Category: CBSE Class 9

RD Sharma Class 9 Solutions Chapter 24 Measures of Central Tendency Ex 24.3

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 24 Measures of Central Tendency Ex 24.3

Other Exercises

• RD Sharma Class 9 Solutions Chapter 24 Measures of Central Tendency Ex 24.1
• RD Sharma Class 9 Solutions Chapter 24 Measures of Central Tendency Ex 24.2
• RD Sharma Class 9 Solutions Chapter 24 Measures of Central Tendency Ex 24.3
• RD Sharma Class 9 Solutions Chapter 24 Measures of Central Tendency Ex 24.4
• RD Sharma Class 9 Solutions Chapter 24 Measures of Central Tendency VSAQS
• RD Sharma Class 9 Solutions Chapter 24 Measures of Central Tendency MCQS

Find the median of the following data (1-8)

Question 1.
83, 37, 70, 29, 45, 63, 41, 70, 34, 54
Solution:
We know that median = \(\frac { 1 }{ 2 } \left[ \frac { n }{ 2 } th\quad term+\left( \frac { n }{ 2 } +1 \right) th\quad term \right] \)
(When n is even)
= \(\frac { n+1 }{ 2 } th\quad term\)
83, 37, 70, 29, 45, 63, 41, 70, 34, 54
Arranging in ascending order, 29, 34, 37, 41, 45, 54, 63, 70, 70, 83
Here n = 10 which an even
Median = \(\frac { 1 }{ 2 } \)[5th term + 6th term]
= \(\frac { 1 }{ 2 } \) (45+54) = \(\frac { 99 }{ 2 } \) = 49.5

Question 2.
133, 73, 89, 108, 94, 104, 94, 85, 100, 120
Solution:
133, 73, 89, 108, 94, 104, 94, 85, 100, 120
Arranging in ascending order, 73, 85, 89, 94, 94, 100, 104, 108, 120, 133
Here n = 10 which is an even
Median = \(\frac { 1 }{ 2 } \)[5th term + 6th term]
= \(\frac { 1 }{ 2 } \) (94+100) = \(\frac { 1 }{ 2 } \) x 194 = 97

Question 3.
31, 38, 27, 28, 36, 25, 35, 40
Solution:
31, 38, 27, 28, 36, 25, 35, 40
Arranging in ascending order, 25, 27, 28, 31, 35, 36, 38, 40
Here n = 8 which is even
Median = \(\frac { 1 }{ 2 } \)[4th term + 5th term]
= \(\frac { 1 }{ 2 } \) (31+35) = \(\frac { 1 }{ 2 } \) x 66 = 33

Question 4.
15, 6, 16, 8, 22, 21, 9, 18, 25
Solution:
15, 6, 16, 8, 22, 21, 9, 18, 25
Arranging in ascending order = 6, 8, 9, 15, 16, 18, 21, 22, 25
Here n = 9 which is odd
Median \(\frac { n+1 }{ 2 } th\quad term\) = \(\frac { 9+1 }{ 2 } th\quad term\) = \(\frac { 10 }{ 2 } th\quad \)
= 5th term = 16

Question 5.
41, 43, 127, 99, 71, 92, 71, 58, 57
Solution:
41, 43, 127, 99, 71, 92, 71, 58, 57
Arranging in ascending order = 41, 43, 57, 58, 71, 71, 92, 99, 127
Here n = 9 which is an odd
Median \(\frac { n+1 }{ 2 } th\quad term\) = \(\frac { 9+1 }{ 2 } th\quad term\) = \(\frac { 10 }{ 2 } th\quad\)
= 5th term = 71

Question 6.
25, 34, 31, 23, 22, 26, 35, 29, 20, 32
Solution:
25, 34, 31, 23, 22, 26, 35, 29, 20, 32
Arranging in ascending order = 20, 22, 23, 25, 26, 29, 31, 32, 34, 35
Here n = 10 which is even
Median = \(\frac { 1 }{ 2 } \left[ \frac { n }{ 2 } th\quad term+\left( \frac { n }{ 2 } +1 \right) th\quad term \right] \)
= \(\frac { 1 }{ 2 } \) [5th term + 6th term]
= \(\frac { 1 }{ 2 } \) (26 + 29) = \(\frac { 1 }{ 2 } \) x 55 = \(\frac { 55 }{ 2 } \) = 27.5

