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

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

Using factor theorem, factorize each of the following polynomials:
Question 1.
x3 + 6x2 + 11x + 6
Solution:
RD Sharma Class 9 Solutions Chapter 6 Factorisation of Polynomials Ex 6.5 Q1.1

Question 2.
x3 + 2x2 – x – 2
Solution:
RD Sharma Class 9 Solutions Chapter 6 Factorisation of Polynomials Ex 6.5 Q2.1
RD Sharma Class 9 Solutions Chapter 6 Factorisation of Polynomials Ex 6.5 Q2.2

Question 3.
x3 – 6x2 + 3x + 10
Solution:
RD Sharma Class 9 Solutions Chapter 6 Factorisation of Polynomials Ex 6.5 Q3.1
RD Sharma Class 9 Solutions Chapter 6 Factorisation of Polynomials Ex 6.5 Q3.2

Question 4.
x4 – 7x3 + 9x2 + x- 10
Solution:
RD Sharma Class 9 Solutions Chapter 6 Factorisation of Polynomials Ex 6.5 Q4.1

Question 5.
3x3 – x2 – 3x + 1
Solution:
RD Sharma Class 9 Solutions Chapter 6 Factorisation of Polynomials Ex 6.5 Q5.1

Question 6.
x3 – 23x2 + 142x – 120        [NCERT]
Solution:
RD Sharma Class 9 Solutions Chapter 6 Factorisation of Polynomials Ex 6.5 Q6.1
RD Sharma Class 9 Solutions Chapter 6 Factorisation of Polynomials Ex 6.5 Q6.2

Question 7.
y3 – 7y + 6
Solution:
Let f(y) = y3 – 7y + 6
RD Sharma Class 9 Solutions Chapter 6 Factorisation of Polynomials Ex 6.5 Q7.1

Question 8.
X3 -10x2 – 53x – 42
Solution:
RD Sharma Class 9 Solutions Chapter 6 Factorisation of Polynomials Ex 6.5 Q8.1
RD Sharma Class 9 Solutions Chapter 6 Factorisation of Polynomials Ex 6.5 Q8.2

Question 9.
y3 – 2y2– 29y – 42
Solution:
RD Sharma Class 9 Solutions Chapter 6 Factorisation of Polynomials Ex 6.5 Q9.1
RD Sharma Class 9 Solutions Chapter 6 Factorisation of Polynomials Ex 6.5 Q9.2

Question 10.
2y3 – 5y2 – 19y + 42
Solution:
RD Sharma Class 9 Solutions Chapter 6 Factorisation of Polynomials Ex 6.5 Q10.1
RD Sharma Class 9 Solutions Chapter 6 Factorisation of Polynomials Ex 6.5 Q10.2

Question 11.
x3 + 132 + 32x + 20      [NCERT]
Solution:
RD Sharma Class 9 Solutions Chapter 6 Factorisation of Polynomials Ex 6.5 Q11.1
RD Sharma Class 9 Solutions Chapter 6 Factorisation of Polynomials Ex 6.5 Q11.2

Question 12.
x3 – 3x2 – 9x – 5 [NCERT]
Solution:
RD Sharma Class 9 Solutions Chapter 6 Factorisation of Polynomials Ex 6.5 Q12.1
RD Sharma Class 9 Solutions Chapter 6 Factorisation of Polynomials Ex 6.5 Q12.2

Question 13.
2y3+ y2 – 2y – 1      [NCERT]
Solution:
RD Sharma Class 9 Solutions Chapter 6 Factorisation of Polynomials Ex 6.5 Q13.1

Question 14.
x3 – 2x2 – x + 2
Solution:
RD Sharma Class 9 Solutions Chapter 6 Factorisation of Polynomials Ex 6.5 Q14.1

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:
RD Sharma Class 9 Solutions Chapter 6 Factorisation of Polynomials Ex 6.5 Q15.1
RD Sharma Class 9 Solutions Chapter 6 Factorisation of Polynomials Ex 6.5 Q15.2
RD Sharma Class 9 Solutions Chapter 6 Factorisation of Polynomials Ex 6.5 Q15.3

Question 16.
x4 – 2x3 – 7x2 + 8x + 12
Solution:
RD Sharma Class 9 Solutions Chapter 6 Factorisation of Polynomials Ex 6.5 Q16.1
RD Sharma Class 9 Solutions Chapter 6 Factorisation of Polynomials Ex 6.5 Q16.2

Question 17.
x4 + 10x3 + 35x2 + 50x + 24
Solution:
RD Sharma Class 9 Solutions Chapter 6 Factorisation of Polynomials Ex 6.5 Q17.1
RD Sharma Class 9 Solutions Chapter 6 Factorisation of Polynomials Ex 6.5 Q17.2

