CBSE Class 9 – Page 16

NCERT Solutions for Class 9 Maths Chapter 10 Areas of Parallelograms and Triangles Ex 10.4

March 14, 2023

NCERT Solutions for Class 9 Maths Chapter 10 Areas of Parallelograms and Triangles Ex 10.4 are part of NCERT Solutions for Class 9 Maths. Here we have given NCERT Solutions for Class 9 Maths Chapter 10 Areas of Parallelograms and Triangles Ex 10.4.

Board	CBSE
Textbook	NCERT
Class	Class 9
Subject	Maths
Chapter	Chapter 10
Chapter Name	Areas of Parallelograms and Triangles
Exercise	Ex 10.4
Number of Questions Solved	8
Category	NCERT Solutions

NCERT Solutions for Class 9 Maths Chapter 10 Areas of Parallelograms and Triangles Ex 10.4

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 10 Areas of Parallelograms and Triangles Ex 10.4 img 1
∴ BE < SC and AF < AD On adding both, we get, BC + AD > BE + AF …(ii)
⇒ BC + AD + AB + CD > BE + AF + AB + CD (Adding AB + CD on both sides)

⇒ 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.

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 10 Areas of Parallelograms and Triangles Ex 10.4 img 3
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.

Question 3.
In figure, ABCD, DCFE and ABFE are parallelograms. Show that ar(ADE) = ax(BCF).
NCERT Solutions for Class 9 Maths Chapter 10 Areas of Parallelograms and Triangles Ex 10.4 img 4
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)

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 10 Areas of Parallelograms and Triangles Ex 10.4 img 5
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 10 Areas of Parallelograms and Triangles Ex 10.4 img 6
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 10 Areas of Parallelograms and Triangles Ex 10.4 img 7

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

Solution:
Join AD and EC. Let x be the side of ∆ ABC. Then
src=”https://farm2.staticflickr.com/1921/31706925298_d4d194aabd_o.png” width=”525″ height=”586″ alt=”NCERT Solutions for Class 9 Maths Chapter 10 Areas of Parallelograms and Triangles 10.4 5b”>

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 10 Areas of Parallelograms and Triangles Ex 10.4 img 13
From Eqs. (i) and (ii), we get, LHS = RHS
i.e., ar(∆APB) x ar(∆CPD) = ar(∆APD) x ar(∆BPC)

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 10 Areas of Parallelograms and Triangles Ex 10.4 img 14
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 10 Areas of Parallelograms and Triangles Ex 10.4 img 15

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 10 Areas of Parallelograms and Triangles Ex 10.4 img 18
(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)]
We hope the NCERT Solutions for Class 9 Maths Chapter 10 Areas of Parallelograms and Triangles Ex 10.4 help you. If you have any query regarding NCERT Solutions for Class 9 Maths Chapter 10 Areas of Parallelograms and Triangles Ex 10.4, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 9 Maths Chapter 10 Areas of Parallelograms and Triangles Ex 10.2

March 14, 2023

NCERT Solutions for Class 9 Maths Chapter 10 Areas of Parallelograms and Triangles Ex 10.2 are part of NCERT Solutions for Class 9 Maths. Here we have given NCERT Solutions for Class 9 Maths Chapter 10 Areas of Parallelograms and Triangles Ex 10.2.

Board	CBSE
Textbook	NCERT
Class	Class 9
Subject	Maths
Chapter	Chapter 10
Chapter Name	Areas of Parallelograms and Triangles
Exercise	Ex 10.2
Number of Questions Solved	6
Category	NCERT Solutions

NCERT Solutions for Class 9 Maths Chapter 10 Areas of Parallelograms and Triangles Ex 10.2

Question 1.
In figure, ABCD is a parallelogram, AE ⊥ DC and CF ⊥ AD. If AB = 16 cm, AE = 8 cm and CF = 10 cm, find AD.
NCERT Solutions for Class 9 Maths Chapter 10 Areas of Parallelograms and Triangles Ex 10.2 img 1
Solution:
We know that,
Area of parallelogram = Base x Altitude
Given, AE = 8 cm CF = 10 cm and AB = 16cm
∴ ar (parallelogramABCD) = DC x AE
= 16 x 8 cm2 (∵ AE = 8 cm)…(i)
and ar (parallelogram ABCD) = AD x CF – AD x 10 ( ∵ CF = 10 cm)
From Eq. (i), we have,
16 x 8 = AD x 10

Question 2.
If E, F, G and H are respectively the mid-points of the sides of a parallelogram ABCD, show that ar (EFGH) = \(\frac { 1 }{ 2 }\) ar (ABCD).
Solution:
Given: E,F, G and H are respectively the mid-points of the sides AB, BC, CD and AD. Joint if, it will parallel to CD and AB.
Now, parallelogram HDCF and triangle HGF stand on the same base HF and lie between the same parallel lines DC and HF.
NCERT Solutions for Class 9 Maths Chapter 10 Areas of Parallelograms and Triangles Ex 10.2 img 3
Similarly,parallelogram ABFH and triangle HEF stand on the same base HF and lie between the same parallel lines HF and AB.
ar (∆HEF) = \(\frac { 1 }{ 2 }\) ar (∆BFH) …(ii)
On adding Eqs. (i) and (ii), we get
ar (∆HGF) + ar(∆HEF) = \(\frac { 1 }{ 2 }\) [ar (HDCF) + ar (ABFH)]
⇒ ar (EFGH) = \(\frac { 1 }{ 2 }\) ar (ABCD)

Question 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:
Given: a parallelogram ABCD. P and Q are any two points lying on the sides DC and AD, respectively.
NCERT Solutions for Class 9 Maths Chapter 10 Areas of Parallelograms and Triangles Ex 10.2 img 4
Now, parallelogram ABCD and ABQC stand on the same base BC and lie between the same parallel lines BC and AD.
∴ ar (∆ BQC) = \(\frac { 1 }{ 2 }\) ar(ABCD) …(i)
Similarly, ∆ APB and parallelogram ABCD stand on the same base AB and lie between the same parallels AB and CD.
∴ ar (∆ APB) = \(\frac { 1 }{ 2 }\) ar (ABCD) ….(ii)
From Eqs. (i) and (ii), we get
ar (∆ APB) = ar (∆ BQC)

Question 4.
In figure, P is a point in the interior of a parallelogram ABCD. Show that
(i) ar (APB) + ar (PCD) = \(\frac { 1 }{ 2 }\) ar (ABCD)
(ii) ar (APD) + ar(PBC) = ar (APB) + ar (PCD)
NCERT Solutions for Class 9 Maths Chapter 10 Areas of Parallelograms and Triangles Ex 10.2 img 5
Solution:
Given: ABCD is a parallelogram. So, AB || CD, AD |[ BC.
(i) Now, draw MPR parallel to AB and CD both and also draw a perpendicular PS on AB.
∵ MR || AB and AM || BR