Question 7.
12, 17, 3, 14, 5, 8, 7, 15
Solution:
12, 17, 3, 14, 5, 8, 7, 15
Arranging in ascending order = 3, 5, 7, 8, 12, 14, 15, 17
Here n = 8 which is odd
Median = \(\frac { 1 }{ 2 } \left[ \frac { n }{ 2 } th\quad term+\left( \frac { n }{ 2 } +1 \right) th\quad term \right] \)
= \(\frac { 1 }{ 2 } \) [4th term + 5th term]
= \(\frac { 1 }{ 2 } \) (8+12) = \(\frac { 1 }{ 2 } \) x 20 = 10

Question 8.
92, 35, 67, 85, 72, 81, 56, 51, 42, 69
Solution:
92, 35, 67, 85, 72, 81, 56, 51, 42. 69
Arranging in ascending order = 35, 42, 51, 56, 67, 69, 72, 81, 85, 92
Here n = 10 which is even
Median = \(\frac { 1 }{ 2 } \left[ \frac { n }{ 2 } th\quad term+\left( \frac { n }{ 2 } +1 \right) th\quad term \right] \)
= \(\frac { 1 }{ 2 } \) [5th term + 6th term]
= \(\frac { 1 }{ 2 } \) (67+69) = \(\frac { 1 }{ 2 } \) x 136 = 68

Question 9.
Numbers 50, 42, 35, 2x + 10, 2x – 8, 12, 11, 8 are written in descending order and their median is 25, find x.
Solution:
50, 42, 35, 2x + 10, 2x – 8, 12, 11, 8 are in descending order
Here n = 8 which is even
Now Median = \(\frac { 1 }{ 2 } \left[ \frac { n }{ 2 } th\quad term+\left( \frac { n }{ 2 } +1 \right) th\quad term \right] \)
= \(\frac { 1 }{ 2 } \) [4th term + 5th term] = \(\frac { 1 }{ 2 } \)[2x + 10 + 2x – 8]
= \(\frac { 1 }{ 2 } \) [4x + 2] = 2x + 1
But median = 25
2x + 1 = 25
=> 2x = 25 – 1 = 24
=> \(\frac { 24 }{ 2 } \) = 12
Hence x = 12

Question 10.
Find the median of the following observations 46, 64, 87, 41, 58, 77, 35, 90, 55, 92, 33. If 92 is replaced by 99 and 41 by 43 in the above data, find the new median?
Solution:
46, 64, 87, 41, 58, 77, 35, 90, 55, 92, 33
Writing in ascending order = 33, 35, 41, 46, 55, 58, 64, 77, 87, 90, 92
Here n = 11 which is odd
Median = \(\frac { n+1 }{ 2 } \) th term
= \(\frac { 11+1 }{ 2 } \) = \(\frac { 12 }{ 2 } \)
= 6th term = 58
By replacing 92 by 93 and 41 by 43, then new order will be
33, 35, 43, 46, 55, 58, 64, 77, 87, 90, 99
Median = 6th term = 58

Question 11.
Find the median of the following data : 41, 43, 127, 99, 61, 92, 71, 58, 57. If 58 is replaced by 85, what will be the new median.
Solution:
41, 43, 127, 99, 61, 92, 71, 58, 57
Arranging in ascending order = 41, 43, 57, 58, 61, 71, 92, 99, 127
Here n = 9 which is odd
Median = \(\frac { n+1 }{ 2 } \) th term = \(\frac { 9+1 }{ 2 } \) th term
= \(\frac { 10 }{ 2 } \) = 5th term = 61
By change 58 by 92, we get new order = 41, 43, 57, 61, 71, 92, 92, 99, 127
Median = 5th term = 71

Question 12.
The weights (in kg) of 15 students are : 31, 35, 27, 29, 32, 43, 37, 41, 34, 28, 36, 44, 45, 42, 30. Find the median. If the weight 44 kg is replaced by 46 kg and 27 kg by 25 kg, find the new median.
Solution:
Weights of 15 students are 31, 35, 27, 29, 32, 43, 37, 41, 34, 28, 36, 44, 45, 42, 30
Writing in ascending order = 27, 28, 29, 30, 31, 32, 34, 35, 36, 37, 41, 42, 43, 44, 45
here n = 15 which is odd
n+1 15+1
Median = \(\frac { n+1 }{ 2 } \) th term = \(\frac { 15+1 }{ 2 } \)
= \(\frac { 16 }{ 2 } \)th term = 8th term = 35 kg
By replacing 44 kg by 46 kg and 27 kg by 25 kg we get new order,
25, 28, 29, 30, 31, 32, 34, 35, 36, 37, 41, 42, 43, 45, 46
Median = 8th term = 35 kg

