NCERT Solutions for Class 9 Maths Chapter 9 Areas of Parallelograms and Triangles Ex 9.4

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NCERT Solutions for Class 9 Maths Chapter 9 Areas of Parallelograms and Triangles Ex 9.4

The topics and sub-topics in NCERT Book Maths Class 9 Chapter 9 Areas of Parallelograms and Triangles:

  • Areas Of Parallelograms And Triangles
  • Introduction
  • Figures On The Same Base And Between The Same Parallels
  • Parallelograms On The Same Base And Between The Same Parallels
  • Triangles On The Same Base And Between The Same Parallels
  • Summary

Formulae Handbook for Class 9 Maths and ScienceEducational Loans in India

NCERT Solutions for Class 9 Maths Chapter 9 Areas of Parallelograms and Triangles Ex 9.4

Ex 9.4 Class 9 Maths Question 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: parallelogram ABCD and rectangle ABEF are on same base AB, and area of both are equal.
In rectangle ABEF, AB = EF and in parallelogram ABCD,
CD = AB ⇒ AB + CD = AB + EF ….(i)
We know that, the perpendicular distance between two parallel sides of a parallelogram is always less than the length of the other parallel sides.
NCERT Solutions for Class 9 Maths Chapter 9 Areas of Parallelograms and Triangles Ex 9.4 A1
∴ BE BE + AF …(ii)
⇒ BC + AD + AB + CD > BE + AF + AB + CD (Adding AB + CD on both sides)
NCERT Solutions for Class 9 Maths Chapter 9 Areas of Parallelograms and Triangles Ex 9.4 A1.1
⇒ AB + BC+ CD + AD > AB + BE + EF + AF [Put the values from Eq. (i)]
Hence, the perimeter of the parallelogram is greater than the perimeter of the rectangle.

Ex 9.4 Class 9 Maths Question 2.
In figure, 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 Introduction’ of this chapter, whether the field of Budhia has been actually divided into three pares of equal area?
[Remark Note that by taking BD = DE = EC, the ∆ ABC is divided into three triangles ABD, ADE and AEC of equal areas. In the sameway, by dividing BC into n equal parts and joining the points of division so obtained to the opposite vertex of BC, you can divide ∆ ABC into n triangles of equal areas.]
Solution:
Given: ABC is a triangle and D and E are two points on BC, such that
NCERT Solutions for Class 9 Maths Chapter 9 Areas of Parallelograms and Triangles Ex 9.4 A2
BD = DE = EC
Let AO be the perpendicular to BC.
∴ ar ( ∆ABD) = (frac { 1 }{ 2 }) x BD x AO
ar (∆ADE) = (frac { 1 }{ 2 }) x DE x AO
and ar(∆AEC) = (frac { 1 }{ 2 }) x EC x AO
Since, BD = DE = EC (Given)
∴ ar(∆ABD) = ar(∆ADE) = ar(∆AEC)
Yes, altitudes of all triangles are same. Budhia has use the result of this question in dividing her land in three equal parts.

Ex 9.4 Class 9 Maths Question 3.
In figure, ABCD, DCFE and ABFE are parallelograms. Show that ar(ADE) = ax(BCF).
NCERT Solutions for Class 9 Maths Chapter 9 Areas of Parallelograms and Triangles Ex 9.4 A3
Solution:
Given: ABCD, DCFE, and ABFE and parallelograms
In ∆ADE and ∆BCF,
AD = BC (∵ ABCD is a parallelogram)
DE – CF (∵ DCFE is a parallelogram)
and AE = BF (∵ ABFE is a parallelogram)
Hence ∆ADE = ∆BCF
∴ ar (∆ADE) = ar (∆BCF)

