*According to the latest update on NCERT Syllabus 2023-24, this chapter has been removed.
Class 9 Maths Chapter 9 deals with the various concepts and theorems regarding parallelograms and triangles. The subject-matter experts at BYJU’S have devised the Class 9 Maths Chapter 9 – Area of Parallelogram and Triangles Exercise 9.3 NCERT Solutions with the aim to help students get acquainted with the concept that was discussed in Section 9.4 of Chapter 9. This section deals with the concept of “Triangles on the Same Base and between the Same Parallels.” The section also explains, “What can you say about the areas of such triangles?”
Based on the NCERT syllabus and guidelines, the solutions contain detailed step-by-step explanations of the questions that are asked in the exercise section of NCERT Solutions for Class 9 Maths Chapter 9. It helps students to be thorough with the concept and also gain practice solving questions.
NCERT Solutions for Class 9 Maths Chapter 9 – Area of Parallelograms and Triangles Exercise 9.3
Access Answers to NCERT Class 9 Maths Chapter 9 – Area of Parallelograms and Triangles Exercise 9.3
Access Other Exercise Solutions of Class 9 Maths Chapter 9 – Area of Parallelograms and Triangles
Exercise 9.1 Solutions 1 Short Type Answer with Reasoning
Exercise 9.2 Solutions 5 Short Type Answers and 1 Long Type Answer
Exercise 9.4 Solutions 4 Short Type Answer, 1 Long Type Answer and 3 Very Long Type Answers
Access Answers to NCERT Class 9 Maths Chapter 9 – Area of Parallelograms and Triangles Exercise 9.3
1. In Fig.9.23, E is any point on the median AD of a ΔABC. Show that ar (ABE) = ar(ACE).
Solution:
Given,
AD is the median of ΔABC. ∴, 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) = ¼ ar(ABC).
Solution:
ar(BED) = (1/2)×BD×DE
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 — (i)
BD = (1/2)BC — (ii)
From (i) and (ii), we get
ar(BED) = (1/2)×(1/2)BC × (1/2)AD
⇒ ar(BED) = (1/2)×(1/2)ar(ABC)
⇒ ar(BED) = ¼ ar(ABC)
3. Show that the diagonals of a parallelogram divide it into four triangles of equal area.
Solution:
O is the midpoint of AC and BD. (Diagonals 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), we get
ar(BOC) = ar(COD) = ar(AOD) = ar(AOB)
Hence, we get the diagonals of a parallelogram that 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 the 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.)
∴ar(AOC) = ar(AOD) — (i)
also,
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)
5. D, E and F are, respectively, the midpoints of the sides BC, CA and AB of a ΔABC.
Show that
(i) BDEF is a parallelogram.
(ii) ar(DEF) = ¼ ar(ABC)
(iii) ar (BDEF) = ½ ar(ABC)
Solution:
(i) In ΔABC,
EF || BC and EF = ½ BC (by the midpoint theorem)
also,
BD = ½ BC (D is the midpoint)
So, BD = EF
also,
BF and DE are parallel and equal to each other.
∴, the pair of opposite sides are equal in length and parallel to each other.
∴ BDEF is a parallelogram.
(ii) Proceeding from the result of (i),
BDEF, DCEF, and AFDE are parallelograms.
A diagonal of a parallelogram divides it into two triangles of equal area.
∴ar(ΔBFD) = ar(ΔDEF) (For parallelogram BDEF) — (i)
also,
ar(ΔAFE) = ar(ΔDEF) (For parallelogram DCEF) — (ii)
ar(ΔCDE) = ar(ΔDEF) (For parallelogram AFDE) — (iii)
From (i), (ii) and (iii)
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) = ¼ 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× ¼ ar(ΔABC)
⇒ ar(parallelogram BDEF) = ½ ar(ΔABC)
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 ⊥ AC and BF ⊥ AC are drawn.
Proof:
(i) In ΔDOE and ΔBOF,
∠DEO = ∠BFO (Perpendiculars)
∠DOE = ∠BOF (Vertically opposite angles)
OD = OB (Given)
∴, ΔDOE ≅ ΔBOF by AAS congruence condition.
∴, DE = BF (By CPCT) — (i)
also, ar(ΔDOE) = ar(ΔBOF) (Congruent triangles) — (ii)
Now,
In ΔDEC and ΔBFA,
∠DEC = ∠BFA (Perpendiculars)
CD = AB (Given)
DE = BF (From i)
∴, ΔDEC ≅ ΔBFA by RHS congruence condition.
∴, ar(ΔDEC) = ar(ΔBFA) (Congruent triangles) — (iii)
Adding (ii) and (iii),
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 get
⇒ar(ΔDOC) + ar(ΔOCB) = ar(ΔAOB) + ar(ΔOCB)
⇒ ar(ΔDCB) = ar(ΔACB)
(iii) When two triangles have the same base and equal areas, the triangles will be in between the same parallel lines
ar(ΔDCB) = ar(ΔACB)
DA || BC — (iv)
For quadrilateral ABCD, one pair of opposite sides are equal (AB = CD), and the other pair of opposite sides are parallel.