Question 18.
2x4 – 7x3 – 13x2 + 63x – 45
Solution:
RD Sharma Class 9 Solutions Chapter 6 Factorisation of Polynomials Ex 6.5 Q18.1
RD Sharma Class 9 Solutions Chapter 6 Factorisation of Polynomials Ex 6.5 Q18.2
RD Sharma Class 9 Solutions Chapter 6 Factorisation of Polynomials Ex 6.5 Q18.3

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RD Sharma Class 9 Solutions Chapter 8 Lines and Angles Ex 8.1

RD Sharma Class 9 Solutions Chapter 8 Lines and Angles Ex 8.1

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 8 Lines and Angles Ex 8.1

Other Exercises

Question 1.
Plot the following points on the graph paper:
(i)  (2, 5)                    
(ii) (4, -3)
(iii) (-5, -7)                  
(iv) (7, -4)
(v) (-3, 2)        
(vi) (7, 0)
(vii) (-4, 0)               
(viii) (0, 7)
(ix) (0, -4)                    
(x) (0, 0)
Solution:
The given points have been plotted on the graph as given below:
RD Sharma Class 9 Solutions Chapter 8 Lines and Angles Ex 8.1 Q1.1

Question 2.
Write the coordinates of each of the following points marked in the graph paper.
RD Sharma Class 9 Solutions Chapter 8 Lines and Angles Ex 8.1 Q2.1
Solution:
The co-ordinates of the points given in the graph are A (3, 1), B (6, 0), C (0, 6), D (-3, 0), E (-4, 3), F (-2, -4), G (0, -5), H (3, -6), P (7, -3).

 

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RS Aggarwal Class 9 Solutions Chapter 6 Coordinate Geometry Ex 6A

RS Aggarwal Class 9 Solutions Chapter 6 Coordinate Geometry Ex 6A

These Solutions are part of RS Aggarwal Solutions Class 9. Here we have given RS Aggarwal Class 9 Solutions Chapter 6 Coordinate Geometry Ex 6A.

Question 1.
Solution:
Co-ordinates of
A are ( – 6, 5)
B are (5, 4)
C are ( – 3, 2)
D are (2, – 2)
E are ( – 1, – 4)
Ans.
RS Aggarwal Class 9 Solutions Chapter 6 Coordinate Geometry Ex 6A Q1.1

Question 2.
Solution:
The given points have been plotted as shown in the adjoining graph.
Where X’OX and YOY’ are the axis:
RS Aggarwal Class 9 Solutions Chapter 6 Coordinate Geometry Ex 6A Q2.1

Question 3.
Solution:
(i) (7, 0) lies on x-axis as its ordinate is (0)
(ii) (0, – 5) lies on y-axis as its abscissa is (0)
(iii) (0, 1) lies on y-axis as its abscissa is (0)
(iv) ( – 4, 0) lies on jc-axis as its ordinate is (0)
Ans.

Question 4.
Solution:
(i) ( – 6, 5) lies in second quadrant because A is of the type (-, +)
(ii) ( – 3, – 2) lies in third quadrant because A is of the type (-, -)
(iii) (2, – 9) lies in fourth quadrant because it is of the type (+, -).

Question 5.
Solution:
In the given equation, .
y = x+1
Put x = 0, then y = 0 + 1 = 1
x = 1, then, y = 1 + 1=2
x = 2, then, y = 2 + 1 = 3
Now, plot the points as given in the table given below on the graph, and join them as shown.
RS Aggarwal Class 9 Solutions Chapter 6 Coordinate Geometry Ex 6A Q5.1

Question 6.
Solution:
In the given equation
y = 3x + 2
Put x = 0,
then y = 3 x 0 + 2 = 0 + 2 = 2
x = 1, then
y = 3 x 1 + 2 = 3 + 2 = 5
and x = – 2, then
y = 3 x ( – 2) + 2 = – 6 + 2 = – 4
RS Aggarwal Class 9 Solutions Chapter 6 Coordinate Geometry Ex 6A Q6.1

Question 7.
Solution:
In the given equation,
y = 5x – 3
Put x = 1,y = 5 x 1 – 3 = 5 – 3 = 2
x = 0 then,
y = 5 x 0 – 3 = 0 – 3 = – 3
and x = 2, then
y = 5 x 2 – 3 = 10 – 3 = 7
RS Aggarwal Class 9 Solutions Chapter 6 Coordinate Geometry Ex 6A Q7.1

Question 8.
Solution:
In the given equation,
RS Aggarwal Class 9 Solutions Chapter 6 Coordinate Geometry Ex 6A Q8.1
y = 3x,
Put x = 0,
then y = 3 x 0 = 0
Put x = 1, then
y = 3 x 1 = 3
Put x = – 1, then
y = 3 ( – 1) = – 3
Now, plot the points as given in the table below
RS Aggarwal Class 9 Solutions Chapter 6 Coordinate Geometry Ex 6A Q8.2
and join them as shown.

