NCERT Solutions Class 9 Maths Chapter 9 Areas of Parallelograms and Triangles: Exercise 3

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

Solution:

Given:

AD is median of ΔABC. 

Therefore, 

It will divide ΔABC into two triangles of equal area.

Therefore,

ar(ABD) = ar(ACD)          -(equation 1)

also,

ED is the median of ΔABC.

Therefore,

ar(EBD) = ar(ECD)          -(equation 2)

Subtracting equation (2) from (1),

ar(ABD) – ar(EBD) = ar(ACD) – ar(ECD)

=> ar(ABE) = ar(ACE)

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

Solution:

ar(BED) = (1/2)×BD×DE

Thus, 

E is the mid-point of AD,

AE = DE

Thus, 

AD is the median on side BC of triangle ABC,

BD = DC

DE = (1/2) AD          -(equation 1)

BD = (1/2)BC          -(equation-2)

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

ar(BED) = (1/2) × (1/2)BC × (1/2)AD

=> ar(BED) = (1/2) × (1/2)ar(ABC)

=> ar(BED) = 1/4 ar(ABC)

Question 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. (as diagonals bisect each other)

In ΔABC, BO is the median.

Therefore,

ar(AOB) = ar(BOC)          -(equation 1)

also,

In ΔBCD, CO is the median.

Therefore,

ar(BOC) = ar(COD)          -(equation 2)

In ΔACD, OD is the median.

Therefore,

ar(AOD) = ar(COD)          -(equation 3)

In ΔABD, AO is the median.

Therefore,

ar(AOD) = ar(AOB).         -(equation4)

From equations (1), (2), (3) and (4), we have,

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

Therefore, 

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

Question 4. In Figure, 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 ΔABC, AO is the median. (CD is bisected by AB at O)

Therefore,

ar(AOC) = ar(AOD)          -(equation 1)

also,

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

Therefore,

ar(BOC) = ar(BOD)          -(equation 2)

Adding equation (1) and (2),

We will get,

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

=> ar(ABC) = ar(ABD)

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

(i) BDEF is a parallelogram.        

(ii) ar(DEF) = 1/4 ar(ABC)

(iii) ar (BDEF) = 1/2 ar(ABC)

Solution:

(i) In ΔABC,

EF || BC and EF = 1/2 BC (by mid point theorem)

also,

BD = 1/2 BC (D is the mid point)

So, 

BD = EF

also,

BF and DE are parallel and equal to each other.

Therefore, 

The pair opposite sides are equal in length and parallel to each other.

Hence, 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.

Therefore,

ar(ΔBFD) = ar(ΔDEF) (For parallelogram BDEF)          -(equation 1)

also,

ar(ΔAFE) = ar(ΔDEF) (For parallelogram DCEF)          -(equation 2)

ar(ΔCDE) = ar(ΔDEF) (For parallelogram AFDE)          -(equation 3)

From equations (1), (2) and (3)

ar(ΔBFD) = ar(ΔAFE) = ar(ΔCDE) = ar(ΔDEF)

=> ar(ΔBFD) + ar(ΔAFE) + ar(ΔCDE) + ar(ΔDEF) = ar(ΔABC)

=> 4 ar(ΔDEF) = ar(ΔABC)

=> ar(DEF) = 1/4 ar(ABC)

(iii) Area (parallelogram BDEF) = ar(ΔDEF) + ar(ΔBDE)

=> ar(parallelogram BDEF) = ar(ΔDEF) + ar(ΔDEF)

=> ar(parallelogram BDEF) = 2× ar(ΔDEF)

=> ar(parallelogram BDEF) = 2× 1/4 ar(ΔABC)

=> ar(parallelogram BDEF) = 1/2 ar(ΔABC)

Question 6. In Figure, 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

Construct,

DE ⊥ AC and BF ⊥ AC.

To proof:

(i) In ΔDOE and ΔBOF,

∠DEO = ∠BFO (Perpendiculars)

∠DOE = ∠BOF (Vertically opposite angles)

OD = OB (Given)

Therefore, 

ΔDOE ≅ ΔBOF by AAS congruence condition.

Therefore, DE = BF (By CPCT)          -(equation 1)

also, 

ar(ΔDOE) = ar(ΔBOF) (Congruent triangles)          -(equation 2)

Now,

In ΔDEC and ΔBFA,

∠DEC = ∠BFA (as they are perpendiculars)

CD = AB (Given)

DE = BF (From equation 1)

Therefore, 

ΔDEC ≅ ΔBFA by RHS congruence condition.

