Category: CBSE Class 9

RD Sharma Class 9 Solutions Chapter 7 Introduction to Euclid’s Geometry Ex 7.1

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 7 Introduction to Euclid’s Geometry Ex 7.1

Other Exercises

• RD Sharma Class 9 Solutions Chapter 7 Introduction to Euclid’s Geometry Ex 7.1
• RD Sharma Class 9 Solutions Chapter 7 Introduction to Euclid’s Geometry Ex 7.2
• RD Sharma Class 9 Solutions Chapter 7 Introduction to Euclid’s Geometry Ex 7.3
• RD Sharma Class 9 Solutions Chapter 7 Introduction to Euclid’s Geometry Ex 7.4
• RD Sharma Class 9 Solutions Chapter 7 Introduction to Euclid’s Geometry VSAQS
• RD Sharma Class 9 Solutions Chapter 7 Introduction to Euclid’s Geometry MCQS

Question 1.
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 = 12
(ii) x-\(\frac { y }{ 2 }\) -5 = 0
(iii) 2x + 3y = 9.35
(iv) 3x = -7y
(v) 2x + 3 = 0
(vi) y – 5 = 0
(vii) 4 = 3x
(viii) y = \(\frac { x }{ 2 }\)
Solution:
(i) -2x + 3y = 12
⇒  -2x + 3y – 12 = 0
Here a -2, b = 3, c = – 12
(ii) x – \(\frac { y }{ 2 }\) -5 = 0
Here a = 1, b =\(\frac { 1 }{ 2 }\) ,c = -5
(iii) 2x + 3y = 9.35
⇒  2x + 3y – 9.35 = 0
Here a = 2, b = 3, c = – 9.35
(iv) 3x = -7y
⇒  3x + 7y + 0 = 0
Here a = 3, b = 7,c = 0
(v) 2x + 3 = 0
⇒ 2x + 0y + 3 = 0
Here a = 2, b = 0, c = 3
(vi) y-5 = 0 ⇒ ox+y-5 = 0
Here a = 0, b = 1, c = -5
(vii) 4 = 3x
⇒ 3x – 4 = 0
⇒ 3x + 0y – 4 = 0
Here a = 3, b = 0, c = -4
(Viii) y= \(\frac { x }{ 2 }\)
⇒  \(\frac { x }{ 2 }\) – y+ 0 = 0
⇒  x-2y + 0 = 0
Here a = 1, y = -2, c = 0

Question 2.
Write each of the following as an equation in two variables.
(i) 2x = -3
(ii) y = 3
(iii) 5x = \(\frac { 7 }{ 2 }\)
(iv) y =\(\frac { 3 }{ 2 }\)x
Solution:
(i) 2x = -3⇒  2x + 3 = 0
⇒ 2x + 0y + 3 = 0
(ii) y= 3 ⇒  y-3=0
⇒  0x+ y-3 = 0
(iii) 5x =\(\frac { 7 }{ 2 }\) ⇒ 10x = 7
⇒  10x + 0y – 7 = 0
(iv) y=\(\frac { 3 }{ 2 }\)x⇒2y = 3x
⇒ 3x – 2y + 0 = 0

Question 3.
The cost of ball pen is ₹5 less than half of the cost of fountain pen. Write this statement as a linear equation in two variables.
Solution:
Let cost of a fountain pen = ₹x
and cost of ball pen = ₹y
∴ According to the condition,
y = \(\frac { x }{ 2 }\) -5
⇒  2y = x – 10
⇒  x – 2y – 10 = 0

Hope given RD Sharma Class 9 Solutions Chapter 7 Introduction to Euclid’s Geometry Ex 7.1 are helpful to complete your math homework.