Question 9.
Solution:
In the given equation, y = – x,
Put x = 1,
then y = – 1
Put x = 2, then
y = – 2
Put x = – 2, then
y = – ( – 2) = 2
RS Aggarwal Class 9 Solutions Chapter 6 Coordinate Geometry Ex 6A Q9.1

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RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14A

RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14A

These Solutions are part of RS Aggarwal Solutions Class 9. Here we have given RS Aggarwal Solutions Class 9 Chapter 14 Statistics Ex 14A.

Other Exercises

Question 1.
Solution:
Statistics is a science which deals with the collection, presentation, analysis and interpretations of numerical data.

Question 2.
Solution:
(i) Numerical facts alone constitute data
(ii) Qualitative characteristics like intelligence, poverty etc. which cannot be measured, numerically, don’t form data.
(iii) Data are an aggregate of facts. A single observation does not form data.
(iv) Data collected for a definite purpose may not be suited for another purpose.
(v) Data is different experiments are comparable.

Question 3.
Solution:
(i) Primary data : The data collected by the investigator himself with a definite plan in mind are called primary data.
(ii) Secondary data : The data collected by some one other than the investigator are called secondary data. The primary data is more reliable and relevant.

Question 4.
Solution:
(i) Variate : Any character which is capable of taking several different values is called a variate or variable.
(ii) Class interval : Each group into which the raw data is condensed, is called a class interval
(iii) Class size : The difference between the true upper limit and the true lower limit of a class is called class size.
(iv) Class Mark : \(\frac { upper\quad limit+lower\quad limit }{ 2 } \) is called a class mark
(v) Class limits : Each class is bounded by two figures which are called class limits which are lower class limit and upper class limit.
(vi) True class limits : In exclusive form, the upper and lower limits of a class are respectively are the true upper limit and true lower limit but in inclusive form, the true lower limit of a class is obtained by subtracting O.S from lower limit of the class and for true limit, adding 0.5 to the upper limit.
(vii) Frequency of a class : The number of times an observation occurs in a class is called its frequency.
(viii) Cumulative frequency of a class : The cumulative frequency corresponding to a class is the sum of all frequencies upto and including that class.

Question 5.
Solution:
The given data can be represent in form of frequency table as given below:
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14A Q5.1

Question 6.
Solution:
The frequency distribution table of the given data is given below :
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14A Q6.1

Question 7.
Solution:
The frequency distribution table of the
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14A Q7.1

Question 8.
Solution:
The frequency table is given below :
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14A Q8.1

Question 9.
Solution:
The frequency table of given data is given below :
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14A Q9.1

Question 10.
Solution:
The frequency distribution table of the given data in given below :
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14A Q10.1

Question 11.
Solution:
The frequency table of the given data:
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14A Q11.1

Question 12.
Solution:
The cumulative frequency of the given table is given below:
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14A Q12.1

Question 13.
Solution:
The given table can be represented in a group frequency table in given below :
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14A Q13.1

Question 14.
Solution:
Frequency table of the given cumulative frequency is given below :
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14A Q14.1

Question 15.
Solution:
A frequency table of the given cumulative frequency table is given below :
RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14A Q15.1

 

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RD Sharma Class 9 Solutions Chapter 6 Factorisation of Polynomials VSAQS

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

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:
RD Sharma Class 9 Solutions Chapter 6 Factorisation of Polynomials VSAQS Q2.1

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

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

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.
RD Sharma Class 9 Solutions Chapter 10 Congruent Triangles Ex 10.3 Q1.1
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.
RD Sharma Class 9 Solutions Chapter 10 Congruent Triangles Ex 10.3 Q2.1
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.
RD Sharma Class 9 Solutions Chapter 10 Congruent Triangles Ex 10.3 Q3.1
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.
RD Sharma Class 9 Solutions Chapter 10 Congruent Triangles Ex 10.3 Q4.1
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°
RD Sharma Class 9 Solutions Chapter 10 Congruent Triangles Ex 10.3 Q5.1
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°
RD Sharma Class 9 Solutions Chapter 10 Congruent Triangles Ex 10.3 Q6.1
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.
RD Sharma Class 9 Solutions Chapter 10 Congruent Triangles Ex 10.3 Q7.1
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°
RD Sharma Class 9 Solutions Chapter 10 Congruent Triangles Ex 10.3 Q8.1
∵ 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.
RD Sharma Class 9 Solutions Chapter 10 Congruent Triangles Ex 10.3 Q9.1
Solution:
In the figure, AB and CD intersect each other at O
∠AOC + ∠BOE = 70°
∠BOD = 40°
RD Sharma Class 9 Solutions Chapter 10 Congruent Triangles Ex 10.3 Q9.2
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
RD Sharma Class 9 Solutions Chapter 10 Congruent Triangles Ex 10.3 Q12.1
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.
RD Sharma Class 9 Solutions Chapter 10 Congruent Triangles Ex 10.3 Q13.1
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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RS Aggarwal Class 9 Solutions Chapter 8 Linear Equations in Two Variables Ex 8A

RS Aggarwal Class 9 Solutions Chapter 8 Linear Equations in Two Variables Ex 8A

These Solutions are part of RS Aggarwal Solutions Class 9. Here we have given RS Aggarwal Class 9 Solutions Chapter 8 Linear Equations in Two Variables Ex 8A.

