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NCERT Solutions for Class 9 Maths Chapter 7 Triangles

Leave a Comment / NCERT Solution, Class 9 / Mr Ahmad / August 10, 2022

NCERT Solutions for Class 9 Maths Chapter 7 Triangles – students will study about the congruence of triangles, rules of congruence, some properties of triangles and the inequalities in triangles in details. All These 5 exercises, in which the students are asked “to-prove” as well as application-level problems. Check previous chapter solutions – NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles

ALSO CHECK – NCERT Solutions for Class 10 Maths

NCERT Solutions Class 9 Maths Chapter 7 Triangles is prepared by our best subject experts teachers group thats help students to understand all the topics easily. These Solutions of NCERT Maths help the students in solving the problems efficiently for the upcoming exams. With the help of these NCERT Solutions for Class 9 Maths, students can understand the complex topics of class 9 Maths.

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NCERT Solutions for Class 9 Maths Chapter 7 Triangles Exercise 7.1

Question 1.
In quadrilateral ACBD, AC = AD and AB bisects ∠ A (see figure). Show that ∆ABC ≅ ∆ABD. What can you say about BC and BD?

Solution:
In quadrilateral ACBD, we have AC = AD and AB being the bisector of ∠A.
Now, In ∆ABC and ∆ABD,
AC = AD (Given)
∠ CAB = ∠ DAB ( AB bisects ∠ CAB)
and AB = AB (Common)
∴ ∆ ABC ≅ ∆ABD (By SAS congruence axiom)
∴ BC = BD (By CPCT)

Question 2.
ABCD is a quadrilateral in which AD = BC and ∠ DAB = ∠ CBA (see figure). Prove that

(i) ∆ABD ≅ ∆BAC
(ii) BD = AC
(iii) ∠ABD = ∠ BAC
Solution:
In quadrilateral ACBD, we have AD = BC and ∠ DAB = ∠ CBA

(i) In ∆ ABC and ∆ BAC,
AD = BC (Given)
∠DAB = ∠CBA (Given)
AB = AB (Common)
∴ ∆ ABD ≅ ∆BAC (By SAS congruence)

(ii) Since ∆ABD ≅ ∆BAC
⇒ BD = AC [By C.P.C.T.]

(iii) Since ∆ABD ≅ ∆BAC
⇒ ∠ABD = ∠BAC [By C.P.C.T.]

Question 3.
AD and BC are equal perpendiculars to a line segment AB (see figure). Show that CD bisects AB.

Solution:
In ∆BOC and ∆AOD, we have
∠BOC = ∠AOD
BC = AD [Given]
∠BOC = ∠AOD [Vertically opposite angles]
∴ ∆OBC ≅ ∆OAD [By AAS congruency]
⇒ OB = OA [By C.P.C.T.]
i.e., O is the mid-point of AB.
Thus, CD bisects AB.

Question 4.
l and m are two parallel lines intersected by another pair of parallel lines p and q (see figure). Show that ∆ABC = ∆CDA.

Solution:
∵ p || q and AC is a transversal,
∴ ∠BAC = ∠DCA …(1) [Alternate interior angles]
Also l || m and AC is a transversal,
∴ ∠BCA = ∠DAC …(2)
[Alternate interior angles]
Now, in ∆ABC and ∆CDA, we have
∠BAC = ∠DCA [From (1)]
CA = AC [Common]
∠BCA = ∠DAC [From (2)]
∴ ∆ABC ≅ ∆CDA [By ASA congruency]

Question 5.
Line l is the bisector of an ∠ A and ∠ B is any point on l. BP and BQ are perpendiculars from B to the arms of LA (see figure). Show that
(i) ∆APB ≅ ∆AQB
(ii) BP = BQ or B is equidistant from the arms ot ∠A.

Solution:
We have, l is the bisector of ∠QAP.
∴ ∠QAB = ∠PAB
∠Q = ∠P [Each 90°]
∠ABQ = ∠ABP
[By angle sum property of A]
Now, in ∆APB and ∆AQB, we have
∠ABP = ∠ABQ [Proved above]
AB = BA [Common]
∠PAB = ∠QAB [Given]
∴ ∆APB ≅ ∆AQB [By ASA congruency]
Since ∆APB ≅ ∆AQB
⇒ BP = BQ [By C.P.C.T.]
i. e., [Perpendicular distance of B from AP]
= [Perpendicular distance of B from AQ]
Thus, the point B is equidistant from the arms of ∠A.

Question 6.
In figure, AC = AE, AB = AD and ∠BAD = ∠EAC. Show that BC = DE.

Solution:
We have, ∠BAD = ∠EAC
Adding ∠DAC on both sides, we have
∠BAD + ∠DAC = ∠EAC + ∠DAC
⇒ ∠BAC = ∠DAE
Now, in ∆ABC and ∆ADE. we have
∠BAC = ∠DAE [Proved above]
AB = AD [Given]
AC = AE [Given]
∴ ∆ABC ≅ ∆ADE [By SAS congruency]
⇒ BC = DE [By C.P.C.T.]

Question 7.
AS is a line segment and P is its mid-point. D and E are points on the same side of AB such that ∠ BAD = ∠ ABE and ∠ EPA = ∠ DPB. (see figure). Show that
(i) ∆DAP ≅ ∆EBP
(ii) AD = BE

Solution:
We have, P is the mid-point of AB.
∴ AP = BP
∠EPA = ∠DPB [Given]
Adding ∠EPD on both sides, we get
∠EPA + ∠EPD = ∠DPB + ∠EPD
⇒ ∠APD = ∠BPE

(i) Now, in ∆DAP and ∆EBP, we have
∠PAD = ∠PBE [ ∵∠BAD = ∠ABE]
AP = BP [Proved above]
∠DPA = ∠EPB [Proved above]
∴ ∆DAP ≅ ∆EBP [By ASA congruency]

(ii) Since, ∆ DAP ≅ ∆ EBP
⇒ AD = BE [By C.P.C.T.]

Question 8.
In right triangle ABC, right angled at C, M is the mid-point of hypotenuse AB. C is joined to M and produced to a point D such that DM = CM. Point D is joined to point B (see figure). Show that
(i) ∆AMC ≅ ∆BMD
(ii) ∠DBC is a right angle
(iii) ∆DBC ≅ ∆ACB

(iv) CM = 1/2 AB
Solution:
Since M is the mid – point of AB.
∴ BM = AM

(i) In ∆AMC and ∆BMD, we have
CM = DM [Given]
∠AMC = ∠BMD [Vertically opposite angles]
AM = BM [Proved above]
∴ ∆AMC ≅ ∆BMD [By SAS congruency]

(ii) Since ∆AMC ≅ ∆BMD
⇒ ∠MAC = ∠MBD [By C.P.C.T.]
But they form a pair of alternate interior angles.
∴ AC || DB
Now, BC is a transversal which intersects parallel lines AC and DB,
∴ ∠BCA + ∠DBC = 180° [Co-interior angles]
But ∠BCA = 90° [∆ABC is right angled at C]
∴ 90° + ∠DBC = 180°
⇒ ∠DBC = 90°

(iii) Again, ∆AMC ≅ ∆BMD [Proved above]
∴ AC = BD [By C.P.C.T.]
Now, in ∆DBC and ∆ACB, we have
BD = CA [Proved above]
∠DBC = ∠ACB [Each 90°]
BC = CB [Common]
∴ ∆DBC ≅ ∆ACB [By SAS congruency]

(iv) As ∆DBC ≅ ∆ACB
DC = AB [By C.P.C.T.]
But DM = CM [Given]
∴ CM = 1/2DC = 1/2AB
⇒ CM = 1/2AB

NCERT Solutions for Class 9 Maths Chapter 7 Triangles Exercise 7.2

Question 1.
In an isosceles triangle ABC, with AB = AC, the bisectors of ∠B and ∠C intersect each other at 0. Join A to 0. Show that
(i) OB = OC
(ii) AO bisects ∠A
Solution:
i) in ∆ABC, we have
AB = AC [Given]
∴ ∠ABC = ∠ACB [Angles opposite to equal sides of a A are equal]

⇒ 12∠ABC = 12∠ACB
or ∠OBC = ∠OCB
⇒ OC = OB [Sides opposite to equal angles of a ∆ are equal]

(ii) In ∆ABO and ∆ACO, we have
AB = AC [Given]
∠OBA = ∠OCA [ ∵12∠B = 12∠C]
OB = OC [Proved above]
∆ABO ≅ ∆ACO [By SAS congruency]
⇒ ∠OAB = ∠OAC [By C.P.C.T.]
⇒ AO bisects ∠A.

Question 2.
In ∆ABC, AD is the perpendicular bisector of BC (see figure). Show that ∆ ABC is an isosceles triangle in which AB = AC.

Solution:
Since AD is bisector of BC.
∴ BD = CD
Now, in ∆ABD and ∆ACD, we have
AD = DA [Common]
∠ADB = ∠ADC [Each 90°]
BD = CD [Proved above]
∴ ∆ABD ≅ ∆ACD [By SAS congruency]
⇒ AB = AC [By C.P.C.T.]
Thus, ∆ABC is an isosceles triangle.

Question 3.
ABC is an isosceles triangle in which altitudes BE and CF are drawn to equal sides AC and AB respectively (see figure). Show that these altitudes are equal.

Solution:
∆ABC is an isosceles triangle.
∴ AB = AC
⇒ ∠ACB = ∠ABC [Angles opposite to equal sides of a A are equal]
⇒ ∠BCE = ∠CBF
Now, in ∆BEC and ∆CFB
∠BCE = ∠CBF [Proved above]
∠BEC = ∠CFB [Each 90°]
BC = CB [Common]
∴ ∆BEC ≅ ∆CFB [By AAS congruency]
So, BE = CF [By C.P.C.T.]

Question 4.
ABC is a triangle in which altitudes BE and CF to sides AC and AB are equal (see figure).

Show that
(i) ∆ABE ≅ ∆ACF
(ii) AB = AC i.e., ABC is an isosceles triangle.
Solution:
(i) In ∆ABE and ∆ACE, we have
∠AEB = ∠AFC
[Each 90° as BE ⊥ AC and CF ⊥ AB]
∠A = ∠A [Common]
BE = CF [Given]
∴ ∆ABE ≅ ∆ACF [By AAS congruency]

(ii) Since, ∆ABE ≅ ∆ACF
∴ AB = AC [By C.P.C.T.]
⇒ ABC is an isosceles triangle.

Question 5.
ABC and DBC are isosceles triangles on the same base BC (see figure). Show that ∠ ABD = ∠ACD.

Solution:
In ∆ABC, we have
AB = AC [ABC is an isosceles triangle]
∴ ∠ABC = ∠ACB …(1)
[Angles opposite to equal sides of a ∆ are equal]
Again, in ∆BDC, we have
BD = CD [BDC is an isosceles triangle]
∴ ∠CBD = ∠BCD …(2)
[Angles opposite to equal sides of a A are equal]
Adding (1) and (2), we have
∠ABC + ∠CBD = ∠ACB + ∠BCD
⇒ ∠ABD = ∠ACD.

Question 6.
∆ABC is an isosceles triangle in which AB = AC. Side BA is produced to D such that AD = AB (see figure). Show that ∠BCD is a right angle.

Solution:
AB = AC [Given] …(1)
AB = AD [Given] …(2)
From (1) and (2), we have
AC = AD
Now, in ∆ABC, we have
∠ABC + ∠ACB + ∠BAC = 180° [Angle sum property of a A]
⇒ 2∠ACB + ∠BAC = 180° …(3)
[∠ABC = ∠ACB (Angles opposite to equal sides of a A are equal)]
Similarly, in ∆ACD,
∠ADC + ∠ACD + ∠CAD = 180°
⇒ 2∠ACD + ∠CAD = 180° …(4)
[∠ADC = ∠ACD (Angles opposite to equal sides of a A are equal)]
Adding (3) and (4), we have
2∠ACB + ∠BAC + 2 ∠ACD + ∠CAD = 180° +180°
⇒ 2[∠ACB + ∠ACD] + [∠BAC + ∠CAD] = 360°
⇒ 2∠BCD +180° = 360° [∠BAC and ∠CAD form a linear pair]
⇒ 2∠BCD = 360° – 180° = 180°
⇒ ∠BCD = 180∘2 = 90°
Thus, ∠BCD = 90°

Question 7.
ABC is a right angled triangle in which ∠A = 90° and AB = AC, find ∠B and ∠C.
Solution:
In ∆ABC, we have AB = AC [Given]
∴ Their opposite angles are equal.

⇒ ∠ACB = ∠ABC
Now, ∠A + ∠B + ∠C = 180° [Angle sum property of a ∆]
⇒ 90° + ∠B + ∠C = 180° [∠A = 90°(Given)]
⇒ ∠B + ∠C= 180°- 90° = 90°
But ∠B = ∠C
∠B = ∠C = 90∘2 = 45°
Thus, ∠B = 45° and ∠C = 45°

Question 8.
Show that the angles of an equilateral triangle are 60° each.
Solution:
In ∆ABC, we have

AB = BC = CA
[ABC is an equilateral triangle]
AB = BC
⇒ ∠A = ∠C …(1) [Angles opposite to equal sides of a A are equal]
Similarly, AC = BC
⇒ ∠A = ∠B …(2)
From (1) and (2), we have
∠A = ∠B = ∠C = x (say)
Since, ∠A + ∠B + ∠C = 180° [Angle sum property of a A]
∴ x + x + x = 180o
⇒ 3x = 180°
⇒ x = 60°
∴ ∠A = ∠B = ∠C = 60°
Thus, the angles of an equilateral triangle are 60° each.

NCERT Solutions for Class 9 Maths Chapter 7 Triangles Exercise 7.3

Question 1.
∆ABC and ∆DBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC (see figure). If AD is extended to intersect BC at P, show that

(i) ∆ABD ≅ ∆ACD
(ii) ∆ABP ≅ ∆ACP
(iii) AP bisects ∠A as well as ∠D
(iv) AP is the perpendicular bisector of BC.
Solution:
(i) In ∆ABD and ∆ACD, we have
AB = AC [Given]
AD = DA [Common]
BD = CD [Given]
∴ ∆ABD ≅ ∆ACD [By SSS congruency]
∠BAD = ∠CAD [By C.P.C.T.] …(1)

(ii) In ∆ABP and ∆ACP, we have
AB = AC [Given]
∠BAP = ∠CAP [From (1)]
∴ AP = PA [Common]
∴ ∆ABP ≅ ∆ACP [By SAS congruency]

(iii) Since, ∆ABP ≅ ∆ACP
⇒ ∠BAP = ∠CAP [By C.P.C.T.]
∴ AP is the bisector of ∠A.
Again, in ∆BDP and ∆CDP,
we have BD = CD [Given]
DP = PD [Common]
BP = CP [ ∵ ∆ABP ≅ ∆ACP]
⇒ A BDP = ACDP [By SSS congruency]
∴ ∠BDP = ∠CDP [By C.P.C.T.]
⇒ DP (or AP) is the bisector of ∠BDC
∴ AP is the bisector of ∠A as well as ∠D.

(iv) As, ∆ABP ≅ ∆ACP
⇒ ∠APS = ∠APC, BP = CP [By C.P.C.T.]
But ∠APB + ∠APC = 180° [Linear Pair]
∴ ∠APB = ∠APC = 90°
⇒ AP ⊥ BC, also BP = CP
Hence, AP is the perpendicular bisector of BC.

Question 2.
AD is an altitude of an isosceles triangle ABC in which AB = AC. Show that
(i) AD bisects BC
(ii) AD bisects ∠A
Solution:
(i) In right ∆ABD and ∆ACD, we have
AB =AC [Given]

∠ADB = ∠ADC [Each 90°]
AD = DA [Common]
∴ ∆ABD ≅ ∆ACD [By RHS congruency]
So, BD = CD [By C.P.C.T.]
⇒ D is the mid-point of BC or AD bisects BC.