∴ ABRM is a parallelogram so
∴ ar (|| gm ABRM) = AB x PS …(i)
and ar (∆ APB) = \(\frac { 1 }{ 2 }\) x AB x PS
ar (∆ APB) = \(\frac { 1 }{ 2 }\) ar (|| gm ABRM)
ar (∆PCD) = \(\frac { 1 }{ 2 }\) ar (|| gm MRCD)
Now, ar (∆APB) + ar (∆ PCD) = \(\frac { 1 }{ 2 }\) ar (|| gm ABRM) + \(\frac { 1 }{ 2 }\) ar (||gm MRCD)
= \(\frac { 1 }{ 2 }\) ar ( || gm ABCD) …(ii)
(ii) Similarly, we can draw a line through P parallel to AD and through the point P draw perpendicular on AD, we cah prove that
ar (∆APD) + ar (∆PBC) = \(\frac { 1 }{ 2 }\) ar (|| gm ABCD) …(iii)
From Eqs. (ii) and (iii), we get
ar (∆APD) + ar (∆PBQ = ar (∆APB) + ar (∆PCD)
Hence proved.

Question 5.
In figure, PQRS and ABRS are parallelograms and X is any point on side BR. Show that
(i) ar (PQRS) = ar (ABRS)
(ii) ar (AXS) = \(\frac { 1 }{ 2 }\) ar (PQRS)
NCERT Solutions for Class 9 Maths Chapter 10 Areas of Parallelograms and Triangles Ex 10.2 img 7
Solution:
Given: PQRS and ABRS both are parallelogram and x is any point on BR.
(i) Here, parallelogram PQRS and ABRS lies on the same base SR and between the same parallel lines SR and PB.
∴ ar (PQRS) = ar (ABRS) …(i)
(ii) Again, in parallelogram ABRS, ∆ AXS and parallelogram lies on the same base AS and between the same parallel lines AS and BR.
∴ ar (∆AXS) = \(\frac { 1 }{ 2 }\)ar (∆BRS) …(ii)
Now, from Eqs. (i) and (ii), we get
ar (∆AXS) = \(\frac { 1 }{ 2 }\) ar (∆ PQRS)

Question 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:
Given: PQRS is a parallelogram and A is any point as RS. Now, join PA and PQ. Thus, the field will be divided into three parts and each part is in the shape of a triangle.
Since, the AAPQ and parallelogram PQRS lie on the same base PQ and between same parallel lines PQ and SR.
NCERT Solutions for Class 9 Maths Chapter 10 Areas of Parallelograms and Triangles Ex 10.2 img 8
∴ ar (∆APQ) = \(\frac { 1 }{ 2 }\) ar (PQRS) ….(i)
Then, remaining
∴ ar (∆ASP) + ar (∆ARQ) = \(\frac { 1 }{ 2 }\) ar (PQRS) ….(ii)
Now, from Eqs. (i) and (ii), we get
ar (∆APQ) = ar (∆ASP) + ar (∆ARQ)
So, farmer has two options.
Either the farmer should sow wheat and pulses in ∆APS and ∆AQR or in ar [∆APQ and (∆APS and ∆AQR)] separately.
We hope the NCERT Solutions for Class 9 Maths Chapter 10 Areas of Parallelograms and Triangles Ex 10.2 help you. If you have any query regarding NCERT Solutions for Class 9 Maths Chapter 10 Areas of Parallelograms and Triangles Ex 10.2, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 9 Maths Chapter 9 Quadrilaterals Ex 9.2

March 14, 2023

NCERT Solutions for Class 9 Maths Chapter 9 Quadrilaterals Ex 9.2 are part of NCERT Solutions for Class 9 Maths. Here we have given NCERT Solutions for Class 9 Maths Chapter 9 Quadrilaterals Ex 9.2.

Board	CBSE
Textbook	NCERT
Class	Class 9
Subject	Maths
Chapter	Chapter 9
Chapter Name	Quadrilaterals
Exercise	Ex 9.2
Number of Questions Solved	7
Category	NCERT Solutions

NCERT Solutions for Class 9 Maths Chapter 9 Quadrilaterals Ex 9.2

Question 1.
ABCD is a quadrilateral in which P, Q, R and S are mid-points of the sides AB, BC, CD and DA (see figure). AC is a diagonal. Show that
(i) SR || AC and SR = \(\frac { 1 }{ 2 }\) AC
(ii) PQ = SR
(iii) PQRS is a parallelogram.
NCERT Solutions for Class 9 Maths Chapter 9 Quadrilaterals Ex 9.2 img 1
Solution:
Given: P, Q, Ft and S are mid-points of the sides.
∴ AP = PB, BQ = CQ
CR = DR and AS = DS
(i) In ∆ADC, we have
S is mid-point of AD and R is mid-point of the DC.
We know that, the line segment joining the mid-points of two sides of a triangle is parallel to the third side.
∴ SB || AC …(i)
Also , SR = \(\frac { 1 }{ 2 }\) AC …(ii)
(ii) Similarly, in ∆ABC, we have
PQ || AC ….(iii)
and PQ = \(\frac { 1 }{ 2 }\) AC ….(iv)
Now, from Eqs. (i) and (iii), we get
SR = \(\frac { 1 }{ 2 }\) AC …..(v)
(iii) Now, from Eqs. (i) and (iii), we get
PQ || SR
and from Eq. (v), PQ = SR
Since, a pair of opposite sides of a quadrilateral PQRS is equal and parallel.
So, PQRS is a parallelogram.
Hence proved.

Question 2.
ABCD is a rhombus and P, Q, R and S are the mid-points of the sides AB, BC, CD and DA, respectively. Show that the quadrilateral PQRS is a rectangle.
Solution:
Given: ABCD is a rhombus and P, Q, R and S are mid-points of AB, BC, CD and DA

NCERT Solutions for Class 9 Maths Chapter 9 Quadrilaterals Ex 9.2 img 2
By mid-point theorem,

∴ PQRS is a parallelogram.
Now, we know that diagonals of a rhombus bisect each other at right angles.
∴ ∠EOF = 90°
Now, RQ || BD (By mid-point theorem)
⇒ RE || OF
Also, SP|| AC [From Eq. (i)]
⇒ FR || OE
∴ OERF is a parallelogram.
So, ∠ ERF = ∠EOF = 90°
(Opposite angle of a quadrilateral is equal)
Thus, PQRS is a parallelogram with ∠R = 90°
Hence, PQRS is a rectangle.