Question 1.
Solution:
(i) x = 5 is the line AB parallel to the y-axis at a distance of 5 units.
(ii) y = – 2 is the line CD parallel to x-axis at a distance of – 2 units.
(iii) x + 6 = 0 => x = – 6 is the line EF parallel to y-axis at a distance of – 6 units.
(iv) x + 7 = 0 => x = – 7 is the line PQ parallel to y-axis at a distance of – 7 units.
(v) y = 0 is the equation of x-axis. The graph of y = 0 is the line X’OX
(vi) x = 0 is the equation of y-axis.The graph of x = 0 is the line YOY’
Ans.
RS Aggarwal Class 9 Solutions Chapter 8 Linear Equations in Two Variables Ex 8A 1

Question 2.
Solution:
In the given equation.
RS Aggarwal Class 9 Solutions Chapter 8 Linear Equations in Two Variables Ex 8A 2
y = 3x
Put x = 1, then y = 3 x 1 = 3
Put x = 2, then y = 3 x 2 = 6
Put x = – 1, then y = 3 ( – 1) = – 3
Now, plot the points (1, 3), (2, 6) and ( – 1, – 3) as given the following table
RS Aggarwal Class 9 Solutions Chapter 8 Linear Equations in Two Variables Ex 8A 2.1
and join them to form a line of the given equation.
Now from x = – 2,
draw a line parallel to y-axis at a distance of x = – 2, meeting the given line at P. From P, draw, a line parallel to x-axis joining y-axis at M, which is y = – 6 Hence, y = – 6 Ans.

Question 3.
Solution:
In the given equation x + 2y – 3 = 0
=> 2y = 3 – x
y = \(\frac { 3-x }{ 2 } \)
put x = 1,then
RS Aggarwal Class 9 Solutions Chapter 8 Linear Equations in Two Variables Ex 8A 3
RS Aggarwal Class 9 Solutions Chapter 8 Linear Equations in Two Variables Ex 8A 3.1
RS Aggarwal Class 9 Solutions Chapter 8 Linear Equations in Two Variables Ex 8A 3.2

Question 4.
Solution:
(i) In the equation y = x
When x = 1, then y = 1
when x = 2, then y = 2
and when x = 3, then y = 3
RS Aggarwal Class 9 Solutions Chapter 8 Linear Equations in Two Variables Ex 8A 4
RS Aggarwal Class 9 Solutions Chapter 8 Linear Equations in Two Variables Ex 8A 4.1
RS Aggarwal Class 9 Solutions Chapter 8 Linear Equations in Two Variables Ex 8A 4.2
RS Aggarwal Class 9 Solutions Chapter 8 Linear Equations in Two Variables Ex 8A 4.3
RS Aggarwal Class 9 Solutions Chapter 8 Linear Equations in Two Variables Ex 8A 4.4
RS Aggarwal Class 9 Solutions Chapter 8 Linear Equations in Two Variables Ex 8A 4.5

Question 5.
Solution:
In the given equation
2x – 3y = 5
RS Aggarwal Class 9 Solutions Chapter 8 Linear Equations in Two Variables Ex 8A 5
RS Aggarwal Class 9 Solutions Chapter 8 Linear Equations in Two Variables Ex 8A 5.1
RS Aggarwal Class 9 Solutions Chapter 8 Linear Equations in Two Variables Ex 8A 5.2

Question 6.
Solution:
In the given equation
2x + y = 6
=> y = 6 – 2x
Put x = 1, then y – 6 – 2 x 1 = 6 – 2 = 4
Put x = 2, then y = 6 – 2 x 2 = 6 – 4 = 2
Put x = 3, then y = 6 – 2 x 3 = 6 – 6 = 0
RS Aggarwal Class 9 Solutions Chapter 8 Linear Equations in Two Variables Ex 8A 6
RS Aggarwal Class 9 Solutions Chapter 8 Linear Equations in Two Variables Ex 8A 6.1

Question 7.
Solution:
In the given equation
3x + 2y = 6
RS Aggarwal Class 9 Solutions Chapter 8 Linear Equations in Two Variables Ex 8A 7
RS Aggarwal Class 9 Solutions Chapter 8 Linear Equations in Two Variables Ex 8A 7.1
RS Aggarwal Class 9 Solutions Chapter 8 Linear Equations in Two Variables Ex 8A 7.2

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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 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

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.
RD Sharma Class 9 Solutions Chapter 9 Triangle and its Angles Ex 9.1 Q1.1
(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.
RD Sharma Class 9 Solutions Chapter 9 Triangle and its Angles Ex 9.1 Q1.2

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:
RD Sharma Class 9 Solutions Chapter 9 Triangle and its Angles Ex 9.1 Q5.1
RD Sharma Class 9 Solutions Chapter 9 Triangle and its Angles Ex 9.1 Q5.2
(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.