Therefore, 

ar(ΔDEC) = ar(ΔBFA) (Congruent triangles)          -(equation 3)

Adding equation (2) and (3),

ar(ΔDOE) + ar(ΔDEC) = ar(ΔBOF) + ar(ΔBFA)

=> ar (DOC) = ar (AOB)

(ii) ar(ΔDOC) = ar(ΔAOB)

Adding ar(ΔOCB) in LHS and RHS, 

we will get,

=> ar(ΔDOC) + ar(ΔOCB) = ar(ΔAOB) + ar(ΔOCB)

=> ar(ΔDCB) = ar(ΔACB)

(iii) When two triangles have same base and equal areas,

the triangles will be in between the same parallel lines

ar(ΔDCB) = ar(ΔACB)

DA || BC          -(equation 4)

For quadrilateral ABCD, one pair of opposite sides are 

equal (AB = CD) and other pair of opposite sides are parallel.

Therefore, 

ABCD is parallelogram.

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

Solution:

ΔDBC and ΔEBC are on the same base BC and are having equal areas.

Therefore, 

They will lie between the same parallel lines.

Therefore, 

DE || BC.

Question 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 that ar(ΔABE) = ar(ΔACF)

Solution:

Given,

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

We have to show that,

ar(ΔABE) = ar(ΔAC)

Proof:

BCYE is a parallelogram as ΔABE and ||gm BCYE are on the same 

base BE and between the same parallel lines BE and AC.

Therefore,

ar(ABE) =  1/2 ar(BCYE)          -(equation 1)

Now,

CF || AB and XY || BC

=> CF || AB and XF || BC

=> BCFX is a parallelogram.

Thus ΔACF and parallelogram BCFX are on the same base CF 

and in-between the same parallel AB and FC. Therefore,

ar (ΔACF)= 1/2 ar (BCFX)          -(equation 2)

But,

Parallelogram BCFX and parallelogram BCYE are on the same

base BC and between the same parallels BC and EF. Therefore,

ar (BCFX) = ar(BCYE)          -(equation 3)

From equations (1), (2) and (3), 

We will get,

ar (ΔABE) = ar(ΔACF)

=> ar(BEYC) = ar(BXFC)

As the parallelograms are on the same base BC and 

in-between the same parallels EF and BC          -(equation 3)

Also,

ΔAEB and parallelogram BEYC are on the same base

BE and in-between the same parallels BE and AC.

=> ar(ΔAEB) = 1/2 ar(BEYC)          -(equation 4)

Similarly,

ΔACF and parallelogram BXFC on the same base CF 

and between the same parallels CF and AB.

=> ar(ΔACF) = 1/2 ar(BXFC)          -(equation 5)

From equations (3), (4) and (5),

ar(ΔABE) = ar(ΔACF).

Question 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 Figure). Show that

ar(ABCD) = ar(PBQR).

[Hint: Join AC and PQ. Now compare ar(ACQ) and ar(APQ).]

Solution:

AC and PQ are joined.

Ar(ΔACQ) = ar(ΔAPQ)(On the same base AQ and between

                                       the same parallel lines AQ and CP)

=> ar(ΔACQ)-ar(ΔABQ) = ar(ΔAPQ)-ar(ΔABQ)

=> ar(△ABC) = ar(△QBP)          -(equation 1)

AC and QP are diagonals ABCD and PBQR.

Therefore,

ar(ABC) = 1/2ar(ABCD)          -(equation 2)

ar(QBP) = 1/2 ar(PBQR)         -(equation 3)

From equation (2) and (3),

1/2 ar(ABCD) = 1/2 ar(PBQR)

=> ar(ABCD) = ar(PBQR)

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

Solution:

ΔDAC and ΔDBC lie on the same base DC and 

between the same parallels AB and CD.

Ar(ΔDAC) = ar(ΔDBC)

=>  ar(ΔDAC) – ar(ΔDOC) = ar(ΔDBC) – ar(ΔDOC)

=> ar(ΔAOD) = ar(ΔBOC)

Question 11. In Figure, ABCDE is a pentagon. A line through B parallel to AC meets DC produced at F.

Show that

(i) ar(ΔACB) = ar(ΔACF)

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

Solution:

(i) ΔACB and ΔACF lie on the same base AC and between the same parallels AC and BF.