(ii) Since, ∆ABD ≅ ∆ACD,
⇒ ∠BAD = ∠CAD [By C.P.C.T.]
So, AD bisects ∠A

Question 3.
Two sides AB and BC and median AM of one triangle ABC are respectively equal to sides PQ and OR and median PN of ∆PQR (see figure). Show that
(i) ∆ABC ≅ ∆PQR
(ii) ∆ABM ≅ ∆PQN

Solution:
In ∆ABC, AM is the median.
∴BM = 12 BC ……(1)
In ∆PQR, PN is the median.
∴ QN = 12QR …(2)
And BC = QR [Given]
⇒ 12BC = 12QR
⇒ BM = QN …(3) [From (1) and (2)]

(i) In ∆ABM and ∆PQN, we have
AB = PQ , [Given]
AM = PN [Given]
BM = QN [From (3)]
∴ ∆ABM ≅ ∆PQN [By SSS congruency]

(ii) Since ∆ABM ≅ ∆PQN
⇒ ∠B = ∠Q …(4) [By C.P.C.T.]
Now, in ∆ABC and ∆PQR, we have
∠B = ∠Q [From (4)]
AB = PQ [Given]
BC = QR [Given]
∴ ∆ABC ≅ ∆PQR [By SAS congruency]

Question 4.
BE and CF are two equal altitudes of a triangle ABC. Using RHS congruence rule, prove that the triangle ABC is isosceles.
Solution:
Since BE ⊥ AC [Given]

∴ BEC is a right triangle such that ∠BEC = 90°
Similarly, ∠CFB = 90°
Now, in right ∆BEC and ∆CFB, we have
BE = CF [Given]
BC = CB [Common hypotenuse]
∠BEC = ∠CFB [Each 90°]
∴ ∆BEC ≅ ∆CFB [By RHS congruency]
So, ∠BCE = ∠CBF [By C.P.C.T.]
or ∠BCA = ∠CBA
Now, in ∆ABC, ∠BCA = ∠CBA
⇒ AB = AC [Sides opposite to equal angles of a ∆ are equal]
∴ ABC is an isosceles triangle.

Question 5.
ABC is an isosceles triangle with AB = AC. Draw AP ⊥ BC to show that ∠B = ∠C.
Solution:
We have, AP ⊥ BC [Given]

∠APB = 90° and ∠APC = 90°
In ∆ABP and ∆ACP, we have
∠APB = ∠APC [Each 90°]
AB = AC [Given]
AP = AP [Common]
∴ ∆ABP ≅ ∆ACP [By RHS congruency]
So, ∠B = ∠C [By C.P.C.T.]

NCERT Solutions for Class 9 Maths Chapter 7 Triangles Exercise 7.4

Ex 7.4 Class 9 Maths Question 1.
Show that in a right angled triangle, the hypotenuse is the longest side.
Solution:
Let us consider ∆ABC such that ∠B = 90°
∴ ∠A + ∠B + ∠C = 180°
⇒ ∠A + 90°-+ ∠C = 180°
⇒ ∠A + ∠C = 90°
⇒∠A + ∠C = ∠B
∴ ∠B > ∠A and ∠B > ∠C

⇒ Side opposite to ∠B is longer than the side opposite to ∠A
i.e., AC > BC.
Similarly, AC > AB.
Therefore, we get AC is the longest side. But AC is the hypotenuse of the triangle. Thus, the hypotenuse 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:
∠ABC + ∠PBC = 180° [Linear pair]
and ∠ACB + ∠QCB = 180° [Linear pair]
But ∠PBC < ∠QCB [Given] ⇒ 180° – ∠PBC > 180° – ∠QCB
⇒ ∠ABC > ∠ACB
The side opposite to ∠ABC > the side opposite to ∠ACB
⇒ AC > AB.

Question 3.
In figure, ∠B <∠ A and ∠C <∠ D. Show that AD < BC.

Solution: Since ∠A > ∠B [Given]
∴ OB > OA …(1)
[Side opposite to greater angle is longer]
Similarly, OC > OD …(2)
Adding (1) and (2), we have
OB + OC > OA + OD
⇒ BC > AD

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.

Solution:
Let us join AC.

Now, in ∆ABC, AB < BC [∵ AB is the smallest side of the quadrilateral ABCD] ⇒ BC > AB
⇒ ∠BAC > ∠BCA …(1)
[Angle opposite to longer side of A is greater]
Again, in ∆ACD, CD > AD
[ CD is the longest side of the quadrilateral ABCD]
⇒ ∠CAD > ∠ACD …(2)
[Angle opposite to longer side of ∆ is greater]
Adding (1) and (2), we get
∠BAC + ∠CAD > ∠BCA + ∠ACD
⇒ ∠A > ∠C
Similarly, by joining BD, we have ∠B > ∠D.

Question 5.
In figure, PR > PQ and PS bisect ∠QPR. Prove that ∠PSR >∠PSQ.

Solution:
In ∆PQR, PS bisects ∠QPR [Given]
∴ ∠QPS = ∠RPS
and PR > PQ [Given]
⇒ ∠PQS > ∠PRS [Angle opposite to longer side of A is greater]
⇒ ∠PQS + ∠QPS > ∠PRS + ∠RPS …(1) [∵∠QPS = ∠RPS]
∵ Exterior ∠PSR = [∠PQS + ∠QPS]
and exterior ∠PSQ = [∠PRS + ∠RPS]
[An exterior angle is equal to the sum of interior opposite angles]
Now, from (1), we have
∠PSR = ∠PSQ.

Question 6.
Show that of all line segments drawn from a given point not on it, the perpendicular line segment is the shortest.
Solution:
Let us consider the ∆PMN such that ∠M = 90°

Since, ∠M + ∠N+ ∠P = 180°
[Sum of angles of a triangle is 180°]
∵ ∠M = 90° [PM ⊥ l]
So, ∠N + ∠P = ∠M
⇒ ∠N < ∠M
⇒ PM < PN …(1)
Similarly, PM < PN₁ …(2)
and PM < PN₂ …(3)
From (1), (2) and (3), we have PM is the smallest line segment drawn from P on the line l. Thus, the perpendicular line segment is the shortest line segment drawn on a line from a point not on it.

NCERT Solutions for Class 9 Maths Chapter 7 Triangles Exercise 7.5

Question 1.
ABC is a triangle. Locate a point in the interior of ∆ ABC which is equidistant from all the vertices of ∆ ABC.
Solution:
Let us consider a ∆ABC.
Draw l, the perpendicular bisector of AB.
Draw m, the perpendicular bisector of BC.
Let the two perpendicular bisectors l and m meet at O.
O is the required point which is equidistant from A, B and C.

Note: If we draw a circle with centre O and radius OB or OC, then it will pass through A, B and C. The point O is called circumcentre of the triangle.

Question 2.
In a triangle locate a point in its interior which is equidistant from all the sides of the triangle.
Solution:
Let us consider a ∆ABC.

Draw m, the bisector of ∠C.
Let the two bisectors l and m meet at O.
Thus, O is the required point which is equidistant from the sides of ∆ABC.
Note: If we draw OM ⊥ BC and draw a circle with O as centre and OM as radius, then the circle will touch the sides of the triangle. Point O is called incentre of the triangle.

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.]
Solution:
Let us join A and B, and draw l, the perpendicular bisector of AB.
Now, join B and C, and draw m, the perpendicular bisector of BC. Let the perpendicular bisectors l and m meet at O.
The point O is the required point where the ice cream parlour be set up.
Note: If we join A and C and draw the perpendicular bisector, then it will also meet (or pass through) the point O.

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?

Solution:
It is an activity.
We require 150 equilateral triangles of side 1 cm in the Fig. (i) and 300 equilateral triangles in the Fig. (ii).
∴ The Fig. (ii) has more triangles.

NCERT Solutions for Class 9 Maths Chapter 7 Triangles Read More »

NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles

Leave a Comment / NCERT Solution, Class 9 / Mr Ahmad / August 10, 2022

NCERT Solutions Class 9 Maths Chapter 6 Lines ans Angles – In this topic we learn basic Geometry primarily focusing on the properties of the angles formed i) when two lines intersect each other and ii) when a line intersects two or more parallel lines at distinct points. This chapter falls under the Unit – Geometry and is included for the CBSE Syllabus of Class 9 Maths. Check previous chapter solutions – NCERT Solutions for Class 9 Maths Chapter 5 Introduction Euclids Geometry

ALSO CHECK – NCERT Solutions for Class 10 Maths

NCERT Solutions Class 9 Maths Chapter 6 Lines ans Angles is prepared by our best subject experts teachers group thats help students to understand all the topics easily. These Solutions of NCERT Maths help the students in solving the problems efficiently for the upcoming exams. With the help of these NCERT Solutions for Class 9 Maths, students can understand the complex topics of class 9 Maths.

NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles Exercise 6.1

Question 1
In figure, lines AB and CD intersect at 0. If ∠AOC + ∠BOE = 70° and ∠BOD = 40°, find ∠BOE and reflex ∠COE.

Solution:
Since AB is a straight line,
∴ ∠AOC + ∠COE + ∠EOB = 180°
or (∠AOC + ∠BOE) + ∠COE = 180° or 70° + ∠COE = 180° [ ∵∠AOC + ∠BOE = 70° (Given)]
or ∠COE = 180° – 70° = 110°
∴ Reflex ∠COE = 360° – 110° = 250°
Also, AB and CD intersect at O.
∴∠COA = ∠BOD [Vertically opposite angles]
But ∠BOD = 40° [Given]
∴ ∠COA = 40°
Also, ∠AOC + ∠BOE = 70°
∴ 40° + ∠BOE = 70° or ∠BOE = 70° -40° = 30°
Thus, ∠BOE = 30° and reflex ∠COE = 250°.

Question 2.
In figure, lines XY and MN intersect at 0. If ∠POY = 90° , and a : b = 2 : 3. find c.

Solution:
Since XOY is a straight line.
∴ b+a+∠POY= 180°
But ∠POY = 90° [Given]
∴ b + a = 180° – 90° = 90° …(i)
Also a : b = 2 : 3 ⇒ b = 3𝑎2 …(ii)
Now from (i) and (ii), we get
3𝑎2 + A = 90°
⇒ 5𝑎2 = 90°
⇒ a = 90∘5×2=36∘ = 36°
From (ii), we get
b = 32 x 36° = 54°
Since XY and MN interstect at O,
∴ c = [a + ∠POY] [Vertically opposite angles]
or c = 36° + 90° = 126°
Thus, the required measure of c = 126°.

Question 3.
In figure, ∠PQR = ∠PRQ, then prove that ∠PQS = ∠PRT.

Solution:
ST is a straight line.
∴ ∠PQR + ∠PQS = 180° …(1) [Linear pair]
Similarly, ∠PRT + ∠PRQ = 180° …(2) [Linear Pair]
From (1) and (2), we have
∠PQS + ∠PQR = ∠PRT + ∠PRQ
But ∠PQR = ∠PRQ [Given]
∴ ∠PQS = ∠PRT

Question 4.
In figure, if x + y = w + ⇒, then prove that AOB is a line.

Solution:
Sum of all the angles at a point = 360°
∴ x + y + ⇒ + w = 360° or, (x + y) + (⇒ + w) = 360°
But (x + y) = (⇒ + w) [Given]
∴ (x + y) + (x + y) = 360° or,
2(x + y) = 360°
or, (x + y) = 360∘2 = 180°
∴ AOB is a straight line.

Question 5.
In figure, POQ is a line. Ray OR is perpendicular to line PQ. OS is another ray lying between rays OP and OR. Prove that

Solution:
rara POQ is a straight line. [Given]
∴ ∠POS + ∠ROS + ∠ROQ = 180°
But OR ⊥ PQ
∴ ∠ROQ = 90°
⇒ ∠POS + ∠ROS + 90° = 180°
⇒ ∠POS + ∠ROS = 90°
⇒ ∠ROS = 90° – ∠POS … (1)
Now, we have ∠ROS + ∠ROQ = ∠QOS
⇒ ∠ROS + 90° = ∠QOS
⇒ ∠ROS = ∠QOS – 90° ……(2)
Adding (1) and (2), we have
2 ∠ROS = (∠QOS – ∠POS)
∴ ∠ROS = 12(∠𝑄𝑂𝑆−∠𝑃𝑂𝑆)

Question 6.
It is given that ∠XYZ = 64° and XY is produced to point P. Draw a figure from the given information. If ray YQ bisects ∠ZYP, find ∠XYQ and reflex ∠QYP.
Solution:
XYP is a straight line.

∴ ∠XYZ + ∠ZYQ + ∠QYP = 180°
⇒ 64° + ∠ZYQ + ∠QYP = 180°
[∵ ∠XYZ = 64° (given)]
⇒ 64° + 2∠QYP = 180°
[YQ bisects ∠ZYP so, ∠QYP = ∠ZYQ]
⇒ 2∠QYP = 180° – 64° = 116°
⇒ ∠QYP = 116∘2 = 58°
∴ Reflex ∠QYP = 360° – 58° = 302°
Since ∠XYQ = ∠XYZ + ∠ZYQ
⇒ ∠XYQ = 64° + ∠QYP [∵∠XYZ = 64°(Given) and ∠ZYQ = ∠QYP]
⇒ ∠XYQ = 64° + 58° = 122° [∠QYP = 58°]
Thus, ∠XYQ = 122° and reflex ∠QYP = 302°.

NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles Exercise 6.2

Ex 6.2 Class 9 Maths Question 1.
In figure, find the values of x and y and then show that AB || CD.

Solution:
In the figure, we have CD and PQ intersect at F.

∴ y = 130° …(1)
[Vertically opposite angles]
Again, PQ is a straight line and EA stands on it.
∠AEP + ∠AEQ = 180° [Linear pair]
or 50° + x = 180°
⇒ x = 180° – 50° = 130° …(2)
From (1) and (2), x = y
As they are pair of alternate interior angles.
∴ AB || CD

Question 2.
In figure, if AB || CD, CD || EF and y : z = 3 : 7, find x.

Solution:
AB || CD, and CD || EF [Given]
∴ AB || EF
∴ x = z [Alternate interior angles] ….(1)
Again, AB || CD
⇒ x + y = 180° [Co-interior angles]
⇒ z + y = 180° … (2) [By (1)]
But y : z = 3 : 7
z = 7/3 y = 7/3(180°- z) [By (2)]
⇒ 10z = 7 x 180°
⇒ z = 7 x 180° /10 = 126°
From (1) and (3), we have
x = 126°.

Question 3.
In figure, if AB || CD, EF ⊥ CD and ∠GED = 126°, find ∠AGE, ∠GEF and ∠FGE.

Solution:
AB || CD and GE is a transversal.
∴ ∠AGE = ∠GED [Alternate interior angles]
But ∠GED = 126° [Given]
∴∠AGE = 126°
Also, ∠GEF + ∠FED = ∠GED
or ∠GEF + 90° = 126° [∵ EF ⊥ CD (given)]
x = z [Alternate interior angles]… (1) Again, AB || CD
⇒ x + y = 180° [Co-interior angles]
∠GEF = 126° -90° = 36°
Now, AB || CD and GE is a transversal.
∴ ∠FGE + ∠GED = 180° [Co-interior angles]
or ∠FGE + 126° = 180°
or ∠FGE = 180° – 126° = 54°
Thus, ∠AGE = 126°, ∠GEF=36° and ∠FGE = 54°.

Question 4.
In figure, if PQ || ST, ∠ PQR = 110° and ∠ RST = 130°, find ∠QRS.