 

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Value Based Questions in Science for Class 9 Chapter 3 Atoms and Molecules

Value Based Questions in Science for Class 9 Chapter 3 Atoms and Molecules

These Solutions are part of Value Based Questions in Science for Class 9. Here we have given Value Based Questions in Science for Class 9 Chapter 3 Atoms and Molecules

Question 1.
Kamla prepared aqueous solutions of barium chloride and sodium sulphate. She weighed them separately and then mixed them in a beaker. A white precipitate was immediately formed. She filtered the precipitate, dried it and then weighed it. After reading this narration, answer the following questions :

  1. Will the weight of the precipitate be the same as that of the reactants before mixing ?
  2. If not, what she should have done ?
  3. Which law of chemical combination does this support ?
  4. What is the value based information associated with it ?

Answer:

  1. No, it will not be the same.
  2. She should have weighed the total contents of the beaker after the reaction and not the precipitate alone.
  3. It supports the law of conservation of mass.
  4. Whenever the law of conservation of mass is to be verified in the laboratory, total mass of the reactants and also of products should be taken into account. Moreover, none of the species be allowed to leave the container.

More Resources

Question 2.
In order to verify the law of conservation of mass, a student mixed 6.3 g of sodium carbonate and 15.0 g of ethanoic acid in a conical flask. After experiment, he weighed the flask again. The weight of the residue in the flask was only 18.0 g. He approached the teacher who guided him to carry the experiment in a closed flask with a cork. There was no difference in weight of the flask before and after the experiment.

  1. What was the mistake committed by the student ?
  2. Why did not the two weights match earlier ?
  3. How did the teacher help him ?
  4. What lesson was learnt by the student ?

Answer:

  1. He was carrying the experiment in an open flask.
  2. CO2 gas evolved in the reaction escaped from the flask
    2CH3COOH + Na2CO3 —-> 2CH3COONa + H2O + CO2
  3. Teacher asked him to cork the flask the moment the reactants were mixed.
  4. The student learnt that in future he should not carry the experiment relating to the law of conservation of mass in an open container, particularly when one or more reactants or products are in the gaseous state.

Question 3.
A student was asked by his teacher a verify the law of conservation of mass in the laboratory. He prepared 5% aqueous solutions of NaCl and Na2SO4. He mixed 10 mL of both these solutions in a conical flask. He weighed the flask on a balance. He then stirred the flask with a rod and weighed it after sometime. There was no change in mass. Read this narration and answer the questions given below :

  1. Was the student able to verify the law of conservation of mass ?
  2. If not, what was the mistake committed by him ?
  3. In your opinion, what he should have done ?
  4. What is the value based information associated with this ?

Answer:

  1. No, he could not verify the law of conservation of mass inspite of the fact that there was no change in mass.
  2. No chemical reaction takes place between NaCl and Na2SO4. This means that no reaction actually took place in the flask.
  3. He should have performed the experiment by using aqueous solutions of BaCl2 and Na2SO4. A chemical reaction takes place in this case and a white precipitate of BaSO4 is formed.
  4. While working in the chemistry laboratory, a student must select those chemical substances which actually react with each other. Only then products will be formed.

Question 4.
Dalton was the first scientist to introduce symbols for the known elements. Modern symbols were given by J.J. Berzelius. A symbol in general may be defined as the short hand representation of the name of an element.

  1. How do symbols help in identifying elements ?
  2. Do we use symbols in daily life ?
  3. What values do you attach for using symbols ?

Answer:

  1. All the known elements are identified by their symbols.
    For example, Symbol of calcium = Ca; Symbol of copper = Cu; Symbol of iron= Fe
  2. Yes, these are very common in daily life. For example, all road signs such as diversions, dangerous, zones etc. are indicated by symbols. In playground, umpires, signify the various happenings such as ‘LBW’, ‘Out’ etc. in circket by symbols.
  3. Symbols for road signs save many lives. The names of many complicated compounds are shown by the formulae which are collection of symbols. The chemical composition of all madicines are shown either on the strips or on the bottles by their formulae.

Question 5.
Mole concept is an important tool for dealing with chemical calculations. The elements have atomic masses while compounds have molecular masses or molar masses. Mole is in fact, a collection of Avogadro’s number (NA) of the particles of a substance whether element or compound. The value of Avogadro’s number is 6-022 x 1023.

  1. Why is mole commonly called chemist’s dozen ?
  2. What is the value associated with the term mole ?

Answer:

  1. Just as a dozen represents 12 articles, a mole represents 6.022 x 1023 or Avogadro’s number of particles. Therefore, it has been rightly called chemist’s dozen.
  2. Since particles such as atoms, ions or molecules are very extremely small in size, it is very difficult to identify and express them individually. These are collectively represented as mole. For example, 3.011 x 1023 molecules of CO2 gas are shown as 0.5 mole which is very simple.