Question 13.
The following observations have been arranged in ascending order. If the median of the data is 63, find the value of x: 29, 32, 48, 50, x, x + 2, 72, 78, 84, 95
Solution:
Median = 63
29, 32, 48, 50, x, x + 2, 72, 78, 84, 95
Here n = 10 which is even
median = \(\frac { 1 }{ 2 } \left[ \frac { n }{ 2 } th\quad term+\left( \frac { n }{ 2 } +1 \right) th\quad term \right] \)
= \(\frac { 1 }{ 2 } \left[ \frac { 10 }{ 2 } th\quad term+\left( \frac { 10 }{ 2 } +1 \right) th\quad term \right] \)
= \(\frac { 1 }{ 2 } \) [5th term + 6th term]
= \(\frac { 16 }{ 2 } \) [x+x+2] = \(\frac { 2x + 2 }{ 2 } \) = x + 1
x + 1 = 63 = x = 63 – 1 = 62
Hence x = 62

Hope given RD Sharma Class 9 Solutions Chapter 24 Measures of Central Tendency Ex 24.3 are helpful to complete your math homework.

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RD Sharma Class 9 Solutions Chapter 10 Congruent Triangles VSAQS

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

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.
Define complementary angles.
Solution:
Two angles whose sum is 90°, are called complementary angles.

Question 2.
Define supplementary angles.
Solution:
Two angles whose sum is 180°, are called supplementary angles.

Question 3.
Define adjacent angles.
Solution:
Two angles which have common vertex and one arm common are called adjacent angles.

Question 4.
The complement of an acute angles is…….
Solution:
The complement of an acute angles is an acute angle.

Question 5.
The supplement of an acute angles is………
Solution:
The supplement of an acute angles is a obtuse angle.

Question 6.
The supplement of a right angle is…….
Solution:
The supplement of a right angle is a right angle.

Question 7.
Write the complement of an angle of measure x°.
Solution:
The complement of x° is (90° – x)°

Question 8.
Write the supplement of an angle of measure 2y°.
Solution:
The supplement of 2y° is (180° – 2y)°

Question 9.
If a wheel has six spokes equally spaced, then find the measure of the angle between two adjacent spokes.
Solution:
Total measure of angle around a point = 360°
Number of spokes = 6
∴ Angle between the two adjacent spokes = \(\frac { { 360 }^{ \circ } }{ 6 }\) = 60°

Question 10.
An angle is equal to its supplement. Determine its measure.
Solution:
Let required angle = x°
Then its supplement angle = 180° – x
x = 180° – x
⇒ x + x = 180°
⇒  2x = 180° ⇒  x = \(\frac { { 180 }^{ \circ } }{ 2 }\) = 90°
∴ Required angle = 90°

Question 11.
An angle is equal to five times its complement. Determine its measure.
Solution:
Let required measure of angle = x°
∴  Its complement angle = 90° – x
∴  x = 5(90° – x)
⇒  x = 450° – 5x
⇒  x + 5x = 450°
⇒  6x = 450°
⇒ x = \(\frac { { 450 }^{ \circ } }{ 6 }\) = 75°
∴ Required angle = 75°

Question 12.
How many pairs of adjacent angles are formed when two lines intersect in a point?
Solution:
If two lines AB and CD intersect at a point O, then pairs of two adjacent angles are, ∠AOC and ∠COB, ∠COB and ∠BOD, ∠BOD and DOA, ∠DOA and ∠ZAOC
i.e, 4 pairs

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

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RD Sharma Class 9 Solutions Chapter 24 Measures of Central Tendency MCQS

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 24 Measures of Central Tendency MCQS

Other Exercises

• RD Sharma Class 9 Solutions Chapter 24 Measures of Central Tendency Ex 24.1
• RD Sharma Class 9 Solutions Chapter 24 Measures of Central Tendency Ex 24.2
• RD Sharma Class 9 Solutions Chapter 24 Measures of Central Tendency Ex 24.3
• RD Sharma Class 9 Solutions Chapter 24 Measures of Central Tendency Ex 24.4
• RD Sharma Class 9 Solutions Chapter 24 Measures of Central Tendency VSAQS
• RD Sharma Class 9 Solutions Chapter 24 Measures of Central Tendency MCQS