Question 3.
ABCD is a rectangle and P, Q, R ans S are mid-points of the sides AB, BC, CD and DA, respectively. Show that the quadrilateral PQRS is a rhombus.
Solution:
Given: ABCD is a rectangle.
NCERT Solutions for Class 9 Maths Chapter 9 Quadrilaterals Ex 9.2 img 4
∴ ∠A = ∠B = ∠C= ∠D = 90°
and AD = BC, AB = CD
Also, given P, Q, R and S are mid-points of AB, BC, CD and DA .respectively.
∴ PQ || BD and PQ = \(\frac { 1 }{ 2 }\) BD
In rectangle ABCD,
AC = BD
∴ PQ = SR …(ii)
Now, in ∆ASP and ∆BQP
AP = BP (Given)
AS = BQ (Given)
∠A = ∠B (Given)
∴ ∆ASP ≅ ∆BQP (By SAS)
∴ SP = PQ (By CPCT)…(ii)
Similarly, in ∆RDS and ∆RCQ,
SD = CQ (Given)
DR = RC (Given)
∠C=∠D (Given)
∴ ∆RDS ≅ ∆RCQ (By SAS)
∴ SR = RQ (By CPCT)…(iii)
From Eqs. (i), (ii) and (iii), it is clear that quadrilateral PQRS is a rhombus.

Question 4.
ABCD is a trapezium in which AB | | DC, BD is a diagonal and E is the mid-point of AD. A line is drawn through E parallel to AB intersecting BC at F (see figure). Show that F is the mid-point of BC.
NCERT Solutions for Class 9 Maths Chapter 9 Quadrilaterals Ex 9.2 img 5
Solution:
Given: ABCD is a trapezium in which AB || CD and E is mid-point of AD and EF || AB.
In ∆ABD, we have
EP\\AB

and E is mid-point of AD.
So, by theorem, if a line drawn through the mid-point of one side of a triangle parallel to another side bisect the third side.
∴ P is mid-point of BD.
Similarly, in ∆ BCD, we have,
PF || CD (Given)
and P is mid-point of BD.
So, by converse of mid-point theorem, F is mid-point of CB.

Question 5.
In a parallelogram ABCD, E and F are the mid-points of sides AB and CD respectively (see figure). Show that the line segments AF and EC trisect the diagonal BD.
NCERT Solutions for Class 9 Maths Chapter 9 Quadrilaterals Ex 9.2 img 7
Solution:
Given: ABCD is a parallelogram and E, F are the mid-points of sides AB and CD respectively.
To prove: Line segments AF and EC trisect the diagonal BD.
Proof: Since, ABCD is a parallelogram.
AB || DC
and AB = DC (Opposite sides of a parallelogram)
⇒ AE || FC and \(\frac { 1 }{ 2 }\) AB = \(\frac { 1 }{ 2 }\) DC
⇒ AF || FC and AF = FC
∴ AECF is a parallelogram.
∴ AF || FC
⇒ EQ || AP and FP || CQ
In ∆ BAP, E is the mid-point of AB and EQ || AP, so Q is the mid-point of BP.
(By converse of mid-point theorem)
∴ BQ = PQ ….(i)
Again, in ∆DQC, F is the mid-point of DC and FP || CQ, so P is the mid-point of DQ. (By converse of mid-point theorem)
∴ QP = DP …(ii)
From Eqs. (i) and (ii), we get
BQ = PQ = PD
Hence, CE and AF trisect the diagonal BD.

Question 6.
Show that the line segments joining the mid-points of the opposite sides of a quadrilateral bisect each other.
Solution:
Let ABCD is a quadrilateral and P, Q, R and S are the mid-points of the sides AB, BC, CD and DA, respectively, i.e., AS = SD, AP = BP, BQ = CQ and CR = DR. We have to show that PR and SQ bisect each other i.e., SO = OQ and PO = OR.
NCERT Solutions for Class 9 Maths Chapter 9 Quadrilaterals Ex 9.2 img 8
Now, in ∆ADC, S and R are mid-points of AD and CD.
We know that, the line segment joining the mid-points of two sides of a triangle is parallel to the third side. (By mid-point theorem)
∴ SR || AC and SR = \(\frac { 1 }{ 2 }\) AC …(i)
Similarly, in ∆ ABC, P and Q are mid-points of AB and BC.
∴ PQ || AC and PQ = \(\frac { 1 }{ 2 }\) AC (By mid-point theorem)…(ii)
From Eqs. (i) and (ii), we get
PQ || SR
and PQ = SR = \(\frac { 1 }{ 2 }\) AC
∴ Quadrilateral PQRS is a parallelogram whose diagonals are SQ and PR. Also, we know that diagonals of a parallelogram bisect each other. So, SQ and PR bisect each other.

Question 7.
ABC is a triangle right angled at C. A line through the mid-point M of hypotenuse AB and parallel to BC intersects AC at D. Show that
(i) D is the mid-point of AC
(ii) MD ⊥ AC
(iii) CM = MA = \(\frac { 1 }{ 2 }\) AB
Solution:
Given: ABC is a right angled triangle.
∠C = 90°
NCERT Solutions for Class 9 Maths Chapter 9 Quadrilaterals Ex 9.2 img 9
and M is the mid-point of AB.
Also, DM || BC
(i) In ∆ ABC, BC || MD and M is mid-point of AB.
∴ D is the mid-point of AC. (By converse of mid-point theorem)
(ii) Since, MD || BC and CD is transversal
∴ ∠ADM = ∠ACB (Corresponding angles)
But ∠ACB = 90°
∴ ∠ADM = 90° ⇒ MD ⊥ AC
(iii) Now, in ∆ ADM and ∆ CDM, we have
DM = MD (Common)
AD = CD (∵ D is mid point of AC)
∴ ∠ADM = ∠MDC (Each equal to 90°)
∴ ∆ ADM = ∆ CDM (By SAS)
∴ CM = AM (By CPCT)…(i)
Also, M is mid-point of AB.
∴ AM – BM = \(\frac { 1 }{ 2 }\) AB ….(ii)
From Eqs. (i) and (ii), we get
CM = AM = \(\frac { 1 }{ 2 }\) AB
Hence proved.
We hope the NCERT Solutions for Class 9 Maths Chapter 9 Quadrilaterals Ex 9.2 help you. If you have any query regarding NCERT Solutions for Class 9 Maths Chapter 9 Quadrilaterals Ex 9.2, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 9 Maths Chapter 8 Linear Equations in Two Variables Ex 8.4

March 14, 2023

NCERT Solutions for Class 9 Maths Chapter 8 Linear Equations in Two Variables Ex 8.4 are part of NCERT Solutions for Class 9 Maths. Here we have given NCERT Solutions for Class 9 Maths Chapter 8 Linear Equations in Two Variables Ex 8.4.

