it.

, BE < BC and AF < AD

โ BC > BE and AD > AF

โ BC+AD > BE+AF โฆ (iii)

Adding (ii) and (iii), we get

AB+DC+BC+AD > AB+EF+BE+AF

โ AB+BC+CD+DA > AB+ BE+EF+FA

โ perimeter of || gm ABCD > perimeter of rectangle ABEF.

, the perimeter of the parallelogram is greater than that of the rectangle.

Hence Proved.

**2. In Fig. 9.30, D and E are two points on BC such that BD = DE = EC.**

**Show that ar (ABD) = ar (ADE) = ar (AEC).**

**Can you now answer the question that you have left in the โIntroductionโ of this chapter, whether the field of Budhia has been actually divided into three parts of equal area?**

**[Remark: Note that by taking BD = DE = EC, the triangle ABC is divided into three triangles ABD, ADE and AEC of equal areas. In the same way, by dividing BC into n equal parts and joining the points of division so obtained to the opposite vertex of BC, you can divide DABC into n triangles of equal areas.]**

Solution:

Given,

BD = DE = EC

To prove,

ar (โณABD) = ar (โณADE) = ar (โณAEC)

Proof,

In (โณABE), AD is median [since, BD = DE, given]

We know that, the median of a triangle divides it into two parts of equal areas

, ar(โณABD) = ar(โณAED) โ(i)

Similarly,

In (โณADC), AE is median [since, DE = EC, given]

,ar(ADE) = ar(AEC) โ(ii)

From the equation (i) and (ii), we get

ar(ABD) = ar(ADE) = ar(AEC)

**3. In Fig. 9.31, ABCD, DCFE and ABFE are parallelograms. Show that ar (ADE) = ar (BCF).**

Solution:

Given,

ABCD, DCFE and ABFE are parallelograms

To prove,

ar (โณADE) = ar (โณBCF)

Proof,

In โณADE and โณBCF,

AD = BC [Since, they are the opposite sides of the parallelogram ABCD]

DE = CF [Since, they are the opposite sides of the parallelogram DCFE]

AE = BF [Since, they are the opposite sides of the parallelogram ABFE]

, โณADE โ โณBCF [Using SSS Congruence theorem]

, ar(โณADE) = ar(โณBCF) [ By CPCT]

**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.]**

Solution:

Given:

ABCD is a parallelogram

AD = CQ

To prove:

ar (โณBPC) = ar (โณDPQ)

Proof:

In โณADP and โณQCP,

โ APD = โ QPC [Vertically Opposite Angles]

โ ADP = โ QCP [Alternate Angles]

AD = CQ [given]

, โณABO โ โณACD [AAS congruency]

, DP = CP [CPCT]

In โณCDQ, QP is median. [Since, DP = CP]

Since, median of a triangle divides it into two parts of equal areas.

, ar(โณDPQ) = ar(โณQPC) โ(i)

In โณPBQ, PC is median. [Since, AD = CQ and AD = BC โ BC = QC]

Since, median of a triangle divides it into two parts of equal areas.

, ar(โณQPC) = ar(โณBPC) โ(ii)

From the equation (i) and (ii), we get

ar(โณBPC) = ar(โณDPQ)

**5. In Fig.9.33, ABC and BDE are two equilateral triangles such that D is the mid-point of BC. If AE intersects BC at F, show that:**

**(i) ar (BDE) =1/4 ar (ABC)**

**(ii) ar (BDE) = ยฝ ar (BAE)**

**(iii) ar (ABC) = 2 ar (BEC)**

**(iv) ar (BFE) = ar (AFD)**

**(v) ar (BFE) = 2 ar (FED)**

**(vi) ar (FED) = 1/8 ar (AFC)**

Solution:

(i) Assume that G and H are the mid-points of the sides AB and AC respectively.

Join the mid-points with line-segment GH. Here, GH is parallel to third side.

, BC will be half of the length of BC by mid-point theorem.