Mark the correct alternative in each of the following:

Question 1.
Which one of the following is not a measure of central value?
(a) Mean
(b) Range
(c) Median
(d) Mode
Solution:
Range (b)

Question 2.
The mean of n observations is \(\overline { X } \) . If k is added to each observation, then the new mean is
(a) \(\overline { X } \)
(b) \(\overline { X } \) + k
(c) \(\overline { X } \) – k
(d) k\(\overline { X } \)
Solution:
Mean of n observation = \(\overline { X } \)
By adding k to each observation the new mean will be \(\overline { X } \) + k (b)

Question 3.
The mean of n observations is \(\overline { X } \) . If each observation is multiplied by k, the mean of new observations is
(a) k\(\overline { X } \)
(b) \(\frac { \overline { X } }{ k } \)
(c) \(\overline { X } \) + k
(d) \(\overline { X } \) – k
Solution:
Mean of n observations = \(\overline { X } \)
By multiplying each observation by k,
the new mean = k\(\overline { X } \) (a)

Question 4.
The mean of a set of seven numbers is 81. If one of the numbers is discarded, the mean of the remaining numbers is 78. The value of discarded number is
(a) 98
(b) 99
(c) 100
(d) 101
Solution:
Mean of 7 numbers = 81
Total = 7 x 81 = 567
By discarding one number, the mean of the remaining 7 – 1 = 6 numbers = 78
Total = 6 x 78 = 468
Discarded number = 567 – 468 = 99 (b)

Question 5.
For which set of numbers do the mean, median and mode all have the same value?
(a) 2, 2, 2, 2, 4
(b) 1, 3, 3, 3, 5
(c) 1, 1, 2, 5, 6
(d) 1, 1, 1, 2, 5
Solution:
a) In set 2, 2, 2, 2, 4

Mode = 3 as it come in maximum times
This set has mean, median and mode same (b)

Question 6.
For the set of numbers 2, 2, 4, 5 and 12, which of the following statements is true?
(a) Mean = Median
(b) Mean > Mode
(c) Mean < Mode
(d) Mode = Median
Solution:
The given set is 2, 2, 4, 5, 12

Question 7.
If the arithmetic mean of 7, 5, 13, x and 9 is 10, then the value of x is
(a) 10
(b) 12
(c) 14
(d) 16
Solution:
Arithmetic mean of 7, 5, 13, x, 9 is 10

Question 8.
If the mean of five observations x, x + 2, x + 4, x + 6, x + 8, is 11, then the mean of first three observations is
(a) 9
(b) 11
(c) 13
(d) none of these
Solution:
Mean = 11
But mean of x, x + 2, x + 4, x+ 6, x + 8

Question 9.
Mode is
(a) least frequent value
(b) middle most value
(c) most frequent value
(d) none of these
Solution:
Mode is most frequent value (c)

Question 10.
The following is the data of wages per day: 5, 4, 7, 5, 8, 8, 8, 5, 7, 9, 5, 7, 9, 10, 8 The mode of the data is
(a) 7
(b) 5
(c) 8
(d) 10
Solution:
Wages per day
5, 4, 7, 5, 8, 8, 8, 5, 7, 9, 5, 7, 9, 10, 8
=> 4, 5, 5, 5, 5, 7, 7, 7, 8, 8, 8, 8, 9, 9, 10
Here 8 comes in maximum times
Mode = 8 (c)

Question 11.
The median of the following data :
is ,
(a) 0
(b) -1.5
(c) 2
(d) 3.5
Solution:
Arranging in ascending order,
-3, -3, -1, 0, 2, 2, 2, 5, 5, 5, 5, 6, 6, 6

Question 12.
The algebraic sum of the deviations of a set of n values from their mean is
(a) 0
(b) n – 1
(c) n
(d) n + 1
Solution:
The algebraic sum of deviation of a set of n values from that mean

Question 13.
A, B, C are three sets of values of X:
A : 2, 3, 7, 1, 3, 2, 3
B: 7, 5, 9, 12, 5, 3, 8
C: 4, 4, 11, 7 ,2, 3, 4
Which one of the following statements is
correct?
(a) Mean of A = Mode of C
(b) Mean of C = Median of B
(c) Median of B = Mode of A
(d) Mean, Median and Mode of A are equal.
Solution:
Arranging the sets in ascending order
A{2, 3, 7, 1,3,2,3)
= {1, 2, 2, 3, 3, 3, 7)
B = {7, 5, 9, 12, 5, 3, 8)
= {3, 5, 5, 7, 8, 9, 12)
C = {4, 4, 11,7,2,3,4)
= {2, 3, 4, 4, 4, 7, 11)