Board	CBSE
Textbook	NCERT
Class	Class 9
Subject	Maths
Chapter	Chapter 8
Chapter Name	Linear Equations in Two Variables
Exercise	Ex 8.4
Number of Questions Solved	2
Category	NCERT Solutions

NCERT Solutions for Class 9 Maths Chapter 8 Linear Equations in Two Variables Ex 8.4

Question 1.
Give the geometric representations of y = 3 as an equation
(i) in one variable.
(ii) in two variables.
Solution:
The given linear equation is
y=3 …(i)
(i) The representation of the solution on the number line is shown in the figure below, where y = 3 is treated as an equation in one variable.
NCERT Solutions for Class 9 Maths Chapter 8 Linear Equations in Two Variables Ex 8.4 img 1
(ii) We know that, y = 3 can be written as
0 . x + y = 3
which is a linear equation in the variables x and y. This is represented by a line. Now, all the values of x are permissible because 0 . x is always 0.
However, y must satisfy the equation y = 3.
Note that, the graph AB is a line parallel to the x-axis and at a distance of 3 units to the upper side of it.
NCERT Solutions for Class 9 Maths Chapter 8 Linear Equations in Two Variables Ex 8.4 img 2

Question 2.
Give the geometric representations of 2x + 9 = 0 as an equation
(i) in one variable.
(ii) in two variables.
Solution:
The given linear equation is
2x + 9=0
⇒ x = –\(\frac { 9 }{ 2 }\) …. ( i)
(i) The representation of the solution on the number line is shown in the figure below, where x = –\(\frac { 9 }{ 2 }\) is treated as an equation in one variable.
NCERT Solutions for Class 9 Maths Chapter 8 Linear Equations in Two Variables Ex 8.4 img 3
(ii) We know that, 2x + 9= 0 can be written as
2x + 0y + 9 = 0
which is a linear equation in two variables x and y.
This is represented by a line.
Now, all the values of y are permissible because 0 .
y is always 0.
However, x must satisfy the equation 2x + 9 = 0.
Note that, the graph AB is a line parallel to the y-axis and at a distance of – \(\frac { 9 }{ 2 }\) = – 4.5 to the left of it.

We hope the NCERT Solutions for Class 9 Maths Chapter 8 Linear Equations in Two Variables Ex 8.4 help you. If you have any query regarding NCERT Solutions for Class 9 Maths Chapter 8 Linear Equations in Two Variables Ex 8.4, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 9 Maths Chapter 8 Linear Equations in Two Variables Ex 8.3

March 14, 2023

NCERT Solutions for Class 9 Maths Chapter 8 Linear Equations in Two Variables Ex 8.3 are part of NCERT Solutions for Class 9 Maths. Here we have given NCERT Solutions for Class 9 Maths Chapter 8 Linear Equations in Two Variables Ex 8.3.

Board	CBSE
Textbook	NCERT
Class	Class 9
Subject	Maths
Chapter	Chapter 8
Chapter Name	Linear Equations in Two Variables
Exercise	Ex 8.3
Number of Questions Solved	8
Category	NCERT Solutions

NCERT Solutions for Class 9 Maths Chapter 8 Linear Equations in Two Variables Ex 8.3

Question 1.
Draw the graph of each of the following linear equations in two variables
(i) x + y = 4
(ii) x – y = 2
(iii) y = 3x
(iv) 3 = 2x + y
Solution:
(i) x + y = 4
To draw the graph, we need atleast two solutions of the
NCERT Solutions for Class 9 Maths Chapter 8 Linear Equations in Two Variables Ex 8.3 img 1
So, draw the graph by plotting the two points from table and then joining by a line.

(ii) x – y = 2
To draw the graph, we need atleast two solutions of the equation.
NCERT Solutions for Class 9 Maths Chapter 8 Linear Equations in Two Variables Ex 8.3 img 3
So, draw the graph by plotting the two points from table and then joining by a line.

(iii) y = 3x
To draw the graph, we need atleast two solutions of the equation.
NCERT Solutions for Class 9 Maths Chapter 8 Linear Equations in Two Variables Ex 8.3 img 5
So, draw the graph by plotting the two points from table and then by

(iv) 3 = 2x + y
To draw the graph, we need atleast two solutions of the equation.
NCERT Solutions for Class 9 Maths Chapter 8 Linear Equations in Two Variables Ex 8.3 img 7
So, draw the graph by plotting the points from the table and the by joining the same by a line.

Question 2.
Give the equations of two lines passing through (2,14). How many more such lines are there, and why?
Solution:
Here, (2,14) is a solution of a linear equation. One example of such a linear equation is y = 7x, others are x + y- 16.
There are infinitely many lines because there are infinitely many linear equations which are satisfied by the. coordinates of the point (2,14).

Question 3.
If the point (3,4) lies on the graph of the equation 3y – ax – 7, find the value of a.
Solution:
If the point (3, 4) lies on the graph, then it will satisfies the equations.
Hence, 3 (4) – a (3) – 7 = 0
⇒ 12 – 3a – 7 = 0
⇒ 5 – 3a = 0
⇒ 3a = 5
⇒ a = \(\frac { 5 }{ 3 }\)

Question 4.
The taxi fare in a city is as follows :
For the first kilometre, the fare is ₹ 8 and for the subsequent distance it is ₹5 per km. Taking the distance covered as x km and the total fare as ₹y, write a linear equation for this information and draw its graph.
Solution:
Distance covered = x km = 1+ (x – 1) km
Fare for first kilometre = ₹ 8
Fare for next (x-1 ) km = (x – 1) x 5 = 5(x-1)
According to question, total fare = y
NCERT Solutions for Class 9 Maths Chapter 8 Linear Equations in Two Variables Ex 8.3 img 9
Now, plot the points A (0,3) and B (1, 8) on a graph paper and joining them, to form q line AB.

Question 5.
From the choices given below, choose the equation whose graphs are given in Fig. (a) and Fig. (b).
NCERT Solutions for Class 9 Maths Chapter 8 Linear Equations in Two Variables Ex 8.3 img 10

For Fig. (a)
(i) y = x
(ii) x + y = 0
(iii) y = 2x
(iv) 2 + 3y = 7x

For Fig. (b)
(i) y = x + 2
(ii) y = x – 2
(iii) y = – x + 2
(iv) x + 2y = 6
Solution:
In Fig. (a), we observe that the points (-1,1) and (1,-1) passes through the equation x + y = 0
∵ At (-1,1) x + y = -1 + 1 = 0
and at (1, -1) x + y = 1 -1 = 0
In Fig. (b), we observe that the points (-1, 3), (0, 2) and (2, 0) passes through the equation x + y = 2.
∵ At (-1,3), x + y = -1 + (3) = + 2
At ( 0 , 2 ) , x +y = 0+2 = 2
and at ( 2, 0 ) x+ y = 2 + 0= 2

Question 6.
If the work done by a body on application of a constant force is directly proportional to the distance travelled by the body, express this in the form of an equation in two variables and draw the graph of the same by taking the constant force as 5 units. Also, read from the graph the work done when the distance travelled by the body is.
(i) 2 units
(ii) 0 unit.
Solution:
Given: that, work done by a body on application of a constant force is directly proportional to the distance travelled by the body.
i.e., (W) work done ∝ distance (s)
⇒ W = F. s
(Where, F = arbitrary constant which take the value 5 units)
∴ W = 5s …(i)
NCERT Solutions for Class 9 Maths Chapter 8 Linear Equations in Two Variables Ex 8.3 img 11
Now, plot the points O (0,0), A (1,5) and B (2,10) on graph paper and joining them to get a line AB.