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RD Sharma Class 9 Solutions Chapter 6 Factorisation of Polynomials Ex 6.5

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 6 Factorisation of Polynomials Ex 6.5

Other Exercises

• RD Sharma Class 9 Solutions Chapter 6 Factorisation of Polynomials Ex 6.1
• RD Sharma Class 9 Solutions Chapter 6 Factorisation of Polynomials Ex 6.2
• RD Sharma Class 9 Solutions Chapter 6 Factorisation of Polynomials Ex 6.3
• RD Sharma Class 9 Solutions Chapter 6 Factorisation of Polynomials Ex 6.4
• RD Sharma Class 9 Solutions Chapter 6 Factorisation of Polynomials Ex 6.5
• RD Sharma Class 9 Solutions Chapter 6 Factorisation of Polynomials VSAQS
• RD Sharma Class 9 Solutions Chapter 6 Factorisation of Polynomials MCQS

Using factor theorem, factorize each of the following polynomials:
Question 1.
x³ + 6x² + 11x + 6
Solution:

Question 2.
x³ + 2x² – x – 2
Solution:

Question 3.
x³ – 6x² + 3x + 10
Solution:

Question 4.
x⁴ – 7x³ + 9x² + x- 10
Solution:

Question 5.
3x³ – x² – 3x + 1
Solution:

Question 6.
x³ – 23x² + 142x – 120        [NCERT]
Solution:

Question 7.
y³ – 7y + 6
Solution:
Let f(y) = y³ – 7y + 6

Question 8.
X³ -10x² – 53x – 42
Solution:

Question 9.
y³ – 2y²– 29y – 42
Solution:

Question 10.
2y³ – 5y² – 19y + 42
Solution:

Question 11.
x³ + 13² + 32x + 20      [NCERT]
Solution:

Question 12.
x³ – 3x² – 9x – 5 [NCERT]
Solution:

Question 13.
2y³+ y² – 2y – 1      [NCERT]
Solution:

Question 14.
x³ – 2x² – x + 2
Solution:

Question 15.
Factorize each of the following polynomials:
(i) x³ + 13x² + 31x – 45 given that x + 9 is a factor
(ii) 4x³ + 20x² + 33x + 18 given that 2x + 3 is a factor
Solution:

Question 16.
x⁴ – 2x³ – 7x² + 8x + 12
Solution:

Question 17.
x⁴ + 10x³ + 35x² + 50x + 24
Solution:

Question 18.
2x⁴ – 7x³ – 13x² + 63x – 45
Solution:

Hope given RD Sharma Class 9 Solutions Chapter 6 Factorisation of Polynomials Ex 6.5 are helpful to complete your math homework.

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RD Sharma Class 9 Solutions Chapter 6 Factorisation of Polynomials VSAQS

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 6 Factorisation of Polynomials VSAQS

Other Exercises

• RD Sharma Class 9 Solutions Chapter 6 Factorisation of Polynomials Ex 6.1
• RD Sharma Class 9 Solutions Chapter 6 Factorisation of Polynomials Ex 6.2
• RD Sharma Class 9 Solutions Chapter 6 Factorisation of Polynomials Ex 6.3
• RD Sharma Class 9 Solutions Chapter 6 Factorisation of Polynomials Ex 6.4
• RD Sharma Class 9 Solutions Chapter 6 Factorisation of Polynomials Ex 6.5
• RD Sharma Class 9 Solutions Chapter 6 Factorisation of Polynomials VSAQS
• RD Sharma Class 9 Solutions Chapter 6 Factorisation of Polynomials MCQS

Question 1.
Define zero or root of a polynomial.
Solution:
A real number a is a zero or root of a polynomial f(x) if f(α) = 0

Question 2.
If x = \(\frac { 1 }{ 2 }\) is a zero of the polynomial f(x) =  8x³ + ax² – 4x + 2, find the value of a.
Solution:

Question 3.
Write the remainder when the polynomial f(x) = x³+x²-3x + 2is divided by x + 1.
Solution:
f(x) = x³+x²-3x + 2
Let x + 1 = 0, then x = -1
∴ Remainder =(-1)
Now,f(-1) = (-1)³ + (-1)² – 3(-1) + 2
= -1 + 1+ 3 + 2 = 5
∴ Remainder = 5