Mode = 5 {as it comes max times}
(c) Mean of set C = \(\\ \frac { 2+3+4+4+4+7+11 }{ 7 } \)
= \(\\ \frac { 35 }{ 7 } \) = 5
Median = \(\\ \frac { 7+1 }{ 2 } \) th =\(\\ \frac { 8 }{ 2 } \) =4th term = 4
Mode =4 {as it comes max times}
In set A,mean = median = mode = 3 (d)

Question 14.
The empirical relation between mean, mode and median is
(a) Mode = 3 Median — 2 Mean
(b) Mode 2 Median — 3 Mean
(c) Median 3 Mode — 2 Mean
(d) Mean = 3 Median —2 Mode
Solution:
The empirical relations between mean, mode
and median is
Mode = 3 Median — 2 Mean (a)

Question 15.
The mean of a, b, c, d and e is 28. If the mean of a, c, and e is 24, what is the mean of b and d?
(a) 31
(b) 32
(c) 33
(d) 34
Solution:
Mean of a, b, c, d and e = 28
Total of a, b, c, d and e = 28 x 5 = 140
Mean of a, c and e is = 24
Total of a, c, e = 24 x 3 = 72
Total of b and d = 140 – 72 = 68
Mean = \(\\ \frac { 68 }{ 2 } \) = 34 (d)

Hope given RD Sharma Class 9 Solutions Chapter 24 Measures of Central Tendency MCQS 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.

RD Sharma Class 9 Solutions Chapter 24 Measures of Central Tendency Ex 24.4

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 24 Measures of Central Tendency Ex 24.4

Other Exercises

• RD Sharma Class 9 Solutions Chapter 24 Measures of Central Tendency Ex 24.1
• RD Sharma Class 9 Solutions Chapter 24 Measures of Central Tendency Ex 24.2
• RD Sharma Class 9 Solutions Chapter 24 Measures of Central Tendency Ex 24.3
• RD Sharma Class 9 Solutions Chapter 24 Measures of Central Tendency Ex 24.4
• RD Sharma Class 9 Solutions Chapter 24 Measures of Central Tendency VSAQS
• RD Sharma Class 9 Solutions Chapter 24 Measures of Central Tendency MCQS

Question 1.
Find out the mode of the following marks obtained by 15 students in a class:
Marks : 4, 6, 5, 7, 9, 8, 10, 4, 7, 6, 5, 9, 8, 7, 7.
Solution:
Marks obtained are in ascending order,
4, 4, 5, 5, 6, 6, 7, 7, 7, 7, 8, 8, 9, 9, 10
Here we see that 7 is the number which is maximum times i.e. 4 times
Mode = 7

Question 2.
Find the mode for the following data:
125, 175, 225, 125, 225, 175, 325, 125, 375, 225, 125
Solution:
Arranging in ascending order,
125, 125, 125, 125, 175, 175, 225, 225, 225, 325, 375
We see that, 125 is the number which is in maximum times
Mode = 125

Question 3.
Find the mode for the following series:
7.5, 7.3, 7.2, 7.2, 7.4, 7.7, 7.7, 7.5, 7.3, 7.2, 7.6, 7.2
Solution:
Arranging in ascending order,
7.2, 7.2, 7.2, 7.2, 7.3, 7.3, 7.4, 7.5, 7.5, 7.6, 7.7, 7.7
We see that 7.2 comes in maximum times
Mode = 7.2

Question 4.
Find the mode of the following data in each case:
(i) 14, 25, 14, 28, 18, 17, 18, 14, 23, 22, 14, 18
(ii) 7, 9, 12, 13, 7, 12, 15, 7, 12, 7, 25, 18, 7
Solution:
(i) 14, 25, 14, 28, 18, 17, 18, 14, 23, 22, 14, 18
Arranging in ascending order,
14, 14, 14,. 14, 17, 18, 18, 18, 22, 23, 25, 28
Here we see that 14 comes in maximum times
Mode = 14
(ii) 7, 9, 12, 13, 7, 12, 15, 7, 12, 7, 25, 18, 7
Arranging in order,
7, 7, 7, 7, 7, 9, 12, 12, 12, 13, 15, 18, 25
Here we see that 7 comes in maximum times
Mode = 7

Question 5.
The demand of different shirt sizes, as obtained by a survey, is given below:

Find the modal shirt sizes, as observed from the survey.
Solution:
From the given data

From above, we see that
Modal size is 39 as it has maximum times persons

Hope given RD Sharma Class 9 Solutions Chapter 24 Measures of Central Tendency Ex 24.4 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.