Solution:
Draw a line EF parallel to ST through R.

Since PQ || ST [Given]
and EF || ST [Construction]
∴ PQ || EF and QR is a transversal
⇒ ∠PQR = ∠QRF [Alternate interior angles] But ∠PQR = 110° [Given]
∴∠QRF = ∠QRS + ∠SRF = 110° …(1)
Again ST || EF and RS is a transversal
∴ ∠RST + ∠SRF = 180° [Co-interior angles] or 130° + ∠SRF = 180°
⇒ ∠SRF = 180° – 130° = 50°
Now, from (1), we have ∠QRS + 50° = 110°
⇒ ∠QRS = 110° – 50° = 60°
Thus, ∠QRS = 60°.

Question 5.
In figure, if AB || CD, ∠APQ = 50° and ∠PRD = 127°, find x and y.

Solution:
We have AB || CD and PQ is a transversal.
∴ ∠APQ = ∠PQR
[Alternate interior angles]
⇒ 50° = x [ ∵ ∠APQ = 50° (given)]
Again, AB || CD and PR is a transversal.
∴ ∠APR = ∠PRD [Alternate interior angles]
⇒ ∠APR = 127° [ ∵ ∠PRD = 127° (given)]
⇒ ∠APQ + ∠QPR = 127°
⇒ 50° + y = 127° [ ∵ ∠APQ = 50° (given)]
⇒ y = 127°- 50° = 77°
Thus, x = 50° and y = 77°.

Question 6.
In figure, PQ and RS are two mirrors placed parallel to each other. An incident ray AB strikes the mirror PQ at B, the reflected ray moves along the path BC and strikes the mirror RS at C and again reflects back along CD. Prove that AB || CD.

Solution:
Draw ray BL ⊥PQ and CM ⊥ RS

∵ PQ || RS ⇒ BL || CM
[∵ BL || PQ and CM || RS]
Now, BL || CM and BC is a transversal.
∴ ∠LBC = ∠MCB …(1) [Alternate interior angles]
Since, angle of incidence = Angle of reflection
∠ABL = ∠LBC and ∠MCB = ∠MCD
⇒ ∠ABL = ∠MCD …(2) [By (1)]
Adding (1) and (2), we get
∠LBC + ∠ABL = ∠MCB + ∠MCD
⇒ ∠ABC = ∠BCD
i. e., a pair of alternate interior angles are equal.
∴ AB || CD.

NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles Exercise 6.3

Question 1.
In figure, sides QP and RQ of ∆PQR are produced to points S and T, respectively. If ∠SPR = 135° and ∠PQT = 110°, find ∠PRQ.

Solution:
We have, ∠TQP + ∠PQR = 180°
[Linear pair]
⇒ 110° + ∠PQR = 180°
⇒ ∠PQR = 180° – 110° = 70°
Since, the side QP of ∆PQR is produced to S.
⇒ ∠PQR + ∠PRQ = 135°
[Exterior angle property of a triangle]
⇒ 70° + ∠PRQ = 135° [∠PQR = 70°]
⇒ ∠PRQ = 135° – 70° ⇒ ∠PRQ = 65°

Question 2.
In figure, ∠X = 62°, ∠XYZ = 54°, if YO and ZO are the bisectors of ∠XYZ and ∠XZY respectively of ∆XYZ, find ∠OZY and ∠YOZ.

Solution:
In ∆XYZ, we have ∠XYZ + ∠YZX + ∠ZXY = 180°
[Angle sum property of a triangle]
But ∠XYZ = 54° and ∠ZXY = 62°
∴ 54° + ∠YZX + 62° = 180°
⇒ ∠YZX = 180° – 54° – 62° = 64°
YO and ZO are the bisectors of ∠XYZ and ∠XZY respectively.
∴ ∠OYZ = 1/2∠𝑋𝑌𝑍 = 1/2(54°) = 27°
and ∠OZY = 1/2∠𝑌𝑍𝑋 = 1/2(64°) = 32°
Now, in ∆OYZ, we have
∠YOZ + ∠OYZ + ∠OZY = 180°
[Angle sum property of a triangle]
⇒ ∠YOZ + 27° + 32° = 180°
⇒ ∠YOZ = 180° -27° – 32° = 121°
Thus, ∠OZY = 32° and ∠YOZ = 121°

Question 3.
In figure, if AB || DE, ∠BAC = 35° and ∠CDE = 539 , find ∠DCE.

Solution:
AB || DE and AE is a transversal.
So, ∠BAC = ∠AED
[Alternate interior angles]
and ∠BAC = 35° [Given]
∴ ∠AED = 35°
Now, in ∆CDE, we have ∠CDE + ∠DEC + ∠DCE = 180°
{Angle sum property of a triangle]
∴ 53° + 35° + ∠DCE =180°
[∵ ∠DEC = ∠AED = 35° and∠CDE = 53° (Given)]
⇒ ∠DCE = 180° – 53° – 35° = 92°
Thus, ∠DCE = 92°

Question 4.
In figure, if lines PQ and RS intersect at point T, such that ∠ PRT = 40°, ∠ RPT = 95° and ∠TSQ = 75°, find ∠ SQT.

Solution:
In ∆PRT, we have ∠P + ∠R + ∠PTR = 180°
[Angle sum property of a triangle]
⇒ 95° + 40° + ∠PTR =180°
[ ∵ ∠P = 95°, ∠R = 40° (given)]
⇒ ∠PTR = 180° – 95° – 40° = 45°
But PQ and RS intersect at T.
∴ ∠PTR = ∠QTS
[Vertically opposite angles]
∴ ∠QTS = 45° [ ∵ ∠PTR = 45°]
Now, in ∆ TQS, we have ∠TSQ + ∠STQ + ∠SQT = 180°
[Angle sum property of a triangle]
∴ 75° + 45° + ∠SQT = 180° [ ∵ ∠TSQ = 75° and ∠STQ = 45°]
⇒ ∠SQT = 180° – 75° – 45° = 60°
Thus, ∠SQT = 60°

Question 5.
In figure, if PQ ⊥ PS, PQ||SR, ∠SQR = 2S° and ∠QRT = 65°, then find the values of x and y.

Solution:
In ∆ QRS, the side SR is produced to T.
∴ ∠QRT = ∠RQS + ∠RSQ
[Exterior angle property of a triangle]
But ∠RQS = 28° and ∠QRT = 65°
So, 28° + ∠RSQ = 65°
⇒ ∠RSQ = 65° – 28° = 37°
Since, PQ || SR and QS is a transversal.
∴ ∠PQS = ∠RSQ = 37°
[Alternate interior angles]
⇒ x = 37°
Again, PQ ⊥ PS ⇒ AP = 90°
Now, in ∆PQS,
we have ∠P + ∠PQS + ∠PSQ = 180°
[Angle sum property of a triangle]
⇒ 90° + 37° + y = 180°
⇒ y = 180° – 90° – 37° = 53°
Thus, x = 37° and y = 53°

Question 6.
In figure, the side QR of ∆PQR is produced to a point S. If the bisectors of ∠PQR and ∠PRS meet at point T, then prove that

Solution:
In ∆PQR, side QR is produced to S, so by exterior angle property,
∠PRS = ∠P + ∠PQR
⇒ 1/2∠PRS = 1/2∠P + 1/2∠PQR
⇒ ∠TRS = 1/2∠P + ∠TQR …(1)
[∵ QT and RT are bisectors of ∠PQR and ∠PRS respectively.]
Now, in ∆QRT, we have
∠TRS = ∠TQR + ∠T …(2)
[Exterior angle property of a triangle]
From (1) and (2),
we have ∠TQR + 1/2∠P = ∠TQR + ∠T
⇒ 1/2∠P = ∠T
⇒ 1/2∠QPR = ∠QTR or ∠QTR = 1/2∠QPR

NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles Read More »

NCERT Solutions for Class 9 All Subjects

Leave a Comment / Class 9, NCERT Solution / Mr Ahmad / August 10, 2022

Class 9 is a foundation of your board exam and it is decide the future of higher education so students who are in class 9 must be focused on concept of respective subject. you also need to practice all subject syllabus thoroughly as it helps you score good marks in the exam and also NCERT Solutions for Class 9 help you to prepared competitive exams and like Olympiad, NTSE etc. So you need to strong command over NCERT Book Solutions for Class 9.

ALSO CHECK – NCERT Solutions for Class 10 Maths

These NCERT Solutions for Class 9 cover all the topics included in the NCERT textbook-like Number System, Coordinate Geometry, Polynomials, Euclid’s Geometry, Quadrilaterals, Triangles, Circles, Constructions, Surface Areas and Volumes, Statistics, Probability, etc.

ALSO CHECK – Download the free Class 10th Notes here

Student those who are searching for NCERT Solutions for Class 9 Maths, Science, Social Science, English, and Hindi then this the right place where you can find chapter-wise NCERT solutions for class 9. All these solution are curated by our expert teacher group, with the help of these Solutions of NCERT Books for Class 9, students can practise all types of questions from the chapters. It will boost your skill also help them in building a foundation for higher-level classes.

NCERT Solutions for Class 9 All Subjects Read More »

NCERT Solutions for Class 9 Maths Chapter 5 Introduction Euclids Geometry

Leave a Comment / NCERT Solution, Class 9 / Mr Ahmad / July 31, 2022

NCERT Solutions Class 9 Maths Chapter 5 Introduction to Euclids Geometry – We study about Euclid’s approach to geometry and tries to link it with the present-day geometry. Introduction to Euclid’s Geometry provides the students with a method of defining common geometrical shapes and terms.

ALSO CHECK – NCERT Solutions for Class 10 Maths

NCERT Solutions for Class 9 Maths Chapter 5 Introduction to Euclids Geometry is prepared by our best subject experts teachers group thats help students to understand all the topics easily. These Solutions of NCERT Maths help the students in solving the problems efficiently for the upcoming exams. With the help of these NCERT Solutions for Class 9 Maths, students can understand the complex topics of class 9 Maths.

ALSO CHECK – Download the free Class 10th Notes here

NCERT Solutions for Class 9 Maths Chapter 5 Introduction to Euclid Geometry Exercise 5.1

Question 1.
Which of the following statements are true and which are false? Give reasons for your answers.
(i) Only one line can pass through a single point.
(ii) There are an infinite number of lines which pass through two distinct points.
(iii) A terminated line can be produced indefinitely on both the sides.
(iv) If two circles are equal, then their radii are equal.
(v) In figure, if AB – PQ and PQ = XY, then AB = XY.

Solution:
(i) False
Reason : If we mark a point O on the surface of a paper. Using pencil and scale, we can draw infinite number of straight lines passing
through O.

(ii) False
Reason : In the following figure, there are many straight lines passing through P. There are many lines, passing through Q. But there is one and only one line which is passing through P as well as Q.

(iii) True
Reason: A line that is terminated can be indefinitely produced on both sides as a line can be extended on both its sides infinitely. Hence, the statement mentioned is True.

(iv) True
Reason : The radii of two circles are equal when the two circles are equal. The circumference and the centre of both the circles coincide; and thus, the radius of the two circles should be equal. Hence, the statement mentioned is True.

(v) True
Reason : According to Euclid’s 1^st axiom- “Things which are equal to the same thing are also equal to one another”. Hence, the statement mentioned is True.

Question 2.
Give a definition for each of the following terms. Are there other terms that need to be defined first? What are they and how might you define them?
(i) Parallel lines
(ii) Perpendicular lines
(iii) Line segment
(iv) Radius of a circle
(v) Square
Solution:
Yes, we need to have an idea about the terms like point, line, ray, angle, plane, circle and quadrilateral, etc. before defining the required terms.
Definitions of the required terms are given below:

(i) Parallel Lines:
Two lines l and m in a plane are said to be parallel, if they have no common point and we write them as l ॥ m.

(ii) Perpendicular Lines:
Two lines p and q lying in the same plane are said to be perpendicular if they form a right angle and we write them as p ⊥ q.

(iii) Line Segment:
A line segment is a part of line and having a definite length. It has two end-points. In the figure, a line segment is shown having end points A and B. It is written as 𝐴𝐵⎯⎯⎯⎯⎯⎯⎯⎯ or 𝐵𝐴⎯⎯⎯⎯⎯⎯⎯⎯.

(iv) Radius of a circle :
The distance from the centre to a point on the circle is called the radius of the circle. In the figure, P is centre and Q is a point on the circle, then PQ is the radius.

(v) Square :
A quadrilateral in which all the four angles are right angles and all the four sides are equal is called a square. Given figure, PQRS is a square.

Question 3.
Consider two ‘postulates’ given below
(i) Given any two distinct points A and B, there exists a third point C which is in between A and B.
(ii) There exist atleast three points that are not on the same line.
Do these postulates contain any undefined terms? Are these postulates consistent? Do they follow from Euclid’s postulates? Explain.
Solution:
Yes, these postulates contain undefined terms such as ‘Point and Line’. Also, these postulates are consistent because they deal with two different situations as
(i) says that given two points A and B, there is a point C lying on the line in between them. Whereas
(ii) says that, given points A and B, you can take point C not lying on the line through A and B.
No, these postulates do not follow from Euclid’s postulates, however they follow from the axiom, “Given two distinct points, there is a unique line that passes through them.”

Question 4.
If a point C lies between two points A and B such that AC = BC, then prove that AC = 12 AB, explain by drawing the figure.
Solution:
We have,

AC = BC [Given]
∴ AC + AC = BC + AC
[If equals added to equals then wholes are equal]
or 2AC = AB [∵ AC + BC = AB]
or AC = 12𝐴𝐵

Ex 5.1 Class 9 Maths Question 5.
In question 4, point C is called a mid-point of line segment AB. Prove that every line segment has one and only one mid-point.
Solution:
Let the given line AB is having two mid points ‘C’ and ‘D’.

AC = 12𝐴𝐵 ……(i)
and AD = 12𝐴𝐵 ……(ii)
Subtracting (i) from (ii), we have
AD – AC = 12𝐴𝐵−12𝐴𝐵
or AD – AC = 0 or CD = 0
∴ C and D coincide.
Thus, every line segment has one and only one mid-point.

Question 6.
In figure, if AC = BD, then prove that AB = CD.

Solution:
Given: AC = BD
⇒ AB + BC = BC + CD
Subtracting BC from both sides, we get
AB + BC – BC = BC + CD – BC
[When equals are subtracted from equals, remainders are equal]
⇒ AB = CD

Question 7.
Why is axiom 5, in the list of Euclid’s axioms, considered a ‘universal truth’? (Note that, the question is not about the fifth postulate.
Solution:
The whole is always greater than the part.

For Example: A cake. When it is whole or complete, assume that it measures 2 pounds but when a part from it is taken out and measured, its weigh will be smaller than the previous measurement. So, the fifth axiom of Euclid is true for all the materials in the universe. Hence, Axiom 5, in the list of Euclid’s axioms, is considered a ‘universal truth.

NCERT Solutions for Class 9 Maths Chapter 5 Introduction to Euclid Geometry Exercise 5.2

Question 1.
How would you rewrite Euclid’s fifth postulate so that it would be easier to understand?
Solution:
We can write Euclid’s fifth postulate as ‘Two distinct intersecting lines cannot be parallel to the same line.’

Question 2.
Does Euclid’s fifth postulate imply the existence of parallel lines ? Explain.
Solution:
Yes. If a straight line l falls on two lines m and n such that sum of the interior angles on one side of l is two right angles, then by Euclid’s fifth postulate, lines m and n will not meet on this side of l. Also, we know that the sum of the interior angles on the other side of the line l will be two right angles too. Thus, they will not meet on the other side also.

∴ The lines m and n never meet, i.e, They are parallel.