Question 4.
Find the remainder when x³ + 4x² + 4x – 3 is divided by x.
Solution:
f(x) = x³ + 4x² + 4x – 3
Dividing f(x) by x, we get
Let x = 0, then
f(x) = 0 + 0 + 0 – 3 = -3
∴  Remainder = -3

Question 5.
If x + 1 is a factor of x³ + a, then write the value of a.
Solution:
Let f(x) = x³ + a
∴ x + 1 is a factor of fx)
Let x + 1 = 0
⇒ x = -1
∴ f(-1) = x³ + a
= (-1)³ + a = -1 + a
∴  x + 1 is a factor
∴  Remainder = 0
∴  -1 + a = 0 ⇒  a = 1
Hence a = 1

Question 6.
If f(x) = x⁴-2x³ + 3x² – ax – b when divided by x – 1, the remainder is 6, then find the value of a + b.
Solution:
f(x) = x⁴ – 2x³ + 3x² – ax – b
Dividing f(x) by x – 1, the remainder = 6
Now let x – 1 = 0, then x = 1
∴  f(1) = (1)⁴ – 2(1)³ + 3(1)² -ax 1-b
= 1 -2 + 3-a-b = 2-a-b
∴ Remainder = 6
∴ 2 – a – b = 6  ⇒ a + b = 2 – 6 = -4
Hence a + b = -4

Hope given RD Sharma Class 9 Solutions Chapter 6 Factorisation of Polynomials VSAQS are helpful to complete your math homework.

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RD Sharma Class 9 Solutions Chapter 10 Congruent Triangles Ex 10.3

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 10 Congruent Triangles Ex 10.3

Other Exercises

• RD Sharma Class 9 Solutions Chapter 10 Congruent Triangles Ex 10.1
• RD Sharma Class 9 Solutions Chapter 10 Congruent Triangles Ex 10.2
• RD Sharma Class 9 Solutions Chapter 10 Congruent Triangles Ex 10.3
• RD Sharma Class 9 Solutions Chapter 10 Congruent Triangles Ex 10.4
• RD Sharma Class 9 Solutions Chapter 10 Congruent Triangles VSAQS
• RD Sharma Class 9 Solutions Chapter 10 Congruent Triangles MCQS

Question 1.
In the figure, lines l₁ and l₂ intersect at O, forming angles as shown in the figure. If x = 45, find the values of y, z and n.

Solution:
Two lines l₁ and l₂ intersect each other at O ∠x = 45°
∵ ∠z = ∠x (Vertically opposite angles)
= 45°
But x + y = 180° (Linear pair)
⇒45° + y= 180°
⇒ y= 180°-45°= 135°
But u = y (Vertically opposite angles)
∴ u = 135°
Hence y = 135°, z = 45° and u = 135°

Question 2.
In the figure, three coplanar lines intersect at a point O, forming angles as shown in the figure. Find the values of x, y, z and u.

Solution:
Three lines AB, CD and EF intersect at O
∠BOD = 90°, ∠DOF = 50°
∵ AB is a line
∴ ∠BOD + ∠DOF + FOA = 180°
⇒ 90° + 50° + u = 180°
⇒ 140° + w = 180°
∴ u= 180°- 140° = 40°
But x = u (Vertically opposite angles)
∴ x = 40°
Similarly, y = 50° and z = 90°
Hence x = 40°, y = 50°, z = 90° and u = 40°

Question 3.
In the figure, find the values of x, y and z.

Solution:
Two lines l₁ and l₂ intersect each other at O
∴ Vertically opposite angles are equal,
∴ y = 25° and x = z
Now 25° + x = 180° (Linear pair)
⇒ x= 180°-25°= 155°
∴ z = x = 155°
Hence x = 155°, y = 25°, z = 155°

Question 4.
In the figure, find the value of x.