Ex 9.4 Class 9 Maths Question 4.
In figure, 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) = ax(DPQ).[Hint Join AC.]
NCERT Solutions for Class 9 Maths Chapter 9 Areas of Parallelograms and Triangles Ex 9.4 A4
Solution:
Given: ABCD is a parallelogram and AD || CQ, and AQ = CQ. Join the line segment AC.
Now, ∆ APC and ∆ BPC lie on the same base PC and between the same parallels PC and AB, therefore
ar(∆ APC) = ar(∆ BPC) …(i)
AD = CQ and AD || CQ (Given)
NCERT Solutions for Class 9 Maths Chapter 9 Areas of Parallelograms and Triangles Ex 9.4 A4.1
Thus, in quadrilateral ACQD, one pair of opposite sides is equal and parallel.
∴ ADQC is a parallelogram.
We know that, diagonals of a parallelogram bisect each other.
∴ CP = DP and AP = PQ ….(ii)
In ∆ APC and ∆ DPQ, we have
NCERT Solutions for Class 9 Maths Chapter 9 Areas of Parallelograms and Triangles Ex 9.4 A4.2

NCERT Solutions for Class 9 Maths Chapter 9 Areas of Parallelograms and Triangles Ex 9.4 A4.3

More Resources for CBSE Class 9

Ex 9.4 Class 9 Maths Question 5.
In figure, ABC and BDE are two equilateral triangles such that D is the mid-point of BC. If AE intersects BC at F, Show that
NCERT Solutions for Class 9 Maths Chapter 9 Areas of Parallelograms and Triangles Ex 9.4 A5
[Hint Join EC and AD. Show that BE || AC and DE || AB, etc.]
Solution:
Join AD and EC. Let x be the side of ∆ ABC. Then
NCERT Solutions for Class 9 Maths Chapter 9 Areas of Parallelograms and Triangles Ex 9.4 A5.1

NCERT Solutions for Class 9 Maths Chapter 9 Areas of Parallelograms and Triangles Ex 9.4 A5.2

NCERT Solutions for Class 9 Maths Chapter 9 Areas of Parallelograms and Triangles Ex 9.4 A5.3

NCERT Solutions for Class 9 Maths Chapter 9 Areas of Parallelograms and Triangles Ex 9.4 A5.4

Ex 9.4 Class 9 Maths Question 6.
Diagonals AC and BD of a quadrilateral ABCD intersect each other at P. Show that
ar(APB) x ar(CPD) = ar(APD) x ar(BPC).
[Hint From A and C, draw perpendiculars to BD.]
Solution:
Given: ABCD is a quadrilateral whose diagonals intersect at P.
Draw two perpendiculars AE and CF from A and Con BD, respectively. Now,
LHS = ar (∆ APB) x ar (∆ CPD)
NCERT Solutions for Class 9 Maths Chapter 9 Areas of Parallelograms and Triangles Ex 9.4 A6

NCERT Solutions for Class 9 Maths Chapter 9 Areas of Parallelograms and Triangles Ex 9.4 A6.1
From Eqs. (i) and (ii), we get, LHS = RHS
i.e., ar(∆APB) x ar(∆CPD) = ar(∆APD) x ar(∆BPC)

Ex 9.4 Class 9 Maths Question 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
NCERT Solutions for Class 9 Maths Chapter 9 Areas of Parallelograms and Triangles Ex 9.4 A7
Solution:
(i) Given: P and Q are mid-points of AB and BC. Also, R is mid-point of AP.
Since, P and 0 are the mid-points of AB and BC, respectively.
∴ PQ || AC and PQ = (frac { 1 }{ 2 }) AC (By mid-point theorem)
Draw RM || AC || PQ
Also, draw QG ⊥ RM and MH ⊥ AC
∵ PQ || RM || AC and PR = RA
∴ QM = MC
In ∆ QGM and ∆ MHC,
NCERT Solutions for Class 9 Maths Chapter 9 Areas of Parallelograms and Triangles Ex 9.4 A7.1

NCERT Solutions for Class 9 Maths Chapter 9 Areas of Parallelograms and Triangles Ex 9.4 A7.2

NCERT Solutions for Class 9 Maths Chapter 9 Areas of Parallelograms and Triangles Ex 9.4 A7.3

NCERT Solutions for Class 9 Maths Chapter 9 Areas of Parallelograms and Triangles Ex 9.4 A7.4