โด GH =1/2 BC and GH || BD

โด GH = BD = DC and GH || BD (Since, D is the mid-point of BC)

Similarly,

GD = HC = HA

HD = AG = BG

, ฮABC is divided into 4 equal equilateral triangles ฮBGD, ฮAGH, ฮDHC and ฮGHD

We can say that,

ฮBGD = ยผ ฮABC

Considering, ฮBDG and ฮBDE

BD = BD (Common base)

Since both triangles are equilateral triangle, we can say that,

BG = BE

DG = DE

, ฮBDG ฮBDE [By SSS congruency]

, area (ฮBDG) = area (ฮBDE)

ar (ฮBDE) = ยผ ar (ฮABC)

Hence proved

(ii)

ar(ฮBDE) = ar(ฮAED) (Common base DE and DE||AB)

ar(ฮBDE)โar(ฮFED) = ar(ฮAED)โar (ฮFED)

ar(ฮBEF) = ar(ฮAFD) โฆ(i)

Now,

ar(ฮABD) = ar(ฮABF)+ar(ฮAFD)

ar(ฮABD) = ar(ฮABF)+ar(ฮBEF) [From equation (i)]

ar(ฮABD) = ar(ฮABE) โฆ(ii)

AD is the median of ฮABC.

ar(ฮABD) = ยฝ ar (ฮABC)

= (4/2) ar (ฮBDE)

= 2 ar (ฮBDE)โฆ(iii)

From (ii) and (iii), we obtain

2 ar (ฮBDE) = ar (ฮABE)

ar (BDE) = ยฝ ar (BAE)

Hence proved

(iii) ar(ฮABE) = ar(ฮBEC) [Common base BE and BE || AC]

ar(ฮABF) + ar(ฮBEF) = ar(ฮBEC)

From eq^{n} (i), we get,

ar(ฮABF) + ar(ฮAFD) = ar(ฮBEC)

ar(ฮABD) = ar(ฮBEC)

ยฝ ar(ฮABC) = ar(ฮBEC)

ar(ฮABC) = 2 ar(ฮBEC)

Hence proved

(iv) ฮBDE and ฮAED lie on the same base (DE) and are in-between the parallel lines DE and AB.

โดar (ฮBDE) = ar (ฮAED)

Subtracting ar(ฮFED) from L.H.S and R.H.S,

We get,

โดar (ฮBDE)โar (ฮFED) = ar (ฮAED)โar (ฮFED)

โดar (ฮBFE) = ar(ฮAFD)

Hence proved

(v) Assume that h is the height of vertex E, corresponding to the side BD in ฮBDE.

Also assume that H is the height of vertex A, corresponding to the side BC in ฮABC.

While solving Question (i),

We saw that,

ar (ฮBDE) = ยผ ar (ฮABC)

While solving Question (iv),

We saw that,

ar (ฮBFE) = ar (ฮAFD).

โดar (ฮBFE) = ar (ฮAFD)

= 2 ar (ฮFED)

Hence, ar (ฮBFE) = 2 ar (ฮFED)

Hence proved

(vi) ar (ฮAFC) = ar (ฮAFD) + ar(ฮADC)

= 2 ar (ฮFED) + (1/2) ar(ฮABC) [using (v)

= 2 ar (ฮFED) + ยฝ [4ar(ฮBDE)] [Using result of Question (i)]

= 2 ar (ฮFED) +2 ar(ฮBDE)

Since, ฮBDE and ฮAED are on the same base and between same parallels

= 2 ar (ฮFED) +2 ar (ฮAED)

= 2 ar (ฮFED) +2 [ar (ฮAFD) +ar (ฮFED)]

= 2 ar (ฮFED) +2 ar (ฮAFD) +2 ar (ฮFED) [From question (viii)]

= 4 ar (ฮFED) +4 ar (ฮFED)

โar (ฮAFC) = 8 ar (ฮFED)

โar (ฮFED) = (1/8) ar (ฮAFC)

Hence proved

**6. Diagonals AC and BD of a quadrilateral ABCD intersect each other at P. Show that**

**ar (APB)รar (CPD) = ar (APD)รar (BPC).**

**[Hint : From A and C, draw perpendiculars to BD.]**

Solution:

Given:

The diagonal AC and BD of the quadrilateral ABCD, intersect each other at point E.