Solution:
∵ EF and CD intersect each other at O
∴ Vertically opposite angles are equal,
∴ ∠1 = 2x
AB is a line
3x + ∠1 + 5x = 180° (Angles on the same side of a line)
⇒ 3x + 2x + 5x = 180°
⇒ 10x = 180° ⇒ x = \(\frac { { 180 }^{ \circ } }{ 10 }\)  = 18°
Hence x = 18°

Question 5.
If one of the four angles formed by two intersecting lines is a right angle, then show that each of the four angles is a right angle.
Solution:
Given : Two lines AB and CD intersect each other at O. ∠AOC = 90°

To prove: ∠AOD = ∠BOC = ∠BOD = 90°
Proof : ∵ AB and CD intersect each other at O
∴ ∠AOC = ∠BOD and ∠BOC = ∠AOD (Vertically opposite angles)
∴ But ∠AOC = 90°
∴ ∠BOD = 90°
∴ ∠AOC + ∠BOC = 180° (Linear pair)
⇒ 90° + ∠BOC = 180°
∴ ∠BOC = 180° -90° = 90°
∴ ∠AOD = ∠BOC = 90°
∴ ∠AOD = ∠BOC = ∠BOD = 90°

Question 6.
In the figure, rays AB and CD intersect at O.
(i) Determine y when x = 60°
(ii) Determine x when y = 40°

Solution:
In the figure,
AB is a line
∴ 2x + y = 180° (Linear pair)
(i) If x = 60°, then
2 x 60° + y = 180°
⇒ 120° +y= 180°
∴ y= 180°- 120° = 60°
(ii) If y = 40°, then
2x + 40° = 180°
⇒ 2x = 180° – 40° = 140°
⇒ x= \(\frac { { 140 }^{ \circ } }{ 2 }\)  =70°
∴ x = 70°

Question 7.
In the figure, lines AB, CD and EF intersect at O. Find the measures of ∠AOC, ∠COF, ∠DOE and ∠BOF.

Solution:
Three lines AB, CD and EF intersect each other at O
∠AOE = 40° and ∠BOD = 35°
(i) ∠AOC = ∠BOD (Vertically opposite angles)
= 35°
AB is a line
∴ ∠AOE + ∠DOE + ∠BOD = 180°
⇒ 40° + ∠DOE + 35° = 180°
⇒ 75° + ∠DOE = 180°
⇒ ∠DOE = 180°-75° = 105°
But ∠COF = ∠DOE (Vertically opposite angles)
∴ ∠COF = 105°
Similarly, ∠BOF = ∠AOE (Vertically opposite angles)
⇒ ∠BOF = 40°
Hence ∠AOC = 35°, ∠COF = 105°, ∠DOE = 105° and ∠BOF = 40°

Question 8.
AB, CD and EF are three concurrent lines passing through the point O such that OF bisects ∠BOD. If ∠BOF = 35°, find ∠BOC and ∠AOD.
Solution:
AB, CD and EF intersect at O. Such that OF is the bisector of
∠BOD ∠BOF = 35°

∵ OF bisects ∠BOD,
∴ ∠DOF = ∠BOF = 35° (Vertically opposite angles)
∴ ∠BOD = 35° + 35° = 70°
But ∠BOC + ∠BOD = 180° (Linear pair)
⇒ ∠BOC + 70° = 180°
⇒ ∠BOC = 180°-70°= 110°
But ∠AOD = ∠BOC (Vertically opposite angles)
= 110°
Hence ∠BOC = 110° and ∠AOD =110°

Question 9.
In the figure, lines AB and CD intersect at O. If ∠AOC + ∠BOE = 70° and ∠BOD = 40°, find ∠BOE and reflex ∠COE.