NCERT Solutions for Class 9 Maths Chapter 5 Introduction Euclids Geometry Read More »

NCERT Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variable

Leave a Comment / NCERT Solution, Class 9 / Mr Ahmad / July 30, 2022

NCERT Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variable – we learn linear equation in two variables, i.e., ax + by + c = 0. Students will also learn to plot the graph of a linear equation in two variables. These are the important point that must be remember.

Important Points

An equation of the form ax + by + c = 0, where a, b and c are real numbers, such that a and b are not both zero, is called a linear equation in two variables.
A linear equation in two variables has infinitely many solutions.
The graph of every linear equation in two variables is a straight line.
x = 0 is the equation of the y-axis and y = 0 is the equation of the x-axis.
The graph of x = a is a straight line parallel to the y-axis.
The graph of y = a is a straight line parallel to the x-axis.
An equation of the type y = mx represents a line passing through the origin.

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NCERT Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variable is prepared by our best subject experts teachers group thats help students to understand all the topics easily. These Solutions of NCERT Maths help the students in solving the problems efficiently for the upcoming exams. With the help of these NCERT Solutions for Class 9 Maths, students can understand the complex topics of class 9 Maths.

ALSO CHECK – Download the free Class 10th Notes here

NCERT Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables Exercise – 4.1

Question 1.
The cost of a notebook is twice the cost of a pen. Write a linear equation in two variables to represent this statement. (Take the cost of a notebook to be Rs- x and that of a pen to be Rs- y).
Solution:
Let the cost of a notebook = Rs- x
and the cost of a pen = Rs- y
According to the condition, we have
[Cost of a notebook] =2 x [Cost of a pen]
i. e„ (x) = 2 x (y) or, x = 2y
or, x – 2y = 0
Thus, the required linear equation is x – 2y = 0.

Question 2
Express the following linear equations in the form ax + by + c = 0 and indicate the values of a, b and c in each case:
(i) 2x + 3y = 9.35⎯⎯⎯
(ii) 𝑥−𝑦5−10=0
(iii) – 2x + 3y = 6
(iv) x = 3y
(v) 2x = -5y
(vi) 3x + 2 = 0
(vii) y – 2 = 0
(viii) 5 = 2x
Solution:
(i) We have 2x + 3y = 9.35⎯⎯⎯
or (2)x + (3)y + (−9.35⎯⎯⎯ ) = 0
Comparing it with ax + by +c= 0, we geta = 2,
b = 3 and c= –9.35⎯⎯⎯ .

(ii) We have 𝑥−𝑦5−10=0
or x + (- 15) y + (10) = 0
Comparing it with ax + by + c = 0, we get
a =1, b =- 15 and c= -10

(iii) Wehave -2x + 3y = 6 or (-2)x + (3)y + (-6) = 0
Comparing it with ax – 4 – by + c = 0,we get a = -2, b = 3 and c = -6.

(iv) We have x = 3y or (1)x + (-3)y + (0) = 0 Comparing it with ax + by + c = 0, we get a = 1, b = -3 and c = 0.
(v) We have 2x = -5y or (2)x + (5)y + (0) = 0 Comparing it with ax + by + c = 0, we get a = 2, b = 5 and c = 0.

(vi) We have 3x + 2 = 0 or (3)x + (0)y + (2) = 0 Comparing it with ax + by + c = 0, we get a = 3, b = 0 and c = 2.
(vii) We have y – 2 = 0 or (0)x + (1)y + (-2) = 0 Comparing it with ax + by + c = 0, we get a = 0, b = 1 and c = -2.
(viii) We have 5 = 2x ⇒ 5 – 2x = 0
or -2x + 0y + 5 = 0
or (-2)x + (0)y + (5) = 0
Comparing it with ax + by + c = 0, we get a = -2, b = 0 and c = 5.

NCERT Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables Exercise – 4.2

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:
Option (iii) is true because for every value of x, we get a corresponding value of y and vice-versa in the given equation.
Hence, given linear equation has an 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
When x = 0, 2(0) + y = 7 ⇒ y = 7
∴ Solution is (0, 7)
When x =1, 2(1) + y = 7 ⇒ y = 7 – 2 ⇒ y = 5
∴ Solution is (1, 5)
When x = 2, 2(2) + y =7y = 7 – 4 ⇒ y = 3
∴ Solution is (2, 3)
When x = 3, 2(3) + y = 7y = 7 – 6 ⇒ y = 1
∴ Solution is (3, 1).

(ii) πx + y = 9
When x = 0, π(0) + y = 9 ⇒ y = 9 – 0 ⇒ y = 9
∴ Solution is (0, 9)
When x = 1, π(1) + y = 9 ⇒ y = 9 – π
∴ Solution is (1, (9 – π))
When x = 2, π(2) + y = 9 ⇒ y = 9 – 2π
∴ Solution is (2, (9 – 2π))
When x = -1,π(-1) + y = 9 ⇒ y = 9 + π
∴ Solution is (-1, (9 + π))

(iii) x = 4y
When x = 0, 4y = 1 ⇒ y = 0
∴ Solution is (0, 0)
When x = 1, 4y = 1 ⇒ y = 14
∴ Solution is (1,14 )
When x = 4, 4y = 4 ⇒ y = 1
∴ Solution is (4, 1)
When x = 4, 4y = 4 ⇒ y = -1
∴ Solution is (-4, -1)

Question 3
Check which of the following are solutions 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) (0,2) means x = 0 and y = 2
Puffing x = 0 and y = 2 in x – 2y = 4, we get
L.H.S. = 0 – 2(2) = -4.
But R.H.S. = 4
∴ L.H.S. ≠ R.H.S.
∴ x =0, y =2 is not a solution.

(ii) (2, 0) means x = 2 and y = 0
Putting x = 2 and y = 0 in x – 2y = 4, we get
L.H:S. 2 – 2(0) = 2 – 0 = 2.
But R.H.S. = 4
∴ L.H.S. ≠ R.H.S.
∴ (2,0) is not a solution.

(iii) (4, 0) means x = 4 and y = 0
Putting x = 4 and y = o in x – 2y = 4, we get
L.H.S. = 4 – 2(0) = 4 – 0 = 4 =R.H.S.
∴ L.H.S. = R.H.S.
∴ (4, 0) is a solution.

(iv) (√2, 4√2) means x = √2 and y = 4√2
Putting x = √2 and y = 4√2 in x – 2y = 4, we get
L.H.S. = √2 – 2(4√2) = √2 – 8√2 = -7√2
But R.H.S. = 4
∴ L.H.S. ≠ R.H.S.
∴ (√2 , 4√2) is not a solution.

(v) (1, 1)means x =1 and y = 1
Putting x = 1 and y = 1 in x – 2y = 4, we get
LH.S. = 1 – 2(1) = 1 – 2 = -1. But R.H.S = 4
∴ LH.S. ≠ R.H.S.
∴ (1, 1) is not a solution.

Question 4
Find the value of k, if x = 2, y = 1 ¡s a solution of the equation 2x + 3y = k.
Solution:
We have 2x + 3y = k
putting x = 2 and y = 1 in 2x+3y = k,we get
2(2) + 3(1) ⇒ k = 4 + 3 – k ⇒ 7 = k
Thus, the required value of k is 7.

NCERT Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables Exercise – 4.3

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 ⇒ y = 4 – x
If we have x = 0, then y = 4 – 0 = 4
x = 1, then y =4 – 1 = 3
x = 2, then y = 4 – 2 = 2
∴ We get the following table:

Plot the ordered pairs (0, 4), (1,3) and (2,2) on the graph paper. Joining these points, we get a straight line AB as shown.

Thus, the line AB is the required graph of x + y = 4

(ii) x – y = 2 ⇒ y = x – 2
If we have x = 0, then y = 0 – 2 = -2
x = 1, then y = 1 – 2 = -1
x = 2, then y = 2 – 2 = 0
∴ We get the following table:

Plot the ordered pairs (0, -2), (1, -1) and (2, 0) on the graph paper. Joining these points, we get a straight line PQ as shown.

Thus, the ime is the required graph of x – y = 2

(iii) y = 3x
If we have x = 0,
then y = 3(0) ⇒ y = 0
x = 1, then y = 3(1) = 3
x= -1, then y = 3(-1) = -3
∴ We get the following table:

Plot the ordered pairs (0, 0), (1, 3) and (-1, -3) on the graph paper. Joining these points, we get a straight line LM as shown.

Thus, the line LM is the required graph of y = 3x.

(iv) 3 = 2x + y ⇒ y = 3 – 2x
If we have x = 0, then y = 3 – 2(0) = 3
x = 1,then y = 3 – 2(1) = 3 – 2 = 1
x = 2,then y = 3 – 2(2) = 3 – 4 = -1
∴ We get the following table:

Plot the ordered pairs (0, 3), (1, 1) and (2, – 1) on the graph paper. Joining these points, we get a straight line CD as shown.

Thus, the line CD is the required graph of 3 = 2x + y.

Question 2
Give the equations of two lines passing through (2, 14). How many more such lines are there, and why?
Solution:
We know that infinite number of lines passes through a point.

Equation of 2 lines passing through (2,14) should be in such a way that it satisfies the point.

Let the equation be, 7x = y

7x–y = 0

When x = 2 and y = 14

(7×2)-14 = 0

14–14 = 0

0 = 0

L.H.S = R.H.S

Let another equation be, 4x = y-6

4x-y+6 = 0

When x = 2 and y = 14

(4×2–14+6 = 0

8–14+6 = 0

0 = 0

L.H.S = R.H.S

Since both the equations satisfies the point (2,14), than say that the equations of two lines passing through (2, 14) are 7x = y and 4x = y-6

We know that, infinite number of line passes through one specific point. Since there is only one point (2,14) here, there can be infinite lines that passes through the point.

Question 3
If the point (3, 4) lies on the graph of the equation 3y = ax + 7, find the value of a.
Solution:
The equation of the given line is 3y = ax + 7
∵ (3, 4) lies on the given line.
∴ It must satisfy the equation 3y = ax + 7
We have, (3, 4) ⇒ x = 3 and y = 4.
Putting these values in given equation, we get
3 x 4 = a x 3 + 7
⇒ 12 = 3a + 7
⇒ 3a = 12 – 7 = 5 ⇒ a = 53
Thus, the required value of a is 53

Question 4
The taxi fare In a city Is as follows: For the first kilometre, the fare Is Rs. 8 and for the subsequent distance it is Rs. 5 per km. Taking the distance covered as x km and total fare as Rs.y, write a linear equation for this Information, and draw Its graph.
Solution:
Here, total distance covered = x km and total taxi fare = Rs. y
Fare for 1km = Rs. 8
Remaining distance = (x – 1) km
∴ Fare for (x – 1)km = Rs.5 x(x – 1)
Total taxi fare = Rs. 8 + Rs. 5(x – 1)
According to question,
y = 8 + 5(x – 1) = y = 8 + 5x – 5
⇒ y = 5x + 3,
which is the required linear equation representing the given information.
Graph: We have y = 5x + 3
Wben x = 0, then y = 5(0) + 3 ⇒ y = 3
x = -1, then y = 5(-1) + 3 ⇒ y = -2
x = -2, then y = 5(-2) + 3 ⇒ y = -7
∴ We get the following table:

Now, plotting the ordered pairs (0, 3), (-1, -2) and (-2, -7) on a graph paper and joining them, we get a straight line PQ as shown.

Thus, the line PQ is the required graph of the linear equation y = 5x + 3.

Question 5
From the choices given below, choose the equation whose graphs are given ¡n Fig. (1) and Fig. (2).
For Fig. (1)
(i) y = x
(ii) x + y = 0
(iii) y = 2x
(iv) 2 + 3y = 7x

For Fig. (2)
(i) y = x + 2
(ii) y = x – 2
(iii) y = -x + 2
(iv) x + 2y = 6

Solution:
For Fig. (1), the correct linear equation is x + y = 0
[As (-1, 1) = -1 + 1 = 0 and (1,-1) = 1 + (-1) = 0]
For Fig.(2), the correct linear equation is y = -x + 2
[As(-1,3) 3 = -1(-1) + 2 = 3 = 3 and (0,2)
⇒ 2 = -(0) + 2 ⇒ 2 = 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:
Constant force is 5 units.
Let the distance travelled = x units and work done = y units.
Work done = Force x Distance
⇒ y = 5 x x ⇒ y = 5x
For drawing the graph, we have y = 5x
When x = 0, then y = 5(0) = 0
x = 1, then y = 5(1) = 5
x = -1, then y = 5(-1) = -5
∴ We get the following table:

Ploffing the ordered pairs (0, 0), (1, 5) and (-1, -5) on the graph paper and joining the points, we get a straight line AB as shown.

From the graph, we get
(i) Distance travelled =2 units i.e., x = 2
∴ If x = 2, then y = 5(2) = 10
⇒ Work done = 10 units.

(ii) Distance travelled = 0 unit i.e., x = 0
∴ If x = 0 ⇒ y = 5(0) – 0
⇒ Work done = 0 unit.

Question 7
Yamini and Fatima, two students of Class IX of a school, together contributed Rs. 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 Rs.xand Rs.y.) Draw the graph of the same.
Solution:
Let the contribution of Yamini = Rs. x
and the contribution of Fatima Rs. y
∴ We have x + y = 100 ⇒ y = 100 – x
Now, when x = 0, y = 100 – 0 = 100
x = 50, y = 100 – 50 = 50
x = 100, y = 100 – 100 = 0
∴ We get the following table:

Plotting the ordered pairs (0,100), (50,50) and (100, 0) on a graph paper using proper scale and joining these points, we get a straight line PQ as shown.

Thus, the line PQ is the required graph of the linear equation x + y = 100.

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 = (95 )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) We have
F = (95 )C + 32
When C = 0 , F = (95 ) x 0 + 32 = 32
When C = 15, F = (95 )(-15) + 32= -27 + 32 = 5
When C = -10, F = 95 (-10)+32 = -18 + 32 = 14
We have the following table:

Plotting the ordered pairs (0, 32), (-15, 5) and (-10,14) on a graph paper. Joining these points, we get a straight line AB.

(ii) From the graph, we have 86°F corresponds to 30°C.
(iii) From the graph, we have 95°F corresponds 35°C.
(iv) We have, C = 0
From (1), we get
F = (95)0 + 32 = 32
Also, F = 0
From (1), we get
0 = (95)C + 32 ⇒ −32×59 = C ⇒ C = -17.8
(V) When F = C (numerically)
From (1), we get
F = 95F + 32 ⇒ F – 95F = 32
⇒ −45F = 32 ⇒ F = -40
∴ Temperature is – 40° both in F and C.

NCERT Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables Exercise – 4.4

Question 1
Give the geometric representations of y = 3 as an equation
(i) in one variable
(ii) in two variables
Solution:
(i) y = 3
∵ y = 3 is an equation in one variable, i.e., y only.
∴ y = 3 is a unique solution on the number line as shown below:

(ii) y = 3
We can write y = 3 in two variables as 0.x + y = 3
Now, when x = 1, y = 3
x = 2, y = 3
x = -1, y = 3
∴ We get the following table:

Plotting the ordered pairs (1, 3), (2, 3) and (-1, 3) on a graph paper and joining them, we get aline AB as solution of 0. x + y = 3,
i.e. y = 3.

Question 2
Give the geometric representations of 2x + 9 = 0 as an equation
(i) in one variable
(ii) in two variables
Solution:
(i) 2x + 9 = 0
We have, 2x + 9 = 0 ⇒ 2x = – 9 ⇒ x = −92
which is a linear equation in one variable i.e., x only.
Theref ore, x = −92 is a unique solution on the number line as shown below:

(ii) 2x +9=0
We can write 2x + 9 = 0 in two variables as 2x + 0, y + 9 = 0
or 𝑥=−9−0.𝑦2
∴ When y = 1, x = 𝑥=−9−0(1)2 = −92

Thus, we get the following table:

Now, plotting the ordered pairs (−92,3) ,(−92,3) and (−92,3) on a graph paper and joining them, we get a line PQ as solution of 2x + 9 = 0.