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RS Aggarwal Class 9 Solutions Chapter 10 Area Ex 10A

RS Aggarwal Class 9 Solutions Chapter 10 Area Ex 10A

These Solutions are part of RS Aggarwal Solutions Class 9. Here we have given RS Aggarwal Class 9 Solutions Chapter 10 Area Ex 10A.

Question 1.
Solution:
Given : In the figure, ABCD is a quadrilateral and
AB = CD = 5cm
RS Aggarwal Class 9 Solutions Chapter 10 Area Ex 10A Q1.1

Question 2.
Solution:
In ||gm ABCD,
AB = 10cm, altitude DL = 6cm
and BM is altitude on AD, and BM = 8 cm.
RS Aggarwal Class 9 Solutions Chapter 10 Area Ex 10A Q2.1

Question 3.
Solution:
Diagonals of rhombus are 16cm and 24 cm.
Area = \(\frac { 1 }{ 2 } \) x product of diagonals
= \(\frac { 1 }{ 2 } \) x 1st diagonal x 2nd diagonal
= \(\frac { 1 }{ 2 } \) x 16 x 24
= 192 cm² Ans.

Question 4.
Solution:
Parallel sides of a trapezium are 9cm and 6cm and distance between them is 8cm
RS Aggarwal Class 9 Solutions Chapter 10 Area Ex 10A Q4.1

Question 5.
Solution:
from the figure
(i) In ∆ BCD, ∠ DBC = 90°
RS Aggarwal Class 9 Solutions Chapter 10 Area Ex 10A Q5.1
RS Aggarwal Class 9 Solutions Chapter 10 Area Ex 10A Q5.2
RS Aggarwal Class 9 Solutions Chapter 10 Area Ex 10A Q5.3

Question 6.
Solution:
In the fig, ABCD is a trapezium. AB || DC
AB = 7cm, AD = BC = 5cm.
Distance between AB and DC = 4 cm.
i.e. ⊥AL = ⊥BM = 4cm.
RS Aggarwal Class 9 Solutions Chapter 10 Area Ex 10A Q6.1
RS Aggarwal Class 9 Solutions Chapter 10 Area Ex 10A Q6.2

Question 7.
Solution:
Given : In quad. ABCD. AL⊥BD and CM⊥BD.
To prove : ar(quad. ABCD)
RS Aggarwal Class 9 Solutions Chapter 10 Area Ex 10A Q7.1
RS Aggarwal Class 9 Solutions Chapter 10 Area Ex 10A Q7.2

Question 8.
Solution:
In quad. ABCD, BD is its diagonal and AL⊥BD, CM⊥BD
RS Aggarwal Class 9 Solutions Chapter 10 Area Ex 10A Q8.1

Question 9.
Solution:
Given : ABCD is a trapezium in which AB || DC and its diagonals AC and BD intersect each other at O.
To prove : ar(∆ AOD) = ar(∆ BOC)
RS Aggarwal Class 9 Solutions Chapter 10 Area Ex 10A Q9.1

Question 10.
Solution:
Given : In the figure,
DE || BC.
RS Aggarwal Class 9 Solutions Chapter 10 Area Ex 10A Q10.1

Question 11.
Solution:
Given : In ∆ ABC, D and E are the points on AB and AC such that
ar( ∆ BCE) = ar( ∆ BCD)
To prove : DE || BC.
Proof : (∆ BCE) = ar(∆ BCD)
But these are on the same base BC.
Their altitudes are equal.
Hence DE || BC
Hence proved.

Question 12.
Solution:
Given : In ||gm ABCD, O is any. point inside the ||gm. OA, OB, OC and OD are joined.
RS Aggarwal Class 9 Solutions Chapter 10 Area Ex 10A Q12.1

Question 13.
Solution:
Given : In quad. ABCD.
A line through D, parallel to AC, meets ‘BC produced in P. AP in joined which intersects CD at E.
To prove : ar( ∆ ABP) = ar(quad. ABCD).
Const. Join AC
RS Aggarwal Class 9 Solutions Chapter 10 Area Ex 10A Q13.1
RS Aggarwal Class 9 Solutions Chapter 10 Area Ex 10A Q13.2

Question 14.
Solution:
Given : ∆ ABC and ∆ DBC are on the same base BC with points A and D on , opposite sides of BC and
ar( ∆ ABC) = ar( ∆ DBC).
RS Aggarwal Class 9 Solutions Chapter 10 Area Ex 10A Q14.1
RS Aggarwal Class 9 Solutions Chapter 10 Area Ex 10A Q14.2

Question 15.
Solution:
Given : In ∆ ABC, AD is the median and P is a point on AD
BP and CP are joined
To prove : (i) ar(∆BDP) = ar(∆CDP)
(ii) ar( ∆ ABP) = ar( ∆ ACP)
RS Aggarwal Class 9 Solutions Chapter 10 Area Ex 10A Q15.1