Solution:
In the figure, AB and CD intersect each other at O
∠AOC + ∠BOE = 70°
∠BOD = 40°

AB is a line
∴ ∠AOC + ∠BOE + ∠COE = 180° (Angles on one side of a line)
⇒ 70° + ∠COE = 180°
⇒ ∠COE = 180°-70°= 110°
and ∠AOC = ∠BOD (Vertically opposite angles)
⇒ ∠AOC = 40°
∴ ∠BOE = 70° – 40° = 30°
and reflex ∠COE = 360° – ∠COE
= 360°- 110° = 250°

Question 10.
Which of the following statements are true (T) and which are false (F)?
(i) Angles forming a linear pair are supplementary.
(ii) If two adjacent angles are equal, then each angle measures 90°.
(iii) Angles forming a linear pair can both be acute angles.
(iv) If angles forming a linear pair are equal, then each of these angles is of measure 90°.
Solution:
(i) True.
(ii) False. It can be possible if they are a linear pair.
(iii) False. In a linear pair, if one is acute, then the other will be obtuse.
(iv) True.

Question 11.
Fill in the blanks so as to make the following statements true:
(i) If one angle of a linear pair is acute, then its other angle will be …….. .
(ii) A ray stands on a line, then the sum of the two adjacent angles so formed is ……… .
(iii) If the sum of two adjacent angles is 180°, then the …… arms of the two angles are opposite rays.
Solution:
(i) If one angle of a linear pair is acute, then its other angle will be obtuse.
(ii) A ray stands on a line, then the sum of the two adjacent angles so formed is 180°.
(iii) If the sum of two adjacent angles is 180°, then the uncommon arms of the two angles are opposite rays.

Question 12.
Prove that the bisectors of a pair of vertically opposite angles are in the same straight line.
Solution:
Given : Lines AB and CD intersect each other at O.
OE and OF are the bisectors of ∠AOC and ∠BOD respectively

To prove : OE and OF are in the same line
Proof : ∵ ∠AOC = ∠BOD (Vertically opposite angles)
∵ OE and OF are the bisectors of ∠AOC and ∠BOD
∴ ∠1 = ∠2 and ∠3 = ∠4
⇒ ∠1 = ∠2 = \(\frac { 1 }{ 2 }\) ∠AOC and
∠3 = ∠4 = \(\frac { 1 }{ 2 }\) ∠BOD
∴ ∠1 = ∠2 = ∠3 = ∠4
∵ AOB is a line
∴ ∠BOD + ∠AOD = 180° (Linear pair)
⇒ ∠3 + ∠4 + ∠AOD = 180°
⇒ ∠3 + ∠1 + ∠AOD = 180° (∵ ∠1 = ∠4)
∴ EOF is a straight line

Question 13.
If two straight lines intersect each other, prove that the ray opposite to the bisector of one of the angles thus formed bisects the vertically opposite angle.
Solution:
Given : AB and CD intersect each other at O. OE is the bisector of ∠AOD and EO is produced to F.

To prove : OF is the bisector of ∠BOC
Proof : ∵ AB and CD intersect each other at O
∴ ∠AOD = ∠BOC (Vertically opposite angles)
∵OE is the bisector of ∠AOD
∴ ∠1 = ∠2
∵ AB and EF intersect each other at O
∴∠1 = ∠4 (Vertically opposite angles) Similarly, CD and EF intersect each other at O
∴ ∠2 = ∠3
But ∠1 = ∠2
∴ ∠3 = ∠4
OF is the bisector of ∠BOC

Hope given RD Sharma Class 9 Solutions Chapter 10 Congruent Triangles Ex 10.3 are helpful to complete your math homework.