NCERT Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variable Read More »

NCERT Solutions for Class 9 Maths Chapter 3 Coordinate Geometry

Leave a Comment / NCERT Solution, Class 9 / Mr Ahmad / July 30, 2022

In this NCERT Solutions for Class 9 Maths Chapter 3 Coordinate Geometry includes the concepts of the cartesian plane, coordinates of a point in xy – plane, terms, notations associated with the coordinate plane, including the x-axis, y-axis, x- coordinate, y-coordinate, origin, quadrants and more.

ALSO CHECK – NCERT Solutions for Class 10 Maths

NCERT Solutions for Class 9 Maths Chapter 3 Coordinate Geometry is prepared by our best subject experts teachers group thats help students to understand all the topics easily. These Solutions of NCERT Maths help the students in solving the problems efficiently for the upcoming exams. With the help of these NCERT Solutions for Class 9 Maths, students can understand the complex topics of class 9 Maths. They also focus on formulating the solutions of Maths in such a way that it is easy for the students to understand.

ALSO CHECK – Download the free Class 10th Notes here

NCERT Solutions for Class 9 Maths Chapter 3 Coordinate Geometry Exercise 3.1

Question 1.
How will you describe the position of a table lamp on your study table to another person?
Solution:
To describe the position of a table lamp placed on the table, let us consider the table lamp as P and the table as a plane.
Now choose two perpendicular edges of the table as the axes OX and OY.
Measure the perpendicular distance ‘a’cm of P (lamp) from OY. Measure the perpendicular distance ‘b’ cm of P (lamp) from OX.
Thus, the position of the table lamp P is described by the ordered pair (a, b).

Question 2.
(Street Plan): A city has two main roads which cross each other at the centre of the city. These two roads are along the North-South direction and East-West direction. All other streets of the city run parallel to these roads and are 200 m apart. There are 5 streets in each direction. Using 1 cm = 200 m, draw a model of the city on your notebook. Represent the roads/streets by single lines.
There are many cross-streets in your model. A particular cross-street is made by two streets, one running in the North-South direction and another in the East-West direction. Each cross street is referred to in the following manner: If the 2^nd street running in the North-South direction and 5^th in the East-West direction meet at some crossing, then we will call this cross-street (2,5). Using this convention, find:
(i) how many cross-streets can be referred to as (4,3).
(ii) how many cross-streets can be referred to as (3,4).
Solution:
(i) A unique cross street as shown by the point A(4, 3).
(ii) A unique cross street as shown by the point B(3,4).
The two cross streets are uniquely found because of the two reference lines we have used for locating them.

NCERT Solutions for Class 9 Maths Chapter 3 Coordinate Geometry Exercise 3.2

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) The horizontal line: x – axis and the vertical line: y – axis.
(ii) Each part is called “Quadrant”.
(iii) Origin

Ex 3.2 Class 9 Maths Question 2
See the given figure and write the following:
(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 L.
(viii) The coordinates of the point M.

Solution:
From the figure, we have
(i) The coordinates of B are (-5,2).
(ii) The coordinates of C are (5, -5).
(iii) The point E is identified by the coordinates (-3,-5).
(iv) The point G is identified by the coordinates (2,-4).
(v) The abscissa of the point D is 6.
(vi) The ordinate of the point H is -3.
(vii) The coordinates of the point L are (0,5). (viii) The coordinates of the point M are (-3,0).

NCERT Solutions for Class 9 Maths Chapter 3 Coordinate Geometry Exercise 3.3

Question 1
In which quadrant or on which axis do 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:
The point (-2, 4) is having negative abscissa and positive ordinate.
∴ (-2,4) lies in the 2^nd quadrant.
The point (3, -1) is having positive abscissa and negative ordinate.
∴ (3, -1) lies in the 4^th quadrant.
The point (-1, 0) is having negative abscissa and zero ordinate.
∴ The point (-1, 0) lies on the negative x-axis.
The point (1, 2) is having the abscissa as well as ordinate positive.
∴ Point (1,2) lies in the 1^st quadrant.
The point (-3, -5) is having the abscissa as well as ordinate negative.
∴ Point (-3, -5) lies in the 3rd quadrant.
These points are plotted in the Cartesian plane as shown in the following figure as A(-2, 4); B(3, -1); C(-l, 0); D(l, 2) and E (-3, -5).

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

Solution:
The given points are (-2, 8), (-1, 7), (0, -1.25), (1,3) and (3, -1).
To plot these points:
(i) We draw X’OX and YOY’ as axes.
(ii) We choose suitable units of distance on the axes.
(iii) To plot (-2,8), we start from O, take (-2) units on x-axis and then (+8) units on y – axis. We mark the point as A (-2, 8).
(iv) To plot (-1,7), we start from O, take (-1) units on x-axis and then (+7) units on the y – axis. We mark the point as B(-1, 7).
(v) To plot (0, -1.25), we move along 1.25 units below the x-axis on the y – axis and mark the point as C(0, -1.25).
(vi) To plot (1, 3), we take (+1) unit on the x-axis and then (+3) units on the y – axis. We mark the point as D(1, 3).
(vii) To plot (3, -1), we take (+3) units on the x-axis and then (-1) unit on the y – axis. We mark the point E(3, -1).

NCERT Solutions for Class 9 Maths Chapter 3 Coordinate Geometry Read More »

NCERT Solutions for Class 9 Maths Chapter 2 Polynomials

Leave a Comment / NCERT Solution, Class 9 / Mr Ahmad / July 28, 2022

The second chapter of NCERT Solutions for Class 9 Maths Chapter 2 Polynomials contain the topic that is related with Polynomials, Polynomial is an expression that consists of variables and coefficients, involving the operations of addition, subtraction, multiplication, and non-negative integer exponents of variables. Students will also learn polynomial, degrees, coefficient, zeros and terms of a polynomial.

Important Formulas –

(x + y)² = x² + 2xy + y²

(x – y)² = x² – 2xy + y²

x² – y² = (x + y) (x – y)

(x + y + z)² = x² + y² + z² + 2xy + 2yz + 2zx

(x + y)³ = x³ + y³ + 3xy(x + y)

(x – y)³ = x³ – y³ – 3xy(x – y)

x³ + y³ + z³ – 3xyz = (x + y + z) (x² + y² + z² – xy – yz – zx)

Dividend = (Divisor × Quotient) + Remainder

NCERT Solutions for Class 9 Maths Chapter 2 Polynomials is prepared by our best subject experts teachers group thats help students to understand all the topics easily. These Solutions of NCERT Maths help the students in solving the problems efficiently for the upcoming exams. With the help of these NCERT Solutions for Class 9 Maths, students can understand the complex topics of class 9 Maths. They also focus on formulating the solutions of Maths in such a way that it is easy for the students to understand.

ALSO CHECK – NCERT Solutions for Class 10 Maths

ALSO CHECK – Download the free Class 10th Notes here

NCERT Solutions for Class 9 Maths Chapter 2 Polynomials

Exercise 2.1

Question 1.
Which of the following expressions are polynomials in one variable and which are not? State reasons for your answer.
(i) 4x² – 3x + 7
(ii) y² + √2
(iii) 3 √t + t√2
(iv) y+ 2𝑦
(v) x¹⁰+ y³+t⁵⁰
Solution:
(i) We have 4x² – 3x + 7 = 4x² – 3x + 7x⁰
It is a polynomial in one variable i.e., x
because each exponent of x is a whole number.

(ii) We have y² + √2 = y² + √2y⁰
It is a polynomial in one variable i.e., y
because each exponent of y is a whole number.

(iii) We have 3 √t + t√2 = 3 √t^1/2 + √2.t
It is not a polynomial, because one of the exponents of t is 12,
which is not a whole number.

(iv) We have y + 𝑦+2𝑦 = y + 2.y^-1
It is not a polynomial, because one of the exponents of y is -1,
which is not a whole number.

(v) We have x¹⁰+ y³+ t⁵⁰
Here, exponent of every variable is a whole number, but x¹⁰ + y³ + t⁵⁰ is a polynomial in x, y and t, i.e., in three variables.
So, it is not a polynomial in one variable.

Question 2.
Write the coefficients of x² in each of the following
(i) 2 + x² + x
(ii) 2 – x² + x³
(iii) 𝜋2 x² + x
(iv) √2 x – 1
Solution:
(i) The given polynomial is 2 + x² + x.
The coefficient of x² is 1.
(ii) The given polynomial is 2 – x² + x³.
The coefficient of x² is -1.
(iii) The given polynomial is 𝜋2𝑥2 + x.
The coefficient of x² is 𝜋2.
(iv) The given polynomial is √2 x – 1.
The coefficient of x² is 0.

Question 3.
Give one example each of a binomial of degree 35, and of a monomial of degree 100.
Solution:
(i) Abmomial of degree 35 can be 3x³⁵ -4.
(ii) A monomial of degree 100 can be √2y¹⁰⁰.

Question 4.
Write the degree of each of the following polynomials.
(i) 5x³+4x² + 7x
(ii) 4 – y²
(iii) 5t – √7
(iv) 3
Solution:
(i) The given polynomial is 5x³ + 4x² + 7x.
The highest power of the variable x is 3.
So, the degree of the polynomial is 3.
(ii) The given polynomial is 4- y². The highest
power of the variable y is 2.
So, the degree of the polynomial is 2.
(iii) The given polynomial is 5t – √7 . The highest power of variable t is 1. So, the degree of the polynomial is 1.
(iv) Since, 3 = 3x° [∵ x°=1]
So, the degree of the polynomial is 0.

Question 5.
Classify the following as linear, quadratic and cubic polynomials.
(i) x²+ x
(ii) x – x³
(iii) y + y²+4
(iv) 1 + x
(v) 3t
(vi) r²
(vii) 7x³Solution:
(i) The degree of x² + x is 2. So, it is a quadratic polynomial.
(ii) The degree of x – x³ is 3. So, it is a cubic polynomial.

(iii) The degree of y + y² + 4 is 2. So, it is a quadratic polynomial.

(iv) The degree of 1 + x is 1. So, it is a linear polynomial.
(v) The degree of 3t is 1. So, it is a linear polynomial.
(vi) The degree of r² is 2. So, it is a quadratic polynomial.
(vii) The degree of 7x³ is 3. So, it is a cubic polynomial.

NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Exercise 2.2

Question 1.
Find the value of the polynomial 5x – 4x² + 3 at
(i) x = 0
(ii) x = – 1
(iii) x = 2
Solution:
1et p(x) = 5x – 4x² + 3
(i) p(0) = 5(0) – 4(0)² + 3 = 0 – 0 + 3 = 3
Thus, the value of 5x – 4x² + 3 at x = 0 is 3.
(ii) p(-1) = 5(-1) – 4(-1)² + 3
= – 5x – 4x² + 3 = -9 + 3 = -6
Thus, the value of 5x – 4x² + 3 at x = -1 is -6.
(iii) p(2) = 5(2) – 4(2)² + 3 = 10 – 4(4) + 3
= 10 – 16 + 3 = -3
Thus, the value of 5x – 4x² + 3 at x = 2 is – 3.

Question 2.
Find p (0), p (1) and p (2) for each of the following polynomials.
(i) p(y) = y² – y +1
(ii) p (t) = 2 +1 + 2t² -t³
(iii) P (x) = x³
(iv) p (x) = (x-1) (x+1)
Solution:
(i) Given that p(y) = y² – y + 1.
∴ P(0) = (0)² – 0 + 1 = 0 – 0 + 1 = 1
p(1) = (1)² – 1 + 1 = 1 – 1 + 1 = 1
p(2) = (2)² – 2 + 1 = 4 – 2 + 1 = 3
(ii) Given that p(t) = 2 + t + 2t²– t³
∴p(0) = 2 + 0 + 2(0)²– (0)³
= 2 + 0 + 0 – 0=2
P(1) = 2 + 1 + 2(1)²– (1)³
= 2 + 1 + 2 – 1 = 4
p( 2) = 2 + 2 + 2(2)²– (2)³
= 2 + 2 + 8 – 8 = 4
(iii) Given that p(x) = x³
∴ p(0) = (0)³ = 0, p(1) = (1)³ = 1
p(2) = (2)³ = 8
(iv) Given that p(x) = (x – 1)(x + 1)
∴ p(0) = (0 – 1)(0 + 1) = (-1)(1) = -1
p(1) = (1 – 1)(1 +1) = (0)(2) = 0
P(2) = (2 – 1)(2 + 1) = (1)(3) = 3

Question 3.
Verify whether the following are zeroes of the polynomial, indicated against them.
(i) p(x) = 3x + 1,x = –13
(ii) p (x) = 5x – π, x = 45
(iii) p (x) = x² – 1, x = x – 1
(iv) p (x) = (x + 1) (x – 2), x = – 1,2
(v) p (x) = x², x = 0
(vi) p (x) = 1x + m, x = – 𝑚1
(vii) P (x) = 3x² – 1, x = – 13√,23√
(viii) p (x) = 2x + 1, x = 12
Solution:
(i) We have , p(x) = 3x + 1

(ii) We have, p(x) = 5x – π
∴ 𝑝(−13)=3(−13)+1=−1+1=0
(iii) We have, p(x) = x² – 1
∴ p(1) = (1)² – 1 = 1 – 1=0
Since, p(1) = 0, so x = 1 is a zero of x² -1.
Also, p(-1) = (-1)² -1 = 1 – 1 = 0
Since p(-1) = 0, so, x = -1, is also a zero of x² – 1.

(iv) We have, p(x) = (x + 1)(x – 2)
∴ p(-1) = (-1 +1) (-1 – 2) = (0)(- 3) = 0
Since, p(-1) = 0, so, x = -1 is a zero of (x + 1)(x – 2).
Also, p( 2) = (2 + 1)(2 – 2) = (3)(0) = 0
Since, p(2) = 0, so, x = 2 is also a zero of (x + 1)(x – 2).

(v) We have, p(x) = x²
∴ p(o) = (0)² = 0
Since, p(0) = 0, so, x = 0 is a zero of x².

(vi) We have, p(x) = lx + m

(vii) We have, p(x) = 3x² – 1

(viii) We have, p(x) = 2x + 1
∴ 𝑝(12)=2(12)+1=1+1=2
Since, 𝑝(12) ≠ 0, so, x = 12 is not a zero of 2x + 1.

Question 4.
Find the zero of the polynomial in each of the following cases
(i) p(x)=x+5
(ii) p (x) = x – 5
(iii) p (x) = 2x + 5
(iv) p (x) = 3x – 2
(v) p (x) = 3x
(vi) p (x)= ax, a≠0
(vii) p (x) = cx + d, c ≠ 0 where c and d are real numbers.
Solution:
(i) We have, p(x) = x + 5. Since, p(x) = 0
⇒ x + 5 = 0
⇒ x = -5.
Thus, zero of x + 5 is -5.

(ii) We have, p(x) = x – 5.
Since, p(x) = 0 ⇒ x – 5 = 0 ⇒ x = -5
Thus, zero of x – 5 is 5.

(iii) We have, p(x) = 2x + 5. Since, p(x) = 0
⇒ 2x + 5 =0
⇒ 2x = -5
⇒ x = −52
Thus, zero of 2x + 5 is −52 .