Construction:

From A, draw AM perpendicular to BD

From C, draw CN perpendicular to BD

To Prove,

ar(ฮAED) ar(ฮBEC) = ar (ฮABE) รar (ฮCDE)

Proof,

ar(ฮABE) = ยฝ รBEรAMโฆโฆโฆโฆ.. (i)

ar(ฮAED) = ยฝ รDEรAMโฆโฆโฆโฆ.. (ii)

Dividing eq. ii by i , we get,

ar(AED)/ar(ABE) = DE/BEโฆโฆ.. (iii)

Similarly,

ar(CDE)/ar(BEC) = DE/BE โฆโฆ. (iv)

From eq. (iii) and (iv) , we get

ar(AED)/ar(ABE) = ar(CDE)/ar(BEC)

, ar(ฮAED)รar(ฮBEC) = ar(ฮABE)รar (ฮCDE)

Hence proved.

**7.** **P and Q are respectively the mid-points of sides AB and BC of a triangle ABC and R is the mid-point of AP, show that:**

**(i) ar (PRQ) = ยฝ ar (ARC)**

**(ii) ar (RQC) = (3/8) ar (ABC)**

**(iii) ar (PBQ) = ar (ARC)**

Solution:

(i)

We know that, median divides the triangle into two triangles of equal area,

PC is the median of ABC.

Ar (ฮBPC) = ar (ฮAPC) โฆโฆโฆ.(i)

RC is the median of APC.

Ar (ฮARC) = ยฝ ar (ฮAPC) โฆโฆโฆ.(ii)

PQ is the median of BPC.

Ar (ฮPQC) = ยฝ ar (ฮBPC) โฆโฆโฆ.(iii)

From eq. (i) and (iii), we get,

ar (ฮPQC) = ยฝ ar (ฮAPC) โฆโฆโฆ.(iv)

From eq. (ii) and (iv), we get,

ar (ฮPQC) = ar (ฮARC) โฆโฆโฆ.(v)

P and Q are the mid-points of AB and BC respectively [given]

PQ||AC

and, PA = ยฝ AC

Since, triangles between same parallel are equal in area, we get,

ar (ฮAPQ) = ar (ฮPQC) โฆโฆโฆ.(vi)

From eq. (v) and (vi), we obtain,

ar (ฮAPQ) = ar (ฮARC) โฆโฆโฆ.(vii)

R is the mid-point of AP.

, RQ is the median of APQ.

Ar (ฮPRQ) = ยฝ ar (ฮAPQ) โฆโฆโฆ.(viii)

From (vii) and (viii), we get,

ar (ฮPRQ) = ยฝ ar (ฮARC)

Hence Proved.

(ii) PQ is the median of ฮBPC

ar (ฮPQC) = ยฝ ar (ฮBPC)

= (ยฝ) ร(1/2 )ar (ฮABC)

= ยผ ar (ฮABC) โฆโฆโฆ.(ix)

Also,

ar (ฮPRC) = ยฝ ar (ฮAPC) [From (iv)]

ar (ฮPRC) = (1/2)ร(1/2)ar ( ABC)

= ยผ ar(ฮABC) โฆโฆโฆ.(x)

Add eq. (ix) and (x), we get,

ar (ฮPQC) + ar (ฮPRC) = (1/4)ร(1/4)ar (ฮABC)

ar (quad. PQCR) = ยผ ar (ฮABC) โฆโฆโฆ.(xi)

Subtracting ar (ฮPRQ) from L.H.S and R.H.S,

ar (quad. PQCR)โar (ฮPRQ) = ยฝ ar (ฮABC)โar (ฮPRQ)

ar (ฮRQC) = ยฝ ar (ฮABC) โ ยฝ ar (ฮARC) [From result (i)]

ar (ฮARC) = ยฝ ar (ฮABC) โ(1/2)ร(1/2)ar (ฮAPC)

ar (ฮRQC) = ยฝ ar (ฮABC) โ(1/4)ar (ฮAPC)

ar (ฮRQC) = ยฝ ar (ฮABC) โ(1/4)ร(1/2)ar (ฮABC) [ As, PC is median of ฮABC]

ar (ฮRQC) = ยฝ ar (ฮABC)โ(1/8)ar (ฮABC)

ar (ฮRQC) = [(1/2)-(1/8)]ar (ฮABC)

ar (ฮRQC) = (3/8)ar (ฮABC)