∴, ABCD is parallelogram.
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 also have equal areas.
∴, they will lie between the same parallel lines.
∴, 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 that
ar(ΔABE) = ar(ΔACF)
Solution:
Given,
XY || BC, BE || AC and CF || AB
To show,
ar(ΔABE) = ar(ΔACF)
Proof:
BCYE is a || gm as ΔABE and ||gm BCYE are on the same base BE and between the same parallel lines BE and AC.
∴,ar(ABE) = ½ ar(BCYE) … (1)
Now,
CF || AB and XY || BC
⇒ CF || AB and XF || BC
⇒ BCFX is a || gm
As ΔACF and || gm BCFX are on the same base CF and between the same parallel AB and FC.
∴,ar (ΔACF)= ½ ar (BCFX) … (2)
But,
||gm BCFX and || gm BCYE are on the same base BC and between the same parallels BC and EF.
∴,ar (BCFX) = ar(BCYE) … (3)
From (1), (2) and (3), we get
ar (ΔABE) = ar(ΔACF)
⇒ ar(BEYC) = ar(BXFC)
As the parallelograms are on the same base BC and between the same parallels EF and BC–(iii)
Also,
△AEB and ||gm BEYC are on the same base BE and between the same parallels BE and AC.
⇒ ar(△AEB) = ½ ar(BEYC) — (iv)
Similarly,
△ACF and || gm BXFC on the same base CF and between the same parallels CF and AB.
⇒ ar(△ ACF) = ½ ar(BXFC) — (v)
From (iii), (iv) and (v),
ar(△ABE) = 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).]
Solution:
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.
∴,ar(ABC) = ½ ar(ABCD) — (ii)
ar(QBP) = ½ ar(PBQR) — (iii)
From (ii) and (ii),
½ ar(ABCD) = ½ 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).
Solution:
△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)
Solution:
- △ACB and △ACF lie on the same base AC and between the same parallels AC and BF.
∴ar(△ACB) = ar(△ ACF)
- ar(△ACB) = ar(△ACF)
⇒ ar(△ACB)+ar(ACDE) = ar(△ACF)+ar(ACDE)
⇒ ar(ABCDE) = ar(AEDF)
12. A villager, Itwaari, has a plot of land in 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 an equal amount of land in lieu of the 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 in the shape of a quadrilateral.
To construct,
Join the diagonal BD.
Draw AE parallel to BD.
Join BE, which intersected AD at O.
We get
△BCE is the shape of the original field.
△AOB is the area for constructing the health centre.
△DEO is the land joined to the plot.
To prove,
ar(△DEO) = ar(△AOB)
Proof:
△DEB and △DAB lie on the same base BD, between the same parallels 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.]
Solution:
Given,
ABCD is a trapezium with AB || DC.
XY || AC
Construction,
Join CX
To prove,
ar(ADX) = ar(ACY)
Proof:
ar(△ADX) = ar(△AXC) — (i) (Since they are on the same base AX and between the same parallels AB and CD)
also,
ar(△ AXC)=ar(△ ACY) — (ii) (Since they are on the same base AC and between the same parallels XY and AC.)
(i) and (ii),
ar(△ADX) = ar(△ACY)
14. In Fig.9.28, AP || BQ || CR. Prove that ar(△AQC) = ar(△PBR).
Solution:
Given,
AP || BQ || CR
To prove,
ar(AQC) = ar(PBR)
Proof:
ar(△AQB) = ar(△PBQ) — (i) (Since they are on the same base BQ and between the same parallels AP and BQ.)
also,
ar(△BQC) = ar(△BQR) — (ii) (Since they are 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.
Solution:
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. ∴, they must lie between the same parallel lines.
∴, AB ∥ CD
∴, 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(△DRC) = ar(△DPC)
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) and ar(△ADC) are lying between the same parallel lines.
∴, AB ∥ CD
ABCD is a trapezium.
Similarly,
ar(△DRC) = ar(△DPC).
∴, ar(△DRC) and ar(△DPC) are lying between the same parallel lines.
∴, DC ∥ PR
∴, DCPR is a trapezium.
Exercise 9.3 is based on two main theorems that are discussed in Chapter 9 of the Class 9 textbooks. NCERT Solutions for Class 9 Maths explain these theorems very well. The theorems are as follows:
Theorem 9.2: Two triangles on the same base (or equal bases) and between the same parallels are equal in area
Theorem 9.3: Two triangles having the same base (or equal bases) and equal areas lie between the same parallels.
To explain these concepts, there are easy-to-understand NCERT Solutions for Class 9 Maths, created by our experts to answer various questions of Exercise 9.3 that follow this topic in the textbook. This exercise contains 16 questions, 12 short answer type questions and 4 long answer type questions.
By solving these questions from the NCERT Solutions for Class 9, students will
- Be able to understand the concept thoroughly
- Learn the subject well
- Prepare for the exams
- Gain more confidence
- Get practice solving papers
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