Question 16.
Solution:
Given : In quad. ABCD, diagonals AC and BD intersect each other at O and BO = OD
To prove : ar(∆ ABC) = ar(∆ ADC)
Proof : In ∆ ABD,
RS Aggarwal Class 9 Solutions Chapter 10 Area Ex 10A Q16.1
RS Aggarwal Class 9 Solutions Chapter 10 Area Ex 10A Q16.2

Question 17.
Solution:
In ∆ ABC,D is mid point of BC
and E is midpoint of AD and BE is joined.
RS Aggarwal Class 9 Solutions Chapter 10 Area Ex 10A Q17.1
RS Aggarwal Class 9 Solutions Chapter 10 Area Ex 10A Q17.2

Question 18.
Solution:
Given : In ∆ ABC. D is a point on AB and AD is joined. E is mid point of AD EB and EC are joined.
To prove : ar( ∆ BEC) = \(\frac { 1 }{ 2 } \) ar( ∆ ABC)
Proof : In ∆ ABD,
E is midpoint of AD
BE is its median
ar(∆ EBD) = ar(∆ ABE)
RS Aggarwal Class 9 Solutions Chapter 10 Area Ex 10A Q18.1

Question 19.
Solution:
Given : In ∆ ABC, D is midpoint of BC and E is die midpoint of BO is the midpoint of AE.
To prove that ar( ∆ BOE) = \(\frac { 1 }{ 8 } \) ar(∆ ABC).
RS Aggarwal Class 9 Solutions Chapter 10 Area Ex 10A Q19.1
RS Aggarwal Class 9 Solutions Chapter 10 Area Ex 10A Q19.2

Question 20.
Solution:
Given : In ||gm ABCD, O is any point on diagonal AC.
To prove : ar( ∆ AOB) = ar( ∆ AOD)
Const. Join BD which intersects AC at P
Proof : In ∆ OBD,
P is midpoint of BD
(Diagonals of ||gm bisect each other)
RS Aggarwal Class 9 Solutions Chapter 10 Area Ex 10A Q20.1
RS Aggarwal Class 9 Solutions Chapter 10 Area Ex 10A Q20.2

Question 21.
Solution:
Given : ABCD is a ||gm.
P, Q, R and S are the midpoints of sides AB, BC, CD, DA respectively.
PQ, QR, RS and SP are joined.
RS Aggarwal Class 9 Solutions Chapter 10 Area Ex 10A Q21.1
RS Aggarwal Class 9 Solutions Chapter 10 Area Ex 10A Q21.2
RS Aggarwal Class 9 Solutions Chapter 10 Area Ex 10A Q21.3

Question 22.
Solution:
Given : In pentagon ABCDE,
EG || DA meets BA produced and
CF || DB, meets AB produced.
To prove : ar(pentagon ABCDE) = ar(∆ DGF)
RS Aggarwal Class 9 Solutions Chapter 10 Area Ex 10A Q22.1
RS Aggarwal Class 9 Solutions Chapter 10 Area Ex 10A Q22.2

Question 23.
Solution:
Given ; A ∆ ABC in which AD is the median.
To prove ; ar( ∆ ABD) = ar( ∆ ACD)
Const : Draw AE⊥BC.
Proof : Area of ∆ ABD
RS Aggarwal Class 9 Solutions Chapter 10 Area Ex 10A Q23.1
RS Aggarwal Class 9 Solutions Chapter 10 Area Ex 10A Q23.2

Question 24.
Solution:
Given : A ||gm ABCD in which AC is its diagonal which divides ||gm ABCD in two ∆ ABC and ∆ ADC.
RS Aggarwal Class 9 Solutions Chapter 10 Area Ex 10A Q24.1
RS Aggarwal Class 9 Solutions Chapter 10 Area Ex 10A Q24.2

Question 25.
Solution:
Given : In ∆ ABC,
D is a point on BC such that
BD = \(\frac { 1 }{ 2 } \) DC
AD is joined.
RS Aggarwal Class 9 Solutions Chapter 10 Area Ex 10A Q25.1
RS Aggarwal Class 9 Solutions Chapter 10 Area Ex 10A Q25.2

Question 26.
Solution:
Given : In ∆ ABC, D is a point on BC such that
BD : DC = m : n
AD is joined.
RS Aggarwal Class 9 Solutions Chapter 10 Area Ex 10A Q26.1
RS Aggarwal Class 9 Solutions Chapter 10 Area Ex 10A Q26.2

 

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RD Sharma Class 9 Solutions Chapter 6 Factorisation of Polynomials Ex 6.2