(iii) ar (ฮPRQ) = ยฝ ar (ฮARC) [From result (i)]

2ar (ฮPRQ) = ar (ฮARC) โฆโฆโฆโฆโฆ..(xii)

ar (ฮPRQ) = ยฝ ar (ฮAPQ) [RQ is the median of APQ] โฆโฆโฆ.(xiii)

But, we know that,

ar (ฮAPQ) = ar (ฮPQC) [From the reason mentioned in eq. (vi)] โฆโฆโฆ.(xiv)

From eq. (xiii) and (xiv), we get,

ar (ฮPRQ) = ยฝ ar (ฮPQC) โฆโฆโฆ.(xv)

At the same time,

ar (ฮBPQ) = ar (ฮPQC) [PQ is the median of ฮBPC] โฆโฆโฆ.(xvi)

From eq. (xv) and (xvi), we get,

ar (ฮPRQ) = ยฝ ar (ฮBPQ) โฆโฆโฆ.(xvii)

From eq. (xii) and (xvii), we get,

2ร(1/2)ar(ฮBPQ)= ar (ฮARC)

โน ar (ฮBPQ) = ar (ฮARC)

Hence Proved.

**8. In Fig. 9.34, ABC is a right triangle right angled at A. BCED, ACFG and ABMN are squares on the sides BC, CA and AB respectively. Line segment AX ^ DE meets BC at Y. Show that:**

**(i) ฮMBC โ
ฮABD**

**(ii) ar(BYXD) = 2ar(MBC)**

**(iii) ar(BYXD) = ar(ABMN)**

**(iv) ฮFCB โ
ฮACE**

**(v) ar(CYXE) = 2ar(FCB)**

**(vi) ar(CYXE) = ar(ACFG)**

**(vii) ar(BCED) = ar(ABMN)+ar(ACFG)**

**Note : Result (vii) is the famous Theorem of Pythagoras. You shall learn a simpler proof of this theorem in Class X.**

Solution:

(i) We know that each angle of a square is 90ยฐ. Hence, โ ABM = โ DBC = 90ยบ

โดโ ABM+โ ABC = โ DBC+โ ABC

โดโ MBC = โ ABD

In โMBC and โABD,

โ MBC = โ ABD (Proved above)

MB = AB (Sides of square ABMN)

BC = BD (Sides of square BCED)

โด โMBC โ โABD (SAS congruency)

(ii) We have

โMBC โ โABD

โดar (โMBC) = ar (โABD) โฆ (i)

It is given that AX โฅ DE and BD โฅ DE (Adjacent sides of square BDEC)

โด BD || AX (Two lines perpendicular to same line are parallel to each other)

โABD and parallelogram BYXD are on the same base BD and between the same parallels BD and AX.

Area (โYXD) = 2 Area (โMBC) [From equation (i)] โฆ (ii)

(iii) โMBC and parallelogram ABMN are lying on the same base MB and between same parallels MB and NC.

2 ar (โMBC) = ar (ABMN)

ar (โYXD) = ar (ABMN) [From equation (ii)] โฆ (iii)

(iv) We know that each angle of a square is 90ยฐ.

โดโ FCA = โ BCE = 90ยบ

โดโ FCA+โ ACB = โ BCE+โ ACB

โดโ FCB = โ ACE

In โFCB and โACE,

โ FCB = โ ACE

FC = AC (Sides of square ACFG)

CB = CE (Sides of square BCED)

โFCB โ โACE (SAS congruency)

(v) AX โฅ DE and CE โฅ DE (Adjacent sides of square BDEC) [given]

Hence,

CE || AX (Two lines perpendicular to the same line are parallel to each other)

Consider BACE and parallelogram CYXE

BACE and parallelogram CYXE are on the same base CE and between the same parallels CE and AX.