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RD Sharma Class 9 Solutions Chapter 9 Triangle and its Angles Ex 9.1

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 9 Triangle and its Angles Ex 9.1

Other Exercises

• RD Sharma Class 9 Solutions Chapter 9 Triangle and its Angles Ex 9.1
• RD Sharma Class 9 Solutions Chapter 9 Triangle and its Angles VSAQS

Question 1.
Define the following tenns :
(i) Line segment
(ii) Collinear points
(iii) Parallel lines  
(iv) Intersecting lines
(v) Concurrent lines   
(vi) Ray
(vii) Half-line.
Solution:
(i) A line segment is a part of a line which lies between two points on it and it is denoted as \(\overline { AB }\)   or only by AB. It has two end points and is measureable.

(ii) Three or more points which lie on the same straight line, are called collinear points.
(iii) Two lines which do not intersect each other at any point are called parallel lines.
(iv) If two lines have one point in common, are called intersecting lines.
(v) If two or more lines which pass through a common point are called concurrent lines.
(vi) Ray : A part of a line which has one end point.
(vii) Half line : If A, B, C, be the points on a line l, such that A lies between B and C and we delete the point from line l, two parts of l that remain are each called a half-line.

Question 2.
(i) How many lines can pass through a given point?
(ii) In how many points can two distinct lines at the most intersect?
Solution:
(i) Infinitely many lines can pass through a given point.
(ii) Two distinct lines at the most intersect at one point.

Question 3.
(i) Given two points P and Q, find how many line segments do they determine?
(ii) Name the line segments determined by the three collinear points P, Q and R.
Solution:
(i) Only one line segment can be drawn through two given points P and Q.
(ii) Three collinear points P, Q and R, three lines segments determine : \(\overline { PQ }\) , \(\overline { QR }\)  and \(\overline { PR }\) .

Question 4.
Write the truth value (T/F) of each of the following statements:
(i) Two lines intersect in a point.
(ii) Two lines may intersect in two points.
(iii) A segment has no length.
(iv) Two distinct points always determine a line.
(v) Every ray has a finite length.
(vi) A ray has one end-point only.
(vii) A segment has one end-point only.
(viii) The ray AB is same as ray BA.
(ix) Only a single line may pass through a given point.
(x) Two lines are coincident if they have only one point in common.
Solution:
(i)  False : As two lines do not intersect also any a point.
(ii) False : Two lines intersect at the most one point.
(iii) False : A line segment has definitely length.
(iv) True.
(v) False : Every ray has no definite length.
(vi) True.
(vii) False : A segment has two end point.
(viii)False : Rays AB and BA are two different rays.
(ix) False : Through a given point, infinitely many lines can pass.
(x) False : Two lines are coincident of each and every points coincide each other.

Question 5.
In the figure, name the following:


(i) Five line segments.
(ii) Five rays.
(iii) Four collinear points.
(iv) Two pairs of non-intersecting line segments.
Solution:
From the given figure,
(i) Five line segments are AC, PQ, PR, RS, QS.
(ii) Five rays : \(\xrightarrow { PA }\)  , \(\xrightarrow { RB }\)  , \(\xrightarrow { PB }\)  , \(\xrightarrow { CS }\)  , \(\xrightarrow { DS }\)  .
(iii) Four collinear points are : CDQS, APR, PQL, PRB.
(iv) Two pairs of non-intersecting line segments an AB and CD, AP and CD, AR and CS, PR and QS.

Question 6.
Fill in the blanks so as to make the following statements true:
(i) Two distinct points in a plane determine a _____ line.
(ii) Two distinct_____ in a plane cannot have more than one point in common.
(iii) Given a line and a point, not on the line, there is one and only _____  line which passes through the given point and is_____ to the given line.
(iv) A line separates a plane into ____ parts namely the____  and the____  itself.
Solution:
(i) Two distinct points in a plane determine a unique line.
(ii) Two distinct lines in a plane cannot have more than one point in common.
(iii) Given a line and a point, not on the line, there is one and only perpendicular line which passes through the given point and is perpendicular to the given line.
(iv) A line separates a plane into three parts namely the two half planes, and the one line itself.

Hope given RD Sharma Class 9 Solutions Chapter 9 Triangle and its Angles Ex 9.1 are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.