(iv) We have, p(x) = 3x – 2. Since, p(x) = 0
⇒ 3x – 2 = 0
⇒ 3x = 2
⇒ x = 23
Thus, zero of 3x – 2 is 23

(v) We have, p(x) = 3x. Since, p(x) = 0
⇒ 3x = 0 ⇒ x = 0
Thus, zero of 3x is 0.

(vi) We have, p(x) = ax, a ≠ 0.
Since, p(x) = 0 => ax = 0 => x-0
Thus, zero of ax is 0.

(vii) We have, p(x) = cx + d. Since, p(x) = 0
⇒ cx + d = 0 ⇒ cx = -d ⇒ 𝑥=−𝑑𝑐
Thus, zero of cx + d is −𝑑𝑐

NCERT So1utions for C1ass 9 Maths Chapter 2 Polynomials Exercise 2.3

Question 1.
Find the remainder when x³ + 3x² + 3x + 1 is divided by
(i) x + 1
(ii) x – 12
(iii) x
(iv) x + π
(v) 5 + 2x
Solution:
Let p(x) = x³ + 3x² + 3x +1
(i) The zero of x + 1 is -1.
∴ p(-1) = (-1)3 + 3(-1)2 + 3(-1) +1
= -1 + 3- 3 + 1 = 0
Thus, the required remainder = 0

(ii) The zero of 𝑥−12 is 12

Thus, the required remainder = 278

(iii) The zero of x is 0.
∴ p(0) = (0)³ + 3(0)² + 3(0) + 1
= 0 + 0 + 0 + 1 = 1
Thus, the required remainder = 1.

(iv) The zero of x + π is -π.
p(-π) = (-π)³ + 3(- π)²2 + 3(- π) +1
= -π³ + 3π² + (-3π) + 1
= – π³ + 3π² – 3π +1
Thus, the required remainder is -π³ + 3π² – 3π+1.

(v) The zero of 5 + 2x is −52 .

Thus, the required remainder is −278 .

Question 2.
Find the remainder when x³ – ax² + 6x – a is divided by x – a.
Solution:
We have, p(x) = x³ – ax² + 6x – a and zero of x – a is a.
∴ p(a) = (a)³ – a(a)² + 6(a) – a
= a³ – a³ + 6a – a = 5a
Thus, the required remainder is 5a.

Question 3.
Check whether 7 + 3x is a factor of 3x³+7x.
Solution:
We have, p(x) = 3x³+7x. and zero of 7 + 3x is −73.

Since,( −4909) ≠ 0
i.e. the remainder is not 0.
∴ 3x³ + 7x is not divisib1e by 7 + 3x.
Thus, 7 + 3x is not a factor of 3x³ + 7x.

NCERT So1utions for Class 9 Maths Chapter 2 Polynomials

Exercise 2.4

Question 1.
Determine which of the following polynomials has (x +1) a factor.
(i) x³+x²+x +1
(ii) x⁴ + x³ + x² + x + 1
(iii) x⁴ + 3x³ + 3x² + x + 1
(iv) x³ – x² – (2 +√2 )x + √2
Solution:
The zero of x + 1 is -1.
(i) Let p (x) = x³ + x² + x + 1
∴ p (-1) = (-1)³ + (-1)² + (-1) + 1 .
= -1 + 1 – 1 + 1
⇒ p (- 1) = 0
So, (x+ 1) is a factor of x³ + x² + x + 1.

(ii) Let p (x) = x⁴ + x³ + x² + x + 1
∴ P(-1) = (-1)⁴ + (-1)³ + (-1)² + (-1)+1
= 1 – 1 + 1 – 1 + 1
⇒ P (-1) ≠ 1
So, (x + 1) is not a factor of x⁴ + x³ + x² + x+ 1.

(iii) Let p (x) = x⁴ + 3x³ + 3x² + x + 1 .
∴ p (-1)= (-1)⁴ + 3 (-1)³ + 3 (-1)² + (- 1) + 1
= 1 – 3 + 3 – 1 + 1 = 1
⇒ p (-1) ≠ 0
So, (x + 1) is not a factor of x⁴ + 3x³ + 3x² + x+ 1.

(iv) Let p (x) = x³ – x² – (2 + √2) x + √2
∴ p (- 1) =(- 1)3- (-1)2 – (2 + √2)(-1) + √2
= -1 – 1 + 2 + √2 + √2
= 2√2
⇒ p (-1) ≠ 0
So, (x + 1) is not a factor of x³ – x² – (2 + √2) x + √2.

Question 2.
Use the Factor Theorem to determine whether g (x) is a factor of p (x) in each of the following cases
(i) p (x)= 2x³ + x² – 2x – 1, g (x) = x + 1
(ii) p(x)= x³ + 3x² + 3x + 1, g (x) = x + 2
(iii) p (x) = x³ – 4x² + x + 6, g (x) = x – 3
Solution:
(i) We have, p (x)= 2x³ + x² – 2x – 1 and g (x) = x + 1
∴ p(-1) = 2(-1)³ + (-1)² – 2(-1) – 1
= 2(-1) + 1 + 2 – 1
= -2 + 1 + 2 -1 = 0
⇒ p(-1) = 0, so g(x) is a factor of p(x).

(ii) We have, p(x) x³ + 3x² + 3x + 1 and g(x) = x + 2
∴ p(-2) = (-2)³ + 3(-2)²+ 3(-2) + 1
= -8 + 12 – 6 + 1
= -14 + 13
= -1
⇒ p(-2) ≠ 0, so g(x) is not a factor of p(x).

(iii) We have, = x³ – 4x² + x + 6 and g (x) = x – 3
∴ p(3) = (3)³ – 4(3)² + 3 + 6
= 27 – 4(9) + 3 + 6
= 27 – 36 + 3 + 6 = 0
⇒ p(3) = 0, so g(x) is a factor of p(x).

Question 3.
Find the value of k, if x – 1 is a factor of p (x) in each of the following cases
(i) p (x) = x² + x + k
(ii) p (x) = 2x² + kx + √2
(iii) p (x) = kx² – √2 x + 1
(iv) p (x) = kx² – 3x + k
Solution:
For (x – 1) to be a factor of p(x), p(1) should be equal to 0.

(i) Here, p(x) = x² + x + k
Since, p(1) = (1)² +1 + k
⇒ p(1) = k + 2 = 0
⇒ k = -2.

(ii) Here, p (x) = 2x² + kx + √2
Since, p(1) = 2(1)² + k(1) + √2
= 2 + k + √2 =0
k = -2 – √2 = -(2 + √2)

(iii) Here, p (x) = kx² – √2 x + 1
Since, p(1) = k(1)² – (1) + 1
= k – √2 + 1 = 0
⇒ k = √2 -1

(iv) Here, p(x) = kx² – 3x + k
p(1) = k(1)² – 3(1) + k
= k – 3 + k
= 2k – 3 = 0
⇒ k = 34

Question 4.
Factorise
(i) 12x² – 7x +1
(ii) 2x² + 7x + 3
(iii) 6x² + 5x – 6
(iv) 3x² – x – 4
Solution:
(i) We have,
12x² – 7x + 1 = 12x² – 4x- 3x + 1
= 4x (3x – 1 ) -1 (3x – 1)
= (3x -1) (4x -1)
Thus, 12x² -7x + 3 = (2x – 1) (x + 3)

(ii) We have, 2x² + 7x + 3 = 2x² + x + 6x + 3
= x(2x + 1) + 3(2x + 1)
= (2x + 1)(x + 3)
Thus, 2×2 + 7x + 3 = (2x + 1)(x + 3)

(iii) We have, 6x² + 5x – 6 = 6x² + 9x – 4x – 6
= 3x(2x + 3) – 2(2x + 3)
= (2x + 3)(3x – 2)
Thus, 6x² + 5x – 6 = (2x + 3)(3x – 2)

(iv) We have, 3x² – x – 4 = 3x² – 4x + 3x – 4
= x(3x – 4) + 1(3x – 4) = (3x – 4)(x + 1)
Thus, 3x² – x – 4 = (3x – 4)(x + 1)

Question 5.
Factorise
(i) x³ – 2x² – x + 2
(ii) x³ – 3x² – 9x – 5
(iii) x³ + 13x² + 32x + 20
(iv) 2y³ + y² – 2y – 1
Solution:
(i) We have, x³ – 2x² – x + 2
Rearranging the terms, we have x³ – x – 2x² + 2
= x(x² – 1) – 2(x² -1) = (x² – 1)(x – 2)
= [(x)² – (1)²](x – 2)
= (x – 1)(x + 1)(x – 2)
[∵ (a² – b²) = (a + b)(a-b)]
Thus, x³ – 2x² – x + 2 = (x – 1)(x + 1)(x – 2)

(ii) We have, x³ – 3x² – 9x – 5
= x³ + x² – 4x² – 4x – 5x – 5 ,
= x² (x + 1) – 4x(x + 1) – 5(x + 1)
= (x + 1)(x² – 4x – 5)
= (x + 1)(x² – 5x + x – 5)
= (x + 1)[x(x – 5) + 1(x – 5)]
= (x + 1)(x – 5)(x + 1)
Thus, x³ – 3x² – 9x – 5 = (x + 1)(x – 5)(x +1)

(iii) We have, x³ + 13x² + 32x + 20
= x³ + x² + 12x² + 12x + 20x + 20
= x²(x + 1) + 12x(x +1) + 20(x + 1)
= (x + 1)(x² + 12x + 20)
= (x + 1)(x² + 2x + 10x + 20)
= (x + 1)[x(x + 2) + 10(x + 2)]
= (x + 1)(x + 2)(x + 10)
Thus, x³ + 13x² + 32x + 20
= (x + 1)(x + 2)(x + 10)

(iv) We have, 2y³ + y² – 2y – 1
= 2y³ – 2y² + 3y² – 3y + y – 1
= 2y²(y – 1) + 3y(y – 1) + 1(y – 1)
= (y – 1)(2y² + 3y + 1)
= (y – 1)(2y² + 2y + y + 1)
= (y – 1)[2y(y + 1) + 1(y + 1)]
= (y – 1)(y + 1)(2y + 1)
Thus, 2y³ + y² – 2y – 1
= (y – 1)(y + 1)(2y +1)

NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Exercise 2.5

Question 1.
Use suitable identities to find the following products
(i) (x + 4)(x + 10)
(ii) (x+8) (x -10)
(iii) (3x + 4) (3x – 5)
(iv) (y²+ 32) (y²– 32)
(v) (3 – 2x) (3 + 2x)
Solution:
(i) We have, (x+ 4) (x + 10)
Using identity,
(x+ a) (x+ b) = x² + (a + b) x+ ab.
We have, (x + 4) (x + 10) = x²+(4 + 10) x + (4 x 10)
= x² + 14x+40

(ii) We have, (x+ 8) (x -10)
Using identity,
(x + a) (x + b) = x² + (a + b) x + ab
We have, (x + 8) (x – 10) = x² + [8 + (-10)] x + (8) (- 10)
= x² – 2x – 80

(iii) We have, (3x + 4) (3x – 5)
Using identity,
(x + a) (x + b) = x² + (a + b) x + ab
We have, (3x + 4) (3x – 5) = (3x)² + (4 – 5) x + (4) (- 5)
= 9x² – x – 20

Question 2.
Evaluate the following products without multiplying directly
(i) 103 x 107
(ii) 95 x 96
(iii) 104 x 96
Solution:
(i)We have, 103 x 107 = (100 + 3) (100 + 7)
= ( 100)² + (3 + 7) (100)+ (3 x 7)
[Using (x + a)(x + b) = x² + (a + b)x + ab]
= 10000 + (10) x 100 + 21
= 10000 + 1000 + 21=11021

(ii) We have, 95 x 96 = (100 – 5) (100 – 4)
= ( 100)² + [(- 5) + (- 4)] 100 + (- 5 x – 4)
[Using (x + a)(x + b) = x² + (a + b)x + ab]
= 10000 + (-9) + 20 = 9120
= 10000 + (-900) + 20 = 9120

(iii) We have 104 x 96 = (100 + 4) (100 – 4)
= (100)²-4²
[Using (a + b)(a -b) = a²– b²]
= 10000 – 16 = 9984

Question 3.
Factorise the following using appropriate identities
(i) 9x²+ 6xy + y²
(ii) 4y²-4y + 1
(iii) x² – 𝑦2100
Solution:
(i) We have, 9x² + 6xy + y²
= (3x)² + 2(3x)(y) + (y)²
= (3x + y)²
[Using a² + 2ab + b² = (a + b)²]
= (3x + y)(3x + y)

(ii) We have, 4y² – 4y + 1²
= (2y)² + 2(2y)(1) + (1)²
= (2y -1)²
[Using a² – 2ab + b² = (a- b)²]
= (2y – 1)(2y – 1 )

Question 4.
Expand each of the following, using suitable identity
(i) (x+2y+ 4z)²
(ii) (2x – y + z)²
(iii) (- 2x + 3y + 2z)²
(iv) (3a -7b – c)^z
(v) (- 2x + 5y – 3z)²
(vi) [ 14a –14b + 1] ²
Solution:
We know that
(x + y + z)² = x² + y² + z² + 2xy + 2yz + 2zx

(i) (x + 2y + 4z)²
= x² + (2y)² + (4z)² + 2 (x) (2y) + 2 (2y) (4z) + 2(4z) (x)
= x² + 4y² + 16z² + 4xy + 16yz + 8 zx

(ii) (2x – y + z)² = (2x)² + (- y)² + z² + 2 (2x) (- y)+ 2 (- y) (z) + 2 (z) (2x)
= 4x² + y² + z² – 4xy – 2yz + 4zx

(iii) (- 2x + 3y + 2z)² = (- 2x)² + (3y)² + (2z)² + 2 (- 2x) (3y)+ 2 (3y) (2z) + 2 (2z) (- 2x)
= 4x² + 9y² + 4z² – 12xy + 12yz – 8zx

(iv) (3a -7b- c)²= (3a)² + (- 7b)² + (- c)² + 2 (3a) (- 7b) + 2 (- 7b) (- c) + 2 (- c) (3a)
= 9a² + 49b² + c² – 42ab + 14bc – 6ac

(v)(- 2x + 5y- 3z)² = (- 2x)² + (5y)² + (- 3z)² + 2 (- 2x) (5y) + 2 (5y) (- 3z) + 2 (- 3z) (- 2x)
= 4x² + 25y² + 9z² – 20xy – 30yz + 12zx

NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.5 Q4

Question 5.
Factorise
(i) 4 x² + 9y² + 16z² + 12xy – 24yz – 16xz
(ii) 2x² + y² + 8z² – 2√2xy + 4√2yz – 8xz
Solution:
(i) 4x² + 9y² + 16z² + 12xy – 24yz – 16xz
= (2x)² + (3y)² + (- 4z)² + 2 (2x) (3y) + 2 (3y) (- 4z) + 2 (- 4z) (2x)
= (2x + 3y – 4z)² = (2x + 3y + 4z) (2x + 3y – 4z)

(ii) 2x² + y² + 8z² – 2√2xy + 4√2yz – 8xz
= (- √2x)² + (y)² + (2 √2z)²y + 2(- √2x) (y)+ 2 (y) (2√2z) + 2 (2√2z) (- √2x)
= (- √2x + y + 2 √2z)²
= (- √2x + y + 2 √2z) (- √2x + y + 2 √2z)

Question 6.
Write the following cubes in expanded form

Solution:
We have, (x + y)³ = x³ + y³ + 3xy(x + y) …(1)
and (x – y)³ = x³ – y³ – 3xy(x – y) …(2)

(i) (2x + 1)³ = (2x)³ + (1)³ + 3(2x)(1)(2x + 1) [By (1)]
= 8x³ + 1 + 6x(2x + 1)
= 8x³ + 12x² + 6x + 1