(i) From point B (2,10), draw a line parallel to OY to intersect the x-axis at (2,0) and draw a line parallel to x-axis intersect at C (0, 10).
∴ Work done when the distance travelled by the body is 2 units = 10units.
(ii) Clearly y = 0 when x = 0 so the work done when the distance travelled by the body is 0 unit.

Question 7.
Yamini and Fatima, two students of class IX of a school, together contributed ₹100 towards the Prime Minister’s Relief Fund to help the earthquake victims. Write a linear equation which satisfies this data. (You may take their contributions as ₹ x and ₹ y.) Draw the graph of the same.
Solution:
Let the contributions of Yamini and Fatima together towards the Prime Minister’s Relief Fund to help the earthquake victims are ₹x and ₹ y, respectively.
Then, by given condition,
NCERT Solutions for Class 9 Maths Chapter 8 Linear Equations in Two Variables Ex 8.3 img 13
Here, we plot the points B (0,100) and A (100, 0) on graph paper and join all these points to form a line AB.

Question 8.
In countries like USA and Canada, temperature is measured in Fahrenheit, whereas in countries like India, it is measured in Celsius. Here is a linear equation that converts Fahrenheit to Celsius.
F = (\(\frac { 9 }{ 5 }\) ) c + 32
(i) Draw the graph of the linear equation above using Celsius for x-axis and Fahrenheit for y-axis.
(ii) If the temperature is 30 °C, what is the temperature in Fahrenheit?
(iii) If the temperature is 95°F, what is the temperature in Celsius?
(iv) If the temperature is 0°C, what is the temperature in Fahrenheit and if the temperature is 0°F, what is the temperature in Celsius?
(v) Is there a temperature which is numerically the same in both Fahrenheit and Celsius? If yes, find it.
Solution:
(i) Given, linear equation in Fahrenheit and Celsius is
NCERT Solutions for Class 9 Maths Chapter 8 Linear Equations in Two Variables Ex 8.3 img 15

(ii) If temperature is 30°C i.e., C = 30°C
Then, from Eq. (i), we get
F = \(\frac { 9 }{ 5 }\) x 30 + 32 = 9 x 6 + 32 = 54 + 32 = 86
∴ Temperature in Fahrenheit = 86°F

(iii) If temperature is 95°F i.e., F= 95°F
Then, from Eq. (i),we get
5 x 95 – 9C = 160 ⇒ 9C= 475 – 160= 315 ⇒ C= 35°
∴ Temperature in Celsius = 35°C

(iv) If the temperature is 0°C i.e., C= 0
Then, from Eq. (i), we get
F = \(\frac { 9 }{ 5 }\) x 0 + 32 = -32
∴ Temperature in Fahrenheit = 32°F
If the temperature is 0°F i.e., F = 0
Then, from Eq. (i), we get
5 x 0 – 9C = 160
⇒ C = \(\frac { -160 }{ 9 }\) = – 17.8° C (Approx.)
∴ Temperature in Celsius = -17.8° C

(v) Yes, if we take both temperature are equal i.e., C = F.
Now, from Eq. (i), we get
F = \(\frac { 9 }{ 5 }\) F + 32
⇒ 5F = 9F + 160
⇒ -4F = 160
⇒ F = -40°
∴ F = C = – 40°
We hope the NCERT Solutions for Class 9 Maths Chapter 8 Linear Equations in Two Variables Ex 8.3 help you. If you have any query regarding NCERT Solutions for Class 9 Maths Chapter 8 Linear Equations in Two Variables Ex 8.3, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 9 Maths Chapter 8 Linear Equations in Two Variables Ex 8.2

March 14, 2023

NCERT Solutions for Class 9 Maths Chapter 8 Linear Equations in Two Variables Ex 8.2 are part of NCERT Solutions for Class 9 Maths. Here we have given NCERT Solutions for Class 9 Maths Chapter 8 Linear Equations in Two Variables Ex 8.2.

Board	CBSE
Textbook	NCERT
Class	Class 9
Subject	Maths
Chapter	Chapter 8
Chapter Name	Linear Equations in Two Variables
Exercise	Ex 8.2
Number of Questions Solved	4
Category	NCERT Solutions

NCERT Solutions for Class 9 Maths Chapter 8 Linear Equations in Two Variables Ex 8.2

Question 1.
Which one of the following options is true and why? y = 3x + 5 has
(i) a unique solution
(ii) only two solutions
(iii) infinitely many solutions
Solution:
(iii) A linear equation in two variables has infinitely many solutions.

Question 2.
Write four solutions for each of the following equations
(i) 2x + y = 7
(ii) πx + y = 9
(iii) x = 4y
Solution:
(i) 2x + y = 7
By inspection, x = 2 and y = 3 is a solution because for x = 2, y = 3,
2x + y = 2 x 2 + 3 = 4 + 3 = 7
Now, let us choose x = 0 with this value of x, the given equation reduces to y = 7.
So, x = 0, y = 7 is also a solution of 2x + y = 7. Similarly, taking y = 0,
the given equation reduces to 2x = 7 which has the unique solution x = \(\frac { 7 }{ 2 }\) .
So, x = \(\frac { 7 }{ 2 }\) , y = 0 is a solution of 2x + y = 7.
Finally, let us take x = 1
The given equation now reduces to 2 + y = 7 hose solution is given by y = 5.
Therefore, (1, 5) is also a solution of the given equation.
So, four of the infinitely many solutions of the given equation are (2, 3), (0, 7), ( \(\frac { 7 }{ 2 }\) , 0) and (1,5).

(ii) πx + y = 9
NCERT Solutions for Class 9 Maths Chapter 8 Linear Equations in Two Variables Ex 8.2 img 1
Now, let us choose x = 0 with this value of x,
the given equation reduces to y = 9 which has a unique solution y = 9.
So, x = 0, y = 9 is also a solution of πx + y = 9
Similarly, taking y = 0, the given equation reduces to x = \(\frac { 9 }{ \pi }\) So, x = \(\frac { 9 }{ \pi }\) ,y = 0 is a solution of πx + y = 9as well.
Finally, let us take x = 7 the given equation now reduces to \(\frac { 22 }{ 7 }\) . 7 + y = 9
whose solution is given by y = -13.
Therefore; (7,-13) is also a solution of the given equation.
So, four of the infinitely many solutions of the given equation are
NCERT Solutions for Class 9 Maths Chapter 8 Linear Equations in Two Variables Ex 8.2 img 2
(iii) x = 4y ⇒ x – 4y = 0
By inspection, x = 0, y = 0 is a solution because for x = y = 0, 0 – 4 x 0 = 0 – 0 = 0, it satisfies.
Now, let us choose x = 4 with this value of x,
the given equation reduces to y = 1 which has a unique solution y = 1.
So, x = 4, y = 1 is also a solution , of x – 4y = 0. Similarly, taking y = \(\frac { 1 }{ 2 }\) , the given equation reduces to x = 2.
So , x = 2, y = \(\frac { 1 }{ 2 }\) is a solution x – 4y = 0 as well.
Finally, let us take x = 1, the given equation now reduces to 1 – 4y = 0
whose solution is given by y = \(\frac { 1 }{ 4 }\). Therefore, (1,1/4) is also a solution of the given equation. So, four
of the infinitely many solutions of the given equation are