โดar (โYXE) = 2ar (โACE) โฆ (iv)

We had proved that

โด โFCB โ โACE

ar (โFCB) โ ar (โACE) โฆ (v)

From equations (iv) and (v), we get

ar (CYXE) = 2 ar (โFCB) โฆ (vi)

(vi) Consider BFCB and parallelogram ACFG

BFCB and parallelogram ACFG are lying on the same base CF and between the same parallels CF and BG.

โดar (ACFG) = 2 ar (โFCB)

โดar (ACFG) = ar (CYXE) [From equation (vi)] โฆ (vii)

(vii) From the figure, we can observe that

ar (โCED) = ar (โYXD)+ar (CYXE)

โดar (โCED) = ar (ABMN)+ar (ACFG) [From equations (iii) and (vii)].

Chapter 9 of Class 9 Maths โAreas of Parallelograms and Trianglesโ comes under the unit Menstruation which in total carries 14 marks. From the unit Menstruation there will be 2 mutliple choice questions of 2 marks, 2 short types questions of 6 marks each and 1 long type question of 6 marks which in total makes it 5.

Chapter 9 โ Areas of Parallelograms and Triangles covers topics such as Areas of Parallelograms and Triangles, Figures On The Same Base And Between The Same Parallels as well as Triangles On The Same Base And Between The Same Parallels.

The ninth chapter of Class 9 Maths talks about Areas of Parallelograms and Triangles which introduces students to the basic concepts of Areas of Parallelograms and Triangles, Figures On The Same Base And Between The Same Parallels as well as Triangles On The Same Base And Between The Same Parallels. For Class 9 students these topics are pretty complex and require proper explanations to master. So, the NCERT Solutions for Class 9 Maths Chapter 9 Areas of parallelograms and Triangles exercises will help you to get an idea of the topic and the concepts it covered.

Maths is incredibly important in our daily lives as we are all surrounded by the laws of mathematics and without a good understanding of them, one can face significant problems in life. It is a subject that needs lots of practice and knowledge of different formulas to solve different numerical problems. Class 9 students can now upgrade their maths problem solving skills by practising questions from NCERT Solutions for Class 9 Maths which is provided here. All the questions have been picked from the NCERT textbook of Class 9 Maths so that students can score well in their exam.

We at BYJUโS provide you with the solutions of all the exercises from the NCERT textbook as per the prescribed syllabus, created by our subject experts for the benefit of Class 9 students while preparing for their exam. Thus, by reading our NCERT Solutions for Class 9 Maths Chapter 9 you can easily form a clear understanding of these topics by practising them at your own pace.

Expert tutors have formulated the solutions in a lucid manner to improve the problem solving abilities among the students. For a more clear idea about Areas of Parallelograms and Triangles students can refer to the study materials available at BYJUโS.

## Frequently Asked Questions on NCERT Solutions for Class 9 Maths Chapter 9

### What kind of questions can be expected for exams from NCERT Solutions for Class 9 Maths Chapter 9?

NCERT Solutions for Class 9 Maths Chapter 9 Areas of Parallelogram and Triangles carries 14 marks. From the unit there will be 2 multiple choice questions of 2 marks, 2 short types questions of 6 marks each and 1 long type question of 6 marks which in total makes it 5.

### Is it necessary to learn all the questions provided in NCERT Solutions for Class 9 Maths Chapter 9?

Yes, to attempt all types of questions which come in exams it is necessary to attempt all the questions provided in NCERT Solutions for Class 9 Maths Chapter 9. These solutions are provided in student โ friendly language to help you in solving difficult problems in an easier way. These solutions are formulated by experts at BYJUโS having good knowledge of Maths.

### Is NCERT Solutions for Class 9 Maths Chapter 9 Areas of Parallelogram and Triangles important from an exam perspective?

Yes, all 15 of NCERT Solutions for Class 9 Maths are important from an exam perspective. 9th chapter of NCERT Solutions for Class 9 Maths explains areas of parallelogram and triangles. These solutions are available in free PDF format for easy access. Students can view and download these solutions offline as well online from the BYJUโS website.