(ii) (2a – 3b)³ = (2a)³ – (3b)³ – 3(2a)(3b)(2a – 3b) [By (2)]
= 8a³ – 27b³ – 18ab(2a – 3b)
= 8a³ – 27b³ – 36a²b + 54ab²

NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.5 Q6.1

NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.5 Q6.2

Question 7.
Evaluate the following using suitable identities
(i) (99)³
(ii) (102)³
(iii) (998)³
Solution:
(i) We have, 99 = (100 -1)
∴ 99³ = (100 – 1)³
= (100)³ – 1³ – 3(100)(1)(100 -1)
[Using (a – b)³ = a³ – b³ – 3ab (a – b)]
= 1000000 – 1 – 300(100 – 1)
= 1000000 -1 – 30000 + 300
= 1000300 – 30001 = 970299

(ii) We have, 102 =100 + 2
∴ 102³ = (100 + 2)³
= (100)³ + (2)³ + 3(100)(2)(100 + 2)
[Using (a + b)³ = a³ + b³ + 3ab (a + b)]
= 1000000 + 8 + 600(100 + 2)
= 1000000 + 8 + 60000 + 1200 = 1061208

(iii) We have, 998 = 1000 – 2
∴ (998)³ = (1000-2)³
= (1000)³– (2)³ – 3(1000)(2)(1000 – 2)
[Using (a – b)³ = a³ – b³ – 3ab (a – b)]
= 1000000000 – 8 – 6000(1000 – 2)
= 1000000000 – 8 – 6000000 +12000
= 994011992

Question 8.
Factorise each of the following
(i) 8a³ +b³ + 12a²b+6ab²
(ii) 8a³ -b³-12a²b+6ab²(iii) 27-125a³ -135a+225a²(iv) 64a³ -27b³ -144a²b + 108ab²

Solution:
(i) 8a³ +b³ +12a²b+6ab²
= (2a)³ + (b)³ + 6ab(2a + b)
= (2a)³ + (b)³ + 3(2a)(b)(2a + b)
= (2 a + b)³
[Using a³ + b³ + 3 ab(a + b) = (a + b)³]
= (2a + b)(2a + b)(2a + b)

(ii) 8a³ – b³ – 12o²b + 6ab²
= (2a)³ – (b)³ – 3(2a)(b)(2a – b)
= (2a – b)³
[Using a³ + b³ + 3 ab(a + b) = (a + b)³]
= (2a – b) (2a – b) (2a – b)

(iii) 27 – 125a³ – 135a + 225a²
= (3)³ – (5a)³ – 3(3)(5a)(3 – 5a)
= (3 – 5a)³
[Using a³ + b³ + 3 ab(a + b) = (a + b)³]
= (3 – 5a) (3 – 5a) (3 – 5a)

(iv) 64a³ -27b³ -144a²b + 108ab²
= (4a)³ – (3b)³ – 3(4a)(3b)(4a – 3b)
= (4a – 3b)³
[Using a³ – b³ – 3 ab(a – b) = (a – b)³]
= (4a – 3b)(4a – 3b)(4a – 3b)

NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.5 Q8.1

Question 9.
Verify
(i) x³ + y³ = (x + y)-(x² – xy + y²)
(ii) x³ – y³ = (x – y) (x² + xy + y²)
Solution:
(i) ∵ (x + y)³ = x³ + y³ + 3xy(x + y)
⇒ (x + y)³ – 3(x + y)(xy) = x³ + y³
⇒ (x + y)[(x + y)2-3xy] = x³ + y³
⇒ (x + y)(x² + y² – xy) = x³ + y³
Hence, verified.

(ii) ∵ (x – y)³ = x³ – y³ – 3xy(x – y)
⇒ (x – y)³ + 3xy(x – y) = x³ – y³
⇒ (x – y)[(x – y)² + 3xy)] = x³ – y³
⇒ (x – y)(x² + y² + xy) = x³ – y³
Hence, verified.

Question 10.
Factorise each of the following
(i) 27y³ + 125z³
(ii) 64m³ – 343n³
[Hint See question 9]
Solution:
(i) We know that
x³ + y³ = (x + y)(x² – xy + y²)
We have, 27y³ + 125z³ = (3y)³ + (5z)³
= (3y + 5z)[(3y)² – (3y)(5z) + (5z)²]
= (3y + 5z)(9y² – 15yz + 25z²)

(ii) We know that
x³ – y³ = (x – y)(x² + xy + y²)
We have, 64m³ – 343n³ = (4m)³ – (7n)³
= (4m – 7n)[(4m)² + (4m)(7n) + (7n)²]
= (4m – 7n)(16m² + 28mn + 49n²)

Question 11.
Factorise 27x³ +y³ +z³ -9xyz.
Solution:
We have,
27x³ + y³ + z³ – 9xyz = (3x)³ + (y)³ + (z)³ – 3(3x)(y)(z)
Using the identity,
x³ + y³ + z³ – 3xyz = (x + y + z)(x² + y² + z² – xy – yz – zx)
We have, (3x)³ + (y)³ + (z)³ – 3(3x)(y)(z)
= (3x + y + z)[(3x)³ + y³ + z³ – (3x × y) – (y × 2) – (z × 3x)]
= (3x + y + z)(9x² + y² + z² – 3xy – yz – 3zx)

Question 12.
Verify that
x³ +y³ +z³ – 3xyz = 12 (x + y+z)[(x-y)² + (y – z)² +(z – x)²]
Solution:
R.H.S
= 12(x + y + z)[(x – y)²+(y – z)²+(z – x)²]
= 12 (x + y + 2)[(x² + y² – 2xy) + (y² + z² – 2yz) + (z² + x² – 2zx)]
= 12 (x + y + 2)(x² + y² + y² + z² + z² + x² – 2xy – 2yz – 2zx)
= 12 (x + y + z)[2(x² + y² + z² – xy – yz – zx)]
= 2 x 12 x (x + y + z)(x² + y² + z² – xy – yz – zx)
= (x + y + z)(x² + y² + z² – xy – yz – zx)
= x³ + y³ + z³ – 3xyz = L.H.S.
Hence, verified.

Question 13.
If x + y + z = 0, show that x³ + y³ + z³ = 3 xyz.
Solution:
Since, x + y + z = 0
⇒ x + y = -z (x + y)³ = (-z)³
⇒ x³ + y³ + 3xy(x + y) = -z³
⇒ x³ + y³ + 3xy(-z) = -z³ [∵ x + y = -z]
⇒ x³ + y³ – 3xyz = -z³
⇒ x³ + y³ + z³ = 3xyz
Hence, if x + y + z = 0, then
x³ + y³ + z³ = 3xyz

Question 14.
Without actually calculating the cubes, find the value of each of the following
(i) (- 12)³ + (7)³ + (5)³(ii) (28)³ + (- 15)³ + (- 13)³
Solution:
(i) We have, (-12)³ + (7)³ + (5)³
Let x = -12, y = 7 and z = 5.
Then, x + y + z = -12 + 7 + 5 = 0
We know that if x + y + z = 0, then, x³ + y³ + z³ = 3xyz
∴ (-12)³ + (7)³ + (5)³ = 3[(-12)(7)(5)]
= 3[-420] = -1260

(ii) We have, (28)³ + (-15)³ + (-13)³
Let x = 28, y = -15 and z = -13.
Then, x + y + z = 28 – 15 – 13 = 0
We know that if x + y + z = 0, then x³ + y³ + z³ = 3xyz
∴ (28)³ + (-15)³ + (-13)³ = 3(28)(-15)(-13)
= 3(5460) = 16380

Question 15.
Give possible expressions for the length and breadth of each of the following rectangles, in which their areas are given
(i) Area 25a² – 35a + 12
(ii) Area 35y² + 13y – 12
Solution:
Area of a rectangle = (Length) x (Breadth)
(i) 25a² – 35a + 12 = 25a² – 20a – 15a + 12 = 5a(5a – 4) – 3(5a – 4) = (5a – 4)(5a – 3)
Thus, the possible length and breadth are (5a – 3) and (5a – 4).

(ii) 35y²+ 13y -12 = 35y² + 28y – 15y -12
= 7y(5y + 4) – 3(5y + 4) = (5 y + 4)(7y – 3)
Thus, the possible length and breadth are (7y – 3) and (5y + 4).

Question 16.
What are the possible expressions for the dimensions of the cuboids whose volumes are given below?
(i) Volume 3x² – 12x
(ii) Volume 12ky² + 8ky – 20k
Solution:
Volume of a cuboid = (Length) x (Breadth) x (Height)
(i) We have, 3x² – 12x = 3(x² – 4x)
= 3 x x x (x – 4)
∴ The possible dimensions of the cuboid are 3, x and (x – 4).

(ii) We have, 12ky² + 8ky – 20k
= 4[3ky² + 2ky – 5k] = 4[k(3y² + 2y – 5)]
= 4 x k x (3y² + 2y – 5)
= 4k[3y² – 3y + 5y – 5]
= 4k[3y(y – 1) + 5(y – 1)]
= 4k[(3y + 5) x (y – 1)]
= 4k x (3y + 5) x (y – 1)
Thus, the possible dimensions of the cuboid are 4k, (3y + 5) and (y -1).

NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Read More »

NCERT Solution for Class 9 Maths Chapter 1 Number System

Leave a Comment / NCERT Solution, Class 9 / Mr Ahmad / June 29, 2022

NCERT Solutions for Class 9 Maths Chapter 1 Number Systems is prepared by our best subject experts teachers group thats help students to understand all the topics easily. These Solutions of NCERT Maths help the students in solving the problems efficiently for the upcoming exams. With the help of these NCERT Solutions for Class 9 Maths, students can understand the complex topics of class 9 Maths. They also focus on formulating the solutions of Maths in such a way that it is easy for the students to understand.

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These NCERT Solutions for Class 9 Maths Chapter 1 helps students to understand rest of the chapter as well as upcoming Class 10 Maths problems. Based on these NCERT Solutions, students can practise and prepare for their upcoming any board exams that is based on NCERT Syllabus. So all you need to do stay focus and practising more an more questions.

NCERT Solutions for Class 9 Maths Chapter 1 Number Systems

Exercise 1.1 Page: 5

Question 1.
Is zero a rational number? Can you write it in the form 𝑝𝑞,where p and q are integers and q ≠0?
Solution:
We know that, a number is said to be rational if it can be written in the form p/q , where p and q are integers and q ≠ 0.

Taking the case of ‘0’,

Zero can be written in the form 0/1, 0/2, 0/3 … as well as , 0/1, 0/2, 0/3 ..

Since it satisfies the necessary condition, we can conclude that 0 can be written in the p/q form, where q can either be positive or negative number.

Hence, 0 is a rational number.

Question 2.
Find six rational numbers between 3 and 4.
Solution:
Let q_i be the rational number between 3
and 4, where j = 1 to 6.
∴ Six rational numbers are as follows:

Thus, the six rational numbers between 3 and 4 are

Question 3.
Find five rational numbers between 35 and 45.
Solution:
Since, we need to find five rational numbers, therefore, multiply numerator and denominator by 6.

Question 4.
State whether the following statements are true or false. Give reasons for your answers.
(i) Every natural number is a whole number.
(ii) Every integer is a whole number.
(iii) Every rational number is a whole number.
Solution:
(i) True
∵ The collection of all natural numbers and 0 is called whole numbers.
(ii) False
∵ Negative integers are not whole numbers.
(iii) False
∵ Rational numbers are of the form p/q, q ≠ 0 and q does not divide p completely that are not whole numbers.

Exercise 1.2 Page: 8

Question 1.
State whether the following statements are true or false. Justify your answers.
(i) Every irrational number is a real number.
(ii) Every point on the number line is of the form √m , where m is a natural number.
(iii) Every real number is an irrational number.
Solution:
(i) True
Because all rational numbers and all irrational numbers form the group (collection) of real numbers.
(ii) False
Because negative numbers cannot be the square root of any natural number.
(iii) False
Because rational numbers are also a part of real numbers.

Question 2.
Are the square roots of all positive integers irrational? If not, give an example of the square root of a number that is a rational number.
Solution:
No, the square roots of all positive integers are not irrational.

For example,

√4 = 2 is rational.

√9 = 3 is rational.

Hence, the square roots of positive integers 4 and 9 are not irrational. ( 2 and 3, respectively).

Question 3.
Show how √5 can be represented on the number line.
Solution:
Draw a number line and take point O and A on it such that OA = 1 unit. Draw BA ⊥ OA as BA = 1 unit. Join OB = √2 units.
Now draw BB₁ ⊥ OB such that BB₁ =1 unit. Join OB₁ = √3 units.
Next, draw B₁B₂⊥ OB₁such that B₁B₂ = 1 unit.
Join OB₂ = units.
Again draw B₂B₃ ⊥OB₂ such that B₂B₃ = 1 unit.
Join OB₃ = √5 units.

Take O as centre and OB₃ as radius, draw an arc which cuts the number line at D.
Point D
represents √5 on the number line.

Question 4.
Take a large sheet of paper and construct the ‘square root spiral’ in the following fashion. Start with a point O and draw a line segment OP1 of unit length. Draw a line segment P1P2 perpendicular to OP₁ of unit length (see Fig. 1.9). Now draw a line segment P₂P₃perpendicular to OP₂. Then draw a line segment P₃P₄ perpendicular to OP₃. Continuing in Fig. 1.9 :

Constructing this manner, you can get the line segment P_n-1Pn by square root spiral drawing a line segment of unit length perpendicular to OP_n-1. In this manner, you will have created the points P₂, P₃,….,Pn,… ., and joined them to create a beautiful spiral depicting √2, √3, √4, …

Solution:

Step 1: Mark a point O on the paper. Here, O will be the center of the square root spiral.

Step 2: From O, draw a straight line, OA, of 1cm horizontally.

Step 3: From A, draw a perpendicular line, AB, of 1 cm.

Step 4: Join OB. Here, OB will be of √2

Step 5: Now, from B, draw a perpendicular line of 1 cm and mark the end point C.

Step 6: Join OC. Here, OC will be of √3

Step 7: Repeat the steps to draw √4, √5, √6….

Exercise 1.3 Page: 14

Question 1.
Write the following in decimal form and say what kind of decimal expansion each has

(i) 36/100

Solution:

= 0.36 (Terminating)

(ii)1/11

Solution:

= 4.125 (Terminating)

(iv) 3/13

Solution:

(v) 2/11

Solution:

(vi) 329/400

Solution:

= 0.8225 (Terminating)

Question 2.
You know that 1/7 = 0.142857. Can you predict what the decimal expansions of 2/7, 3/7, 4/7, 5/7, 6/7 are, without actually doing the long division? If so, how?

Solution:

Thus, without actually doing the long division we can predict the decimal expansions of the given rational numbers.

Question 3.
Express the following in the form 𝑝𝑞 where p and q are integers and q ≠ 0.
(i) 0.6¯
(ii) 0.47¯
(iii) 0.001⎯⎯⎯⎯⎯⎯⎯⎯⎯
Solution:
(i) Let x = 0.6¯ = 0.6666… … (1)
As there is only one repeating digit,
multiplying (1) by 10 on both sides, we get
10x = 6.6666… … (2)
Subtracting (1) from (2), we get
10x – x = 6.6666… -0.6666…
⇒ 9x = 6 ⇒ x = 6/9 = 2/3
Thus, 0.6¯ = 2/3

(ii) Let x = 0.47¯ = 0.4777… … (1)
As there is only one repeating digit, multiplying (1) by lo on both sides, we get
10x = 4.777
Subtracting (1) from (2), we get
10x – x = 4.777…… – 0.4777…….
⇒ 9x = 4.3 ⇒ x = 43/90
Thus, 0.47¯ = 43/90

Solution:

Assume that x = 0.001001…

Then, 1000x = 1.001001…

1000x = 1 + x

999x = 1

x = 1/999

Question 4.
Express 0.99999…. in the form p/q . Are you surprised by your answer? With your teacher and classmates discuss why the answer makes sense.
Solution:
Let x = 0.99999….. …. (i)
As there is only one repeating digit,
multiplying (i) by 10 on both sides, we get
10x = 9.9999 … (ii)
Subtracting (i) from (ii), we get
10x – x = (99999 ) — (0.9999 )
⇒ 9x = 9 ⇒ x = 99 = 1
Thus, 0.9999 =1
As 0.9999… goes on forever, there is no such a big difference between 1 and 0.9999
Hence, both are equal.