Question 3.
Check which of the following are solution of the equation x – 2y = 4 and which are not?
(i) (0, 2)
(ii) (2,0)
(iii) (4,0)
(iv) (√2,4√2)
(v) (1,1)
Solution:
(i) Take x – 2y and put x = 0, y = 2,
we get 0 – 2 x 2 = 0 – 4 = -4 ≠ 4
Hence, (0, 2) is not a solution of x – 2y = 4.

(ii) Take x – 2y and put x = 2, y = 0,
we get 2 – 2 x 0 = 2 – 0 = 2 ≠ 4
Hence, (2, 0) is not a solution of x – 2y = 4.

(iii) Take x – 2y and put x = 4, y = 0;
we get 4 – 2 x 0 = 4 – 0 = 4
Hence, (4, 0) is a solution of x – 2y = 4.

(iv) Take x – 2y and put x = √2, y = 4√2, we get
√2 – 2 x 4√2 = √2 – 8√2 =-7√2 ≠ 4
Hence, (√2,4√2) is not a solution of x – 2y = 4

(v) Take x – 2y and put x = 1, y = 1,
we get 1 – 2 x 1 = 1 – 2 = -1 ≠ 4
Hence, (1,1) is not a solution of x – 2y = 4.

Question 4.
Find the value of k, if x = 2, y = 1 is a solution of the equation 2x + 3 y = k.
Solution:
Take 2x + 3y = k
Put x = 2, y = 1 then we get, 2 x 2 + 3 x 1 = k
⇒ 4 + 3 = k
⇒ k = 7
We hope the NCERT Solutions for Class 9 Maths Chapter 8 Linear Equations in Two Variables Ex 8.2 help you. If you have any query regarding NCERT Solutions for Class 9 Maths Chapter 8 Linear Equations in Two Variables Ex 8.2, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 9 Maths Chapter 7 Heron’s Formula Ex 7.2

March 14, 2023

NCERT Solutions for Class 9 Maths Chapter 7 Heron’s Formula Ex 7.2 are part of NCERT Solutions for Class 9 Maths. Here we have given NCERT Solutions for Class 9 Maths Chapter 7 Heron’s Formula Ex 7.2.

Board	CBSE
Textbook	NCERT
Class	Class 9
Subject	Maths
Chapter	Chapter 7
Chapter Name	Heron’s Formula
Exercise	Ex 7.2
Number of Questions Solved	9
Category	NCERT Solutions

NCERT Solutions for Class 9 Maths Chapter 7 Heron’s Formula Ex 7.2

Question 1.
A park, in the shape of a quadrilateral ABCD, has ∠C = 90°, AB = 9m,BC = 12m,CD = 5m and AD = 8 m.
How much area does it occupy?
Solution:
In right ∆ BCD

NCERT Solutions for Class 9 Maths Chapter 7 Heron's Formula Ex 7.2 img 1

Question 2.
Find the area of a quadrilateral ABCD in which AB = 3 cm, BC = 4 cm, CD = 4 cm, DA = 5 cm and AC = 5 cm.
Solution:
Area of quadrilateral ABCD = Area of ∆ABC+ Area of ∆ACD
NCERT Solutions for Class 9 Maths Chapter 7 Heron's Formula Ex 7.2 img 3
In ∆ ABC,
We have, AB = 3 cm, BC = 4 cm, CA = 5 cm
Therefore, AB² + BC² = 3² + 4² = 9 + 16 = 25 = (5)2 = CA²
Hence, ∆ ABC is a right triangle

Question 3.
Radha made a picture of an aeroplane with coloured paper as shown in figure. Find the total area of the paper used.

Solution:
For part I
It is a triangle with sides 5 cm, 5cm and 1cm
NCERT Solutions for Class 9 Maths Chapter 7 Heron's Formula Ex 7.2 img 6
For part II
It is a rectangle with sides 6.5 cm and 1 cm
∴ Area of part II = 6.5 x 1
(∵ Area of rectangle = Length x Breadth)
= 6.5 cm2

For part III
NCERT Solutions for Class 9 Maths Chapter 7 Heron's Formula Ex 7.2 img 7

Question 4.
A triangle and a parallelogram have the same base and the same area. If the sides of the triangle are 26 cm, 28 cm and 30 cm, and the parallelogram stands on the base 28 cm, find the height of the parallelogram
Solution:
Let ABC be a triangle with sides
AB = 26 cm, BC = 28 cm, CA = 30 cm
NCERT Solutions for Class 9 Maths Chapter 7 Heron's Formula Ex 7.2 img 9

We know that,
Area of parallelogram = Base x Height …(i)
We have,
Area of parallelogram = Area of ∆ ABC (Given)
= 336 cm²
From Eq. (i), we have
Base x Height = 336
⇒ 28 x Height = 336
⇒ Height = \(\frac { 336 }{ 28 }\)
⇒ Height =12cm

Question 5.
A rhombus shaped field has green grass for 18 cows to graze. If each side of the rhombus is 30 m and its longer diagonal is 48 m, how much area of grass field will each cow be getting?
Solution:
Let ABCD be a rhombus.
Area of the rhombus ABCD = 2x area of ∆ ABD …(i)
(Since, in a rhombus diagonals divides two equal parts)
NCERT Solutions for Class 9 Maths Chapter 7 Heron's Formula Ex 7.2 img 11

Question 6.
An umbrella is made by stitching 10 triangular pieces of cloth of two different colours (see figure), each piece measuring 20 cm, 50 cm and 50 cm. How much cloth of each colour is required for the umbrella?
NCERT Solutions for Class 9 Maths Chapter 7 Heron's Formula Ex 7.2 img 13
Solution:
In an umbrella, each triangular piece is an isosceles triangle with sides 50 cm, 50 cm, 20 cm.

Since, there are 10 triangular piece, in those of them 5.5 are of different colours.
Hence, total area of cloth of each colour = 5 x 200√6 cm² = 1000√6 cm²

Question 7.
A kite in the shape of a square with a diagonal 32 cm and an isosceles triangle of base 8 cm and sides 6 cm each is to be made of three different shades as shown in figure. How much paper of each shade has been used in it?
NCERT Solutions for Class 9 Maths Chapter 7 Heron's Formula Ex 7.2 img 15
Solution:
Since, the kite is in the shape of a square

Each diagonal of square =32 cm (Given)
We know that, the diagonals of a square bisect each other at right angle.