Ex 9.4 Class 9 Maths Question 8.
In figure, 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
NCERT Solutions for Class 9 Maths Chapter 9 Areas of Parallelograms and Triangles Ex 9.4 A8
(i) ∆MBC = ∆ABD
(ii) ar(BYXD) = 2 ar(MBC)
(iii) ar(BYXD) = ax(ABMN)
(iv) ∆FCB ≅ ∆ACE
(v) ar(CYXE) = 2 ar(FCB)
(vi) ar(CYXE) = ax(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) In ∆ABD and ∆MBC,
BC = BD (These are the sides of square)
MB = AB
and ∠ MBC = 90° + ∠ ABC
= ∠DBC + ∠ABC
= ∠ABD
∴ ∆MBC = ∆ABD (By SAS rule)

(ii) From part (i), ar(∆ MBC) = ar (∆ ABD) …(i)
But ar(∆ ABD) = (frac { 1 }{ 2 }) ar (BYXD) …(ii)
(∵ ∆ ABD and rectangle BYXD lie on the same base and between same parallel between lines.)
From Eqs. (i) and (ii), we get
ar (∆MBC) = (frac { 1 }{ 2 }) ar (BYXD) .. .(iii)
⇒ ar (BYXD) = 2 ar (∆MBC)

(iii) Now ar (∆MBC) = (frac { 1 }{ 2 }) ar (ABMN) …..(iv)
(∵ ∆MBC and square ABMN lie on the same base MB and between same parallels MB and NC)
From Eqs. (iii) and (iv), we get
ar (BYXD) = ar (ABMN)

(iv) In ∆ ACE and ∆FCS,
AC = FC
and CE = BC (These are the sides of square)
∠ FCB = 90° + ∠ ACB = ∠ BCE + ∠ACB = ∠ACE
So, ∆ FCB = ∆ ACE (By SAS rule)

(v) From Eqs. (iv), ar(AACE) = ar(AFCB) …(vi)
But ar(∆ACE) = (frac { 1 }{ 2 }) ar(CVXE)
(∵ Both lie on the same base CE and between same parallel lines CE and AX.)
From Eqs. (vi) and (vii), we get
ar (∆ACE) = (frac { 1 }{ 2 }) ar (CYXE)
= ar (∆FCB) …(vii)
⇒ ar (CYXE) = (frac { 1 }{ 2 }) ar (∆ FCB)

(vi) Now, ar(AFCB) = (frac { 1 }{ 2 }) ar (ACFG) …(ix)
(∵Both lie on same base CF and between same parallel lines CF and BG)
From Eqs. (viii) and (ix), we get
(frac { 1 }{ 2 }) ar (ALFG) = (frac { 1 }{ 2 }) ar (CYXE)
⇒ ar (ACFG) = ar (CYXE)

(vii)
 Now, ar (BCED) = ar (BYXD) + ar (CYXE)
= ar (ABMN) + ar (ACFG) [From part (iii) and (vi)]

NCERT Solutions for Class 9 Maths Chapter 9 Area of ​​Parallelograms and Triangles (समान्तर चतुर्भुज और त्रिभुजों के क्षेत्रफल) (Hindi Medium) Ex 9.4

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NCERT Solutions for Class 9 Maths

  1. Chapter 1 Number systems
  2. Chapter 2 Polynomials
  3. Chapter 3 Coordinate Geometry
  4. Chapter 4 Linear Equations in Two Variables
  5. Chapter 5 Introduction to Euclid Geometry
  6. Chapter 6 Lines and Angles
  7. Chapter 7 Triangles
  8. Chapter 8 Quadrilaterals
  9. Chapter 9 Areas of Parallelograms and Triangles
  10. Chapter 10 Circles
  11. Chapter 11 Constructions
  12. Chapter 12 Heron’s Formula
  13. Chapter 13 Surface Areas and Volumes
  14. Chapter 14 Statistics
  15. Chapter 15 Probability
  16. Class 9 Maths (Download PDF)

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