Therefore,

ar(ΔACB) = ar(ΔACF)

(ii) ar(ΔACB) = ar(ΔACF)

=> ar(ΔACB)+ar(ΔACDE) = ar(ΔACF)+ar(ΔACDE)

=> 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,

ΔBCE is the shape of Itwari’s original field

ΔAOB is the area for construction of the Health Centre.

ΔDEO is the land joined to the plot.

To prove:

ar(ΔDEO) = ar(ΔAOB)

Proof:

ΔDEB and ΔDAB lie on the same base BD, in-between the same parallels BD and AE.

Therefore,

Ar(ΔDEB) = ar(ΔDAB)

=> ar(ΔDEB) – ar(ΔDOB) = ar(ΔDAB) – ar(ΔDOB)

=> ar(ΔDEO) = ar(ΔAOB)

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

[Hint: Join CX.]

Solution:

Given:

ABCD is a trapezium with,

AB || DC.

XY || AC

To Construct,

Join CX

Prove:

ar(ADX) = ar(ACY)

Proof:

ar(ΔADX) = ar(ΔAXC)         -(equation 1)(Since they are on the same base AX and 

                                                                    in-between the same parallels AB and CD)

also,

ar(ΔAXC)=ar(ΔACY)          -(equation-2)(Since they are on the same base AC and 

                                                                     in-between the same parallels XY and AC)

From equation (1) and (2),

ar(ΔADX) = ar(ΔACY)

Question 14. In Figure AP || BQ || CR. Prove that ar(ΔAQC) = ar(ΔPBR).

Solution:

Given:

AP || BQ || CR

Prove:

ar(AQC) = ar(PBR)

Proof:

ar(ΔAQB) = ar(ΔPBQ)          -(equation 1)(Since they are on the same base BQ and

                                                                      between the same parallels AP and BQ)

also,

ar(ΔBQC) = ar(ΔBQR)          (equation 2)(Since they are on the same base BQ and

                                                                     between the same parallels BQ and CR)

Adding equations (1) and (2),

ar(ΔAQB)+ar(ΔBQC) = ar(ΔPBQ)+ar(ΔBQR)

=> ar(ΔAQC) = ar(ΔPBR)

Question 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.

Solution:

Given:

ar(ΔAOD) = ar(ΔBOC)

Prove:

ABCD is a trapezium.

Proof:

ar(ΔAOD) = ar(ΔBOC)

=> ar(ΔAOD) + ar(ΔAOB) = ar(ΔBOC)+ar(ΔAOB)

=> ar(ΔADB) = ar(ΔACB)

Areas of ΔADB and ΔACB are equal.

Therefore,

They should be lying between the same parallel lines.

Therefore,

AB ∥  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(ΔDRC) = ar(ΔDPC)

ar(ΔBDP) = ar(ΔARC)

Prove:

ABCD and DCPR are trapeziums.

Proof:

ar(ΔBDP) = ar(ΔARC)

⇒ ar(ΔBDP) – ar(ΔDPC) = ar(ΔDRC)

⇒  ar(ΔBDC) = ar(ΔADC)

ar(ΔBDC) = ar(ΔADC).

Therefore, 

ar(ΔBDC) and ar(ΔADC) are lying in-between the same parallel lines.

Hence, AB ∥ CD

ABCD is a trapezium.

Similarly,

ar(ΔDRC) = ar(ΔDPC).

Therefore, 

ar(ΔDRC) and ar(ΔDPC) are lying in-between the same parallel lines.

Hence, DC ∥ PR

Thus, DCPR is a trapezium.

NCERT Solutions for Class 9 Maths Chapter 9 Areas of Parallelograms and Triangles – This article contains detailed Solutions for NCERT Class 9 Maths Chapter 9 Areas of Parallelograms and Triangles curated by the team of subject matter experts at GFG, to help students understand how to solve the NCERT problems in an easy manner.

NCERT Class 9 Maths Chapter 9 Areas of Parallelograms and Triangles covers all the following topics:

• Area of Triangles
• Area of Parallelogram
• Area between the Same Base and Two Parallel Lines

This article provides solutions to all the problems asked in Class 9 Maths Chapter 7 Triangles of your NCERT textbook in a step-by-step manner. Solutions to all the exercises in the NCERT Class 9 Maths Chapter 7 Triangles are regularly revised to check errors and updated according to the latest CBSE Syllabus 2023-24 and guidelines.

All of the problems in Chapter 9 Areas of Parallelograms and Triangles exercises from the NCERT textbook have been covered in the NCERT Solutions for Class 9 Maths.