Hence, paper of I colour has been used = 256 cm²
Paper of II colour has been used = 256 cm²
Paper of III colour has been used = 17.92 cm²

Question 8.
A floral design on a floor is made up of 16 tiles which are triangular, the sides of the triangle being 9 cm, 28 cm and 35 cm (see figure). Find the cost of polishing the tiles at the rate of 50 paise per cm .
NCERT Solutions for Class 9 Maths Chapter 7 Heron's Formula Ex 7.2 img 19
Solution:
Given, the sides of a triangular tiles are 9 cm, 28 cm and 35 cm.
For each triangular tile, we have

Question 9.
A field is in the shape of a trapezium whose parallel sides are 25 m and 10 m. The non-parallel sides are 14 m and 13 m. Find the area of the field.
Solution:
Here, ABCD is a trapezium and AB || DC.
NCERT Solutions for Class 9 Maths Chapter 7 Heron's Formula Ex 7.2 img 21

We hope the NCERT Solutions for Class 9 Maths Chapter 7 Heron’s Formula Ex 7.2 help you. If you have any query regarding NCERT Solutions for Class 9 Maths Chapter 7 Heron’s Formula Ex 7.2, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 9 Maths Chapter 6 Coordinate Geometry Ex 6.3

March 14, 2023

NCERT Solutions for Class 9 Maths Chapter 6 Coordinate Geometry Ex 6.3 are part of NCERT Solutions for Class 9 Maths. Here we have given NCERT Solutions for Class 9 Maths Chapter 6 Coordinate Geometry Ex 6.3.

Board	CBSE
Textbook	NCERT
Class	Class 9
Subject	Maths
Chapter	Chapter 6
Chapter Name	Coordinate Geometry
Exercise	Ex 6.3
Number of Questions Solved	2
Category	NCERT Solutions

NCERT Solutions for Class 9 Maths Chapter 6 Coordinate Geometry Ex 6.3

Question 1.
In which quadrant or on which axis does each of the points (-2,4), (3, – 1), (-1,0), (1, 2) and (-3, – 5) lie? Verify your answer by locating them on the cartesian plane.
Solution:
NCERT Solutions for Class 9 Maths Chapter 6 Coordinate Geometry Ex 6.3 img 1
(i) (-2, 4) lies in II quadrant.
(ii) (3, -1) lies in IV quadrant.
(iii) (-1,0) lies on X-axis.
(iv) (1, 2) lies in I quadrant.
(v) (-3, -5) lies in III quadrant.

Question 2.
Plot the points (x, y) given in the following table on the plane, choosing suitable units of distance on the axes.

Solution:
Let 1 unit = 1 cm, then positions of given points in the cartesian plane are given below.

We hope the NCERT Solutions for Class 9 Maths Chapter 6 Coordinate Geometry Ex 6.3 help you. If you have any query regarding NCERT Solutions for Class 9 Maths Chapter 6 Coordinate Geometry Ex 6.3, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 9 Maths Chapter 6 Coordinate Geometry Ex 6.2

March 14, 2023

NCERT Solutions for Class 9 Maths Chapter 6 Coordinate Geometry Ex 6.2 are part of NCERT Solutions for Class 9 Maths. Here we have given NCERT Solutions for Class 9 Maths Chapter 6 Coordinate Geometry Ex 6.2.

Board	CBSE
Textbook	NCERT
Class	Class 9
Subject	Maths
Chapter	Chapter 6
Chapter Name	Coordinate Geometry
Exercise	Ex 6.2
Number of Questions Solved	2
Category	NCERT Solutions

NCERT Solutions for Class 9 Maths Chapter 6 Coordinate Geometry Ex 6.2

Question 1.
Write the answer of each of the following questions
(i) What is the name of horizontal and the vertical lines drawn to determine the position of any point in the Cartesian plane?
(ii) What is the name of each part of the plane formed by these two lines?
(iii) Write the name of the point where these two lines intersect.
Solution:
(i) Horizontal line is known as X-axis and vertical line is known as y – axis.
(ii) Each part of the plane formed by horizontal and vertical lines is known as quadrant.
(iii) Horizontal and vertical lines intersect at the origin.

Question 2.
See figure, and write the following
NCERT Solutions for Class 9 Maths Chapter 6 Coordinate Geometry Ex 6.2 img 1
(i) The coordinates of B.
(ii) The coordinates of C.
(iii) The point identified by the coordinates (-3, – 5).
(iv) The point identified by the coordinates (2, -4).
(v) The abscissa of the point D.
(vi) The ordinate of the point H.
(vii) The coordinates of the point I.
(viii) The coordinates of the point M.
Solution:
(i) The coordinates of B = (-5, 2) and 6 lies in \(\parallel\) quadrant.
(∵ Abscissa = – 5, Ordinate = 2)
(ii) The coordinates of C = (5, -5).
(iii) The point identified by the coordinates (- 3, -5) is E.
(iv) The point identified by the coordinates (2, – 4) is G.
(v) The abscissa of the point D = 6
(vi) The ordinate of the point H = – 3
(vii) The coordinates of the point L = (0,5) (Lies on Y-axis)
(viii) The coordinates of the point M = (-3,0) (Lies on X-axis)

We hope the NCERT Solutions for Class 9 Maths Chapter 6 Coordinate Geometry Ex 6.2 help you. If you have any query regarding NCERT Solutions for Class 9 Maths Chapter 6 Coordinate Geometry Ex 6.2, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 9 Maths Chapter 5 Triangles Ex 5.4

March 14, 2023

NCERT Solutions for Class 9 Maths Chapter 5 Triangles Ex 5.4 are part of NCERT Solutions for Class 9 Maths. Here we have given NCERT Solutions for Class 9 Maths Chapter 5 Triangles Ex 5.4.

Board	CBSE
Textbook	NCERT
Class	Class 9
Subject	Maths
Chapter	Chapter 5
Chapter Name	Triangles
Exercise	Ex 5.4
Number of Questions Solved	6
Category	NCERT Solutions

NCERT Solutions for Class 9 Maths Chapter 5 Triangles Ex 5.4

Question 1.
Show that in a right angled triangle, the hypotenuse is the longest side.
Solution:
Let ABC be a right angled triangle, such that ∠ ABC = 90°
NCERT Solutions for Class 9 Maths Chapter 5 Triangles Ex 5.4 img 1
We know that,
∠ABC + ∠BCA + ∠CAB = 180° (By A property)
⇒ 90° + ∠BCA + ∠CAB = 180°
⇒ ∠BCA + ∠CAB = 90°
From above, we have ∠ BCA and ∠ CAB are acute angles.
⇒ ∠BCA < 90°
and ∠CAB < 90°
⇒ ∠BCA < ∠ABC
and ∠CAB < ∠ABC
⇒ AB < AC and BC < AC (∵ Side opposite to greater angle is longer)
Hence, the hypotenuse (AC) is the longest side.