RD Sharma Class 9 Solutions Chapter 6 Factorisation of Polynomials Ex 6.2

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 6.2

Other Exercises

Question 1.
If f(x) = 2x3 – 13x2 + 17x + 12, find
(i) f (2)
(ii) f (-3)
(iii) f(0)
Solution:
f(x) = 2x3 – 13x2 + 17x + 12
(i) f(2) = 2(2)3 – 13(2)2 + 17(2) + 12
= 2 x 8-13 x 4+17 x 2+12
= 16-52 + 34 + 12
= 62 – 52
= 10
(ii) f(-3) = 2(-3)3 – 13(-3)2 + 17 x (-3) + 12
= 2 x (-27) – 13 x 9 + 17 x (-3) + 12
= -54 – 117 -51 + 12
= -222 + 12
= -210
(iii) f(0) = 2 x (0)3 – 13(0)2 + 17 x 0 + 12
= 0-0 + 0+ 12 = 12

Question 2.
Verify whether the indicated numbers are zeros of the polynomials corresponding to them in the following cases: [NCERT]
RD Sharma Class 9 Solutions Chapter 6 Factorisation of Polynomials Ex 6.2 Q2.1
Solution:
RD Sharma Class 9 Solutions Chapter 6 Factorisation of Polynomials Ex 6.2 Q2.2
RD Sharma Class 9 Solutions Chapter 6 Factorisation of Polynomials Ex 6.2 Q2.3
RD Sharma Class 9 Solutions Chapter 6 Factorisation of Polynomials Ex 6.2 Q2.4
RD Sharma Class 9 Solutions Chapter 6 Factorisation of Polynomials Ex 6.2 Q2.5
RD Sharma Class 9 Solutions Chapter 6 Factorisation of Polynomials Ex 6.2 Q2.6

Question 3.
If x = 2 is a root of the polynomial f(x) = 2x2-3x + la, find the value of a.
Solution:
p(x) = 2x2 – 3x + 7a
∵ x = 2 is its zero, then
p(0) = 0
∴ p( 2) = 2(2)2 – 3×2 + la = 0
⇒2 x 4-3 x2 + 7a = 0
⇒ 8 – 6 + 7o = 0
⇒2 + 7a = 0
⇒ 7a = -2 ⇒ a =\(\frac { -2 }{ 7 }\)
∴ Hence a = \(\frac { -2 }{ 7 }\)

Question 4.
If x = –\(\frac { 1 }{ 2 }\) is a zero of the polynomial p(x) = 8x3 – ax2 – x + 2, find the value of a.
Solution:
RD Sharma Class 9 Solutions Chapter 6 Factorisation of Polynomials Ex 6.2 Q4.1
RD Sharma Class 9 Solutions Chapter 6 Factorisation of Polynomials Ex 6.2 Q4.2

Question 5.
If x = 0 and x = -1 are the roots of the polynomial f(x) = 2x3 – 3x2 + ax + b, find the value of a and b.
Solution:
f(x) = 2x3 – 3x2 + ax + b
∵ x = 0 and x = -1 are its zeros
∴ f(0) = 0 and f(-1) = 0
Now, f(0) = 0
⇒  2(0)3 – 3(0)2 + a x 0 + b = 0
⇒ 0-0 + 0 + b= 0
∴ b = 0
and f(-1) = 0
⇒ 2(-1)3 – 3(-1)2 + a(-1) + b = 0
⇒  2 x (-1) – 3 x 1 + a x (-1) + b = 0
⇒ -2 -3-a + b = 0
⇒ -2-3-a + 0 = 0
⇒ -5- a = 0=>a =-5
Hence a = -5, b = 0

Question 6.
Find the integral roots of the polynomial f(x) = x3 + 6x2 + 11x + 6.
Solution:
f(x) = x3 + 6x2 + 11x + 6
Construct = 6 = ±1, ±2, +3, ±6
If x = 1, then
f(1) = (1)3 + 6(1)2 + 11 x 1 + 6
= 1+ 6+11+ 6 = 24
∵  f(x) ≠ 0, +0
∴ x = 1 is not its zero
Similarly, f(-1) = (-1)3 + 6(-1)2 + 11(-1) + 6
= -1 + 6 x 1-11+6
=-1+6-11+6
= 12-12 = 0
∴  x = -1 is its zero
f(-2) = (-2)3 + 6(-2)2 + 11 (-2) + 6
= -8 + 24 – 22 + 6
= -30 + 30 = 0
∴ x = -2 is its zero
f(-3) = (-3)3 + 6(-3)2 + 11 (-3) + 6
= -27 + 54 – 33 + 6 = 60 – 60 = 0
∴  x = -3 is its zero
x = -1, -2, -3 are zeros of f(x)
Hence roots of f(x) are -1, -2, -3

Question 7.
Find the rational roots of the polynomial f(x) = 2x3 + x2 – 7x – 6.
Solution:
RD Sharma Class 9 Solutions Chapter 6 Factorisation of Polynomials Ex 6.2 Q7.1
RD Sharma Class 9 Solutions Chapter 6 Factorisation of Polynomials Ex 6.2 Q7.2
RD Sharma Class 9 Solutions Chapter 6 Factorisation of Polynomials Ex 6.2 Q7.3

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