Question 5.
What can the maximum number of digits be in the repeating block of digits in the decimal expansion of 1/17? Perform the division to check your answer.
Solution:
In 1/17, In the divisor is 17.
Since, the number of entries in the repeating block of digits is less than the divisor, then the maximum number of digits in the repeating block is 16.
Dividing 1 by 17, we have

The remainder I is the same digit from which we started the division.

Thus, there are 16 digits in the repeating block in the decimal expansion of 1/17.
Hence, our answer is verified.

Question 6.
Look at several examples of rational numbers in the form 𝑝/𝑞 (q ≠ 0). Where, p and q are integers with no common factors other that 1 and having terminating decimal representations (expansions). Can you guess what property q must satisfy?
Solution:
Let us look decimal expansion of the following terminating rational numbers:

We observe that the prime factorisation of q (i.e. denominator) has only powers of 2 or powers of 5 or powers of both.

Question 7.
Write three numbers whose decimal expansions are non-terminating non-recurring.
Solution:
√2 = 1.414213562 ………..
√3 = 1.732050808 …….
√5 = 2.23606797 …….

Question 8.
Find three different irrational numbers between the rational numbers 57 and 911 .
Solution:
We have,

Three different irrational numbers are:

0.73073007300073000073…
0.75075007300075000075…
0.76076007600076000076…

Question 9.
Classify the following numbers as rational or irrational
(i)√23

Solution:

√23 = 4.79583152331…

Since the number is non-terminating non-recurring therefore, it is an irrational number.

(ii)√225

Solution:

√225 = 15 = 15/1

Since the number can be represented in p/q form, it is a rational number.

(iii) 0.3796

Solution:

Since the number,0.3796, is terminating, it is a rational number.

(iv) 7.478478

Solution:

The number,7.478478, is non-terminating but recurring, it is a rational number.

(v) 1.101001000100001…

Solution:

Since the number,1.101001000100001…, is non-terminating non-repeating (non-recurring), it is an irrational number.

Exercise 1.4 Page: 18

Question 1 – Visualise 3.765 on the number line, using successive magnification.

Solution:

Exercise 1.5 Page: 24

1. Classify the following numbers as rational or irrational:

(i) 2 –√5

Solution:

We know that, √5 = 2.2360679…

Here, 2.2360679…is non-terminating and non-recurring.

Now, substituting the value of √5 in 2 –√5, we get,

2-√5 = 2-2.2360679… = -0.2360679

Since the number, – 0.2360679…, is non-terminating non-recurring, 2 –√5 is an irrational number.

(ii) (3 +√23)- √23

Solution:

(3 +√23) –√23 = 3+√23–√23

= 3

= 3/1

Since the number 3/1 is in p/q form, (3 +√23)- √23 is rational.

(iii) 2√7/7√7

Solution:

2√7/7√7 = ( 2/7)× (√7/√7)

We know that (√7/√7) = 1

Hence, ( 2/7)× (√7/√7) = (2/7)×1 = 2/7

Since the number, 2/7 is in p/q form, 2√7/7√7 is rational.

(iv) 1/√2

Solution:

Multiplying and dividing numerator and denominator by √2 we get,

(1/√2) ×(√2/√2)= √2/2 ( since √2×√2 = 2)

We know that, √2 = 1.4142…

Then, √2/2 = 1.4142/2 = 0.7071..

Since the number , 0.7071..is non-terminating non-recurring, 1/√2 is an irrational number.

(v) 2

Solution:

We know that, the value of = 3.1415

Hence, 2 = 2×3.1415.. = 6.2830…

Since the number, 6.2830…, is non-terminating non-recurring, 2 is an irrational number.

2. Simplify each of the following expressions:

(i) (3+√3)(2+√2)

Solution:

(3+√3)(2+√2 )

Opening the brackets, we get, (3×2)+(3×√2)+(√3×2)+(√3×√2)

= 6+3√2+2√3+√6

(ii) (3+√3)(3-√3 )

Solution:

(3+√3)(3-√3 ) = 3²-(√3)² = 9-3

= 6

(iii) (√5+√2)²

Solution:

(√5+√2)²= √5²+(2×√5×√2)+ √2²

= 5+2×√10+2 = 7+2√10

(iv) (√5-√2)(√5+√2)

Solution:

(√5-√2)(√5+√2) = (√5²-√2²) = 5-2 = 3

3. Recall, π is defined as the ratio of the circumference (say c) of a circle to its diameter, (say d). That is, π =c/d. This seems to contradict the fact that π is irrational. How will you resolve this contradiction?

Solution:

There is no contradiction. When we measure a value with a scale, we only obtain an approximate value. We never obtain an exact value. Therefore, we may not realize whether c or d is irrational. The value of π is almost equal to 22/7 or 3.142857…

4. Represent (√9.3) on the number line.

Solution:

Step 1: Draw a 9.3 units long line segment, AB. Extend AB to C such that BC=1 unit.

Step 2: Now, AC = 10.3 units. Let the centre of AC be O.

Step 3: Draw a semi-circle of radius OC with centre O.

Step 4: Draw a BD perpendicular to AC at point B intersecting the semicircle at D. Join OD.

Step 5: OBD, obtained, is a right angled triangle.

Here, OD 10.3/2 (radius of semi-circle), OC = 10.3/2 , BC = 1

OB = OC – BC

⟹ (10.3/2)-1 = 8.3/2

Using Pythagoras theorem,

We get,

OD²=BD²+OB²

⟹ (10.3/2)² = BD²+(8.3/2)²

⟹ BD^{2 =}(10.3/2)²-(8.3/2)²

⟹ (BD)²= (10.3/2)-(8.3/2)(10.3/2)+(8.3/2)

⟹ BD² = 9.3

⟹ BD = √9.3

Thus, the length of BD is √9.3.

Step 6: Taking BD as radius and B as centre draw an arc which touches the line segment. The point where it touches the line segment is at a distance of √9.3 from O as shown in the figure.

5. Rationalize the denominators of the following:

(i) 1/√7

Solution:

Multiply and divide 1/√7 by √7

(1×√7)/(√7×√7) = √7/7

(ii) 1/(√7-√6)

Solution:

Multiply and divide 1/(√7-√6) by (√7+√6)[1/(√7-√6)]×(√7+√6)/(√7+√6) = (√7+√6)/(√7-√6)(√7+√6)

= (√7+√6)/√7²-√6²[denominator is obtained by the property, (a+b)(a-b) = a²-b²]

= (√7+√6)/(7-6)

= (√7+√6)/1

= √7+√6

(iii) 1/(√5+√2)

Solution:

Multiply and divide 1/(√5+√2) by (√5-√2)[1/(√5+√2)]×(√5-√2)/(√5-√2) = (√5-√2)/(√5+√2)(√5-√2)

= (√5-√2)/(√5²-√2²) [denominator is obtained by the property, (a+b)(a-b) = a²-b²]

= (√5-√2)/(5-2)

= (√5-√2)/3

(iv) 1/(√7-2)

Solution:

Multiply and divide 1/(√7-2) by (√7+2)

1/(√7-2)×(√7+2)/(√7+2) = (√7+2)/(√7-2)(√7+2)

= (√7+2)/(√7²-2²) [denominator is obtained by the property, (a+b)(a-b) = a²-b²]

= (√7+2)/(7-4)

= (√7+2)/3

Exercise 1.6 Page: 26

1. Find:

(i)64^1/2

Solution:

64^1/2 = (8×8)^1/2

= (8²)^½

= 8¹ [⸪2×1/2 = 2/2 =1]

= 8

(ii)32^1/5

Solution:

32^1/5= (2⁵)^1/5

= (2⁵)^⅕

= 2¹ [⸪5×1/5 = 1]

= 2

(iii)125^1/3

Solution:

(125)^1/3 = (5×5×5)^1/3

= (5³)^⅓

= 5¹ (3×1/3 = 3/3 = 1)

= 5

2. Find:

(i) 9^3/2

Solution:

9^3/2 = (3×3)^3/2

= (3²)^3/2

= 3³ [⸪2×3/2 = 3]

=27

(ii) 32^2/5

Solution:

32^2/5= (2×2×2×2×2)^2/5

= (2⁵)^2⁄5

= 2² [⸪5×2/5= 2]

= 4

(iii)16^3/4

Solution:

16^3/4= (2×2×2×2)^3/4

= (2⁴)^3⁄4

= 2³ [⸪4×3/4 = 3]

= 8

(iv) 125^-1/3

125^-1/3= (5×5×5)^-1/3

= (5³)^-1⁄3

= 5^-1 [⸪3×-1/3 = -1]

= 1/5

3. Simplify:

(i) 2^2/3×2^1/5

Solution:

2^2/3×2^1/5= 2^(2/3)+(1/5) [⸪Since, a^m×aⁿ=a^m+n____ Laws of exponents]

= 2^13/15 [⸪2/3 + 1/5 = (2×5+3×1)/(3×5) = 13/15]

(ii) (1/3³)⁷

Solution:

(1/3³)⁷= (3^-3)⁷ [⸪Since,(a^m)ⁿ= a^{m x n}____ Laws of exponents]

= 3^-21

(iii) 11^1/2/11^1/4

Solution:

11^1/2/11^1/4= 11^(1/2)-(1/4)

= 11^1/4 [⸪(1/2) – (1/4) = (1×4-2×1)/(2×4) = 4-2)/8 = 2/8 = ¼ ]

(iv) 7^1/2×8^1/2

Solution:

7^1/2×8^1/2 = (7×8)^1/2 [⸪Since, (a^m×b^m= (a×b)^m ____ Laws of exponents]

= 56^1/2

NCERT Solution for Class 9 Maths Chapter 1 Number System Read More »

NCERT Solutions for Class 9 Maths – Chapter-wise

Leave a Comment / NCERT Solution, Class 9 / Mr Ahmad / June 28, 2022

NCERT Solutions for Class 9 Maths includes solutions to all the questions given in the NCERT textbook for Class 9. There are 15 chapters in class 9 maths. These NCERT Solutions for Class 9 cover all the topics included in the NCERT textbook-like Number System, Coordinate Geometry, Polynomials, Euclid’s Geometry, Quadrilaterals, Triangles, Circles, Constructions, Surface Areas and Volumes, Statistics, Probability, etc.

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Student those who are searching for NCERT Solutions for Class 9 Maths, then this the right place where you can find chapter-wise NCERT solutions for class 9. All these solution are curated by our expert teacher group, with the help of these Solutions of NCERT Books for Class 9 Maths, students can practise all types of questions from the chapters. It will boost your skill also help them in building a foundation for higher-level classes.

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NCERT Solutions Class 9 Maths

Chapter 1 Number System
Chapter 2 Polynomials
Chapter 3 Coordinate Geometry
Chapter 4 Linear Equations in Two Variables
Chapter 5 Introduction to Euclids Geometry
Chapter 6 Lines and Angles
Chapter 7 Triangles
Chapter 8 Quadrilaterals
Chapter 9 Areas of Parallelograms and Triangles
Chapter 10 Circles
Chapter 11 Constructions
Chapter 12 Heron’s Formula
Chapter 13 Surface Areas and Volumes
Chapter 14 Statistics
Chapter 15 Probability

NCERT Solutions Class 9 Maths Chapter 1 Number System

This chapter discusses different topics, including rational numbers and irrational numbers. Students will also be learning the extended version of the number line and how to represent numbers (integers, rational and irrational) on it.

Important Formulas –

Operations on Real Numbers

√(a/b) = √a/√b

√ab = √a √b

(√a + √b) (√a – √b) = a – b

(a + √b) (a – √b) = a² – b

(√a + √b) (√c + √d) = √ac + √ad + √bc + √bd

(√a + √b)² = a + 2√ab + b

Laws of Exponents for Real Numbers

a^m . aⁿ = a^{m + n}

(a^m)ⁿ = a^mn

a^m/aⁿ = a^{m – n}, m > n

a^mb^m = (ab)^m

NCERT Solutions for Class 9 Maths Chapter 2 Polynomials

The second chapter of Class 9 Maths Chapter 2 Polynomials contain the topic that is related with Polynomials, Polynomial is an expression that consists of variables and coefficients, involving the operations of addition, subtraction, multiplication, and non-negative integer exponents of variables. Students will also learn polynomial, degrees, coefficient, zeros and terms of a polynomial.

Important Formulas –

(x + y)² = x² + 2xy + y²

(x – y)² = x² – 2xy + y²

x² – y² = (x + y) (x – y)

(x + y + z)² = x² + y² + z² + 2xy + 2yz + 2zx

(x + y)³ = x³ + y³ + 3xy(x + y)

(x – y)³ = x³ – y³ – 3xy(x – y)

x³ + y³ + z³ – 3xyz = (x + y + z) (x² + y² + z² – xy – yz – zx)

Dividend = (Divisor × Quotient) + Remainder

NCERT Solutions for Class 9 Maths Chapter 3 Coordinate Geometry

In this Chapter 3 Coordinate Geometry includes the concepts of the cartesian plane, coordinates of a point in xy – plane, terms, notations associated with the coordinate plane, including the x-axis, y-axis, x- coordinate, y-coordinate, origin, quadrants and more.

NCERT Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variable

Important Points

An equation of the form ax + by + c = 0, where a, b and c are real numbers, such that a and b are not both zero, is called a linear equation in two variables.
A linear equation in two variables has infinitely many solutions.
The graph of every linear equation in two variables is a straight line.
x = 0 is the equation of the y-axis and y = 0 is the equation of the x-axis.
The graph of x = a is a straight line parallel to the y-axis.
The graph of y = a is a straight line parallel to the x-axis.
An equation of the type y = mx represents a line passing through the origin.

NCERT Solutions for Class 9 Maths Chapter 5 Introduction Euclids Geometry

Important Axioms and Postulates –

Some of Euclid’s axioms are:

(1) Things which are equal to the same thing are equal to one another.

(2) If equals are added to equals, the wholes are equal.

(3) If equals are subtracted from equals, the remainders are equal.

(4) Things which coincide with one another are equal to one another.

(5) The whole is greater than the part.

(6) Things which are double of the same things are equal to one another.

(7) Things which are halves of the same things are equal to one another.

Euclid’s Five Postulates

Postulate 1 – A straight line may be drawn from any one point to any other point.

Postulate 2 – A terminated line can be produced indefinitely.

Postulate 3 – A circle can be drawn with any centre and any radius.

Postulate 4 – All right angles are equal to one another.

Postulate 5 – If a straight line falling on two straight lines makes the interior angles on the same side of it taken together less than two right angles, then the two straight lines, if produced indefinitely, meet on that side on which the sum of angles is less than two right angles.

NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles

In this topic we learn basic Geometry primarily focusing on the properties of the angles formed i) when two lines intersect each other and ii) when a line intersects two or more parallel lines at distinct points.

NCERT Solutions for Class 9 Maths Chapter 7 Triangles

In this Chapter 7 students will study about the congruence of triangles, rules of congruence, some properties of triangles and the inequalities in triangles in details. All These 5 exercises, in which the students are asked “to-prove” as well as application-level problems.

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