Question 2.
In figure, sides AB and AC of ∆ABC are extended to points P and Q respectively. Also, ∠PBC < ∠QCB. Show that AC > AB.
Solution:
We know that,
NCERT Solutions for Class 9 Maths Chapter 5 Triangles Ex 5.4 img 2
∠ACB + ∠QCB = 180° (Linear pair)…(i)
and ∠ABC+ ∠PBC = 180° (Linear pair)…(ii)
From Eqs. (i) and (ii), we have
∠ABC + ∠PBC = ∠ACB + ∠QCB ….(iii)
But ∠PBC < ∠QCB (Given)…(iv) From Eqs. (iii) and (iv), we have ∠ ABC > ∠ ACB
⇒ AC > AB
(∵ Side opposite to greater angle is longer)

Question 3.
In figure, ∠B <∠ A and ∠C <∠ D. Show that AD < BC.
Solution:
NCERT Solutions for Class 9 Maths Chapter 5 Triangles Ex 5.4 img 3

Question 4.
AB and CD are respectively the smallest and longest sides of a quadrilateral ABCD (see figure). Show that ∠ A> ∠C and ∠B > ∠D.
NCERT Solutions for Class 9 Maths Chapter 5 Triangles Ex 5.4 img 4
Solution:
Given ABCD is a quadrilateral. AB is the smallest side and CD is the longest side.
To prove ∠A > ∠C and ∠B > ∠D
Construction Join A to C and B to D.
Proof In ∆ABC, we have AB is the smallest side.
∴ AB < BC
⇒ ∠5 < ∠1
NCERT Solutions for Class 9 Maths Chapter 5 Triangles Ex 5.4 img 5
(∵ Angle opposite to longer side is greater) …(i)
In ∆ADC, we have CD is the largest side.

Question 5.
In figure, PR > PQ and PS bisect ∠QPR. Prove that
∠PSR >∠PSQ.
NCERT Solutions for Class 9 Maths Chapter 5 Triangles Ex 5.4 img 7
Solution:
In ∆ PQR, we have
PR > PQ (Given)
⇒ ∠ PQR > ∠ PRQ …(i)
(∵ Angle opposite to longer side is greater)

Question 6.
Show that of all line segments drawn from a given point not on it, the perpendicular line segment is the shortest.
Solution:
Given x is a line and A is a point not lying on x. AB ⊥ x, Cis any point on x other than B.

To prove AB∠C
⇒ AC> AB
(∵ the Side opposite to greater angle is longer)
⇒ AB < AC
Hence, the perpendicular line segment is the shortest.
We hope the NCERT Solutions for Class 9 Maths Chapter 5 Triangles Ex 5.4 help you. If you have any query regarding NCERT Solutions for Class 9 Maths Chapter 5 Triangles Ex 5.4, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 9 Maths Chapter 5 Triangles Ex 5.5

March 14, 2023

NCERT Solutions for Class 9 Maths Chapter 5 Triangles Ex 5.5 are part of NCERT Solutions for Class 9 Maths. Here we have given NCERT Solutions for Class 9 Maths Chapter 5 Triangles Ex 5.5.

Board	CBSE
Textbook	NCERT
Class	Class 9
Subject	Maths
Chapter	Chapter 5
Chapter Name	Triangles
Exercise	Ex 5.5
Number of Questions Solved	4
Category	NCERT Solutions

NCERT Solutions for Class 9 Maths Chapter 5 Triangles Ex 5.5

Question 1.
ABC is a triangle. Locate a point in the interior of ∆ ABC which is equidistant from all the vertices of ∆ ABC.
Solution:
Suppose OM and ON be the perpendicular bisectors of sides BC and AC of ∆ ABC.
NCERT Solutions for Class 9 Maths Chapter 5 Triangles Ex 5.5 img 1
So, O is equidistant from two endpoints 0 and C of line segment BC as O lies on the perpendicular bisector of BC. Similarly, O is equidistant from C and A Hence, O be an orthocentre of ∆ABC.

Question 2.
In a triangle locate a point in its interior which is equidistant from all the sides of the triangle.
Solution:
Suppose BN and CM be the bisectors of ∠ ABC and ∠ ACB, respectively intersect AC and AB at N and M, respectively.
NCERT Solutions for Class 9 Maths Chapter 5 Triangles Ex 5.5 img 2
Since, O lies on the bisector BN of ∠ ABC, so O will be equidistant from BA and BC. Again, O lies on the bisector CM of ∠ ACB.
So, O will be equidistant from CA and BC. Thus, O will be equidistant from AB, BC and CA Hence, O be a circumcentre of ∆ABC.

Question 3.
In a huge park, people are concentrated at three points (see figure)
A: where these are different slides and swings for children.
B: near which a man-made lake is situated.
C: which is near to a large parking and exist.
Where should an ice-cream parlor be set? up so that maximum number of persons can approach it?
[Hint The parlour should be equidistant from A, B and C.]
NCERT Solutions for Class 9 Maths Chapter 5 Triangles Ex 5.5 img 3
Solution:
The ice-cream parlor should be equidistant from A B and C for which the point of intersection of perpendicular bisectors of AB, BC, and CA should be situated.
So, O is the required point which is equidistant from A B and C.

Question 4.
Complete the hexagonal and star shaped Rangolies [see Fig. (i) and (ii)] by filling them with as many equilateral triangles of side 1 cm as you can. Count the number of triangles in each case. Which has more triangles?
NCERT Solutions for Class 9 Maths Chapter 5 Triangles Ex 5.5 img 4
Solution:
We first divide the hexagon into six equilateral triangles of side 5cm as follow.

We take one triangle from six equilateral triangles as shown above and make as many equilateral triangles of one side 1 cm as shown in the figure.
NCERT Solutions for Class 9 Maths Chapter 5 Triangles Ex 5.5 img 6
The number of equilateral triangles of side 1 cm = 1 + 3 + 5 + 7 + 9 = 25
So, the total number of triangles in the hexagon = 6x 25 = 150
To find the number of triangles in the Fig. (ii), we adopt the same procedure.
So, the number of triangles in the Fig. (ii) = 12 x 25 = 30Q Hence, Fig. (ii) has more triangles.
We hope the NCERT Solutions for Class 9 Maths Chapter 5 Triangles Ex 5.5 help you. If you have any query regarding NCERT Solutions for Class 9 Maths Chapter 5 Triangles Ex 5.5, drop a comment below and we will get back to you at the earliest.

Category: CBSE Class 9

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NCERT Solutions for Class 9 Maths Chapter 5 Triangles Ex 5.4

NCERT Solutions for Class 9 Maths Chapter 5 Triangles Ex 5.5