RD Sharma Class 9 Solutions Chapter 11 Co-ordinate Geometry VSAQS

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 11 Co-ordinate Geometry VSAQS

Other Exercises

• RD Sharma Class 9 Solutions Chapter 11 Co-ordinate Geometry Ex 11.1
• RD Sharma Class 9 Solutions Chapter 11 Co-ordinate Geometry Ex 11.2
• RD Sharma Class 9 Solutions Chapter 11 Co-ordinate Geometry VSAQS
• RD Sharma Class 9 Solutions Chapter 11 Co-ordinate Geometry MCQS

Question 1.
Define a triangle.
Solution:
A figure bounded by three lines segments in a plane is called a triangle.

Question 2.
Write the sum of the angles of an obtuse triangle.
Solution:
The sum of angles of an obtuse triangle is 180°.

Question 3.
In ∆ABC, if ∠B = 60°, ∠C = 80° and the bisectors of angles ∠ABC and ∠ACB meet at a point O, then find the measure of ∠BOC.
Solution:
In ∆ABC, ∠B = 60°, ∠C = 80°
OB and OC are the bisectors of ∠B and ∠C
∵ ∠A + ∠B + ∠C = 180° (Sum of angles of a triangle)
⇒ ∠A + 60° + 80° = 180°
⇒ ∠A + 140° = 180°
∴ ∠A = 180°- 140° = 40°

= 90° + – x 40° = 90° + 20° = 110°

Question 4.
If the angles of a triangle are in the ratio 2:1:3. Then find the measure of smallest angle.
Solution:
Sum of angles of a triangle = 180°
Ratio in the angles = 2 : 1 : 3
Let first angle = 2x
Second angle = x
and third angle = 3x
∴ 2x + x + 3x = 180° ⇒ 6x = 180°
∴ x = \(\frac { { 180 }^{ \circ } }{ 6 }\)  = 30°
∴ First angle = 2x = 2 x 30° = 60°
Second angle = x = 30°
and third angle = 3x = 3 x 30° = 90°
Hence angles are 60°, 30°, 90°

Question 5.
State exterior angle theorem.
Solution:
Given : In ∆ABC, side BC is produced to D

To prove : ∠ACD = ∠A + ∠B
Proof: In ∆ABC,
∠A + ∠B + ∠ACB = 180° …(i) (Sum of angles of a triangle)
and ∠ACD + ∠ACB = 180° …(ii) (Linear pair)
From (i) and (ii)
∠ACD + ∠ACB = ∠A + ∠B + ∠ACB
∠ACD = ∠A + ∠B
Hence proved.

Question 6.
The sum of two angles of a triangle is equal to its third angle. Determine the measure of the third angle.
Solution:
In ∆ABC,
∠A + ∠C = ∠B

But ∠A + ∠B + ∠C = 180° (Sum of angles of a triangle)
∴ ∠B + ∠A + ∠C = 180°
⇒ ∠B + ∠B = 180°
⇒ 2∠B = 180°
⇒ ∠B = \(\frac { { 180 }^{ \circ } }{ 2 }\)  = 90°
∴ Third angle = 90°

Question 7.
In the figure, if AB || CD, EF || BC, ∠BAC = 65° and ∠DHF = 35°, find ∠AGH.

Solution:
Given : In figure, AB || CD, EF || BC ∠BAC = 65°, ∠DHF = 35°

∵ EF || BC
∴ ∠A = ∠ACH (Alternate angle)
∴ ∠ACH = 65°
∵∠GHC = ∠DHF
(Vertically opposite angles)
∴ ∠GHC = 35°
Now in ∆GCH,
Ext. ∠AGH = ∠GCH + ∠GHC
= 65° + 35° = 100°

Question 8.
In the figure, if AB || DE and BD || FG such that ∠FGH = 125° and ∠B = 55°, find x and y.

Solution:
In the figure, AB || DF, BD || FG

∠FGH = 125° and ∠B = 55°
∠FGH + FGE = 180° (Linear pair)
⇒ 125° + y – 180°
⇒ y= 180°- 125° = 55°
∵ BA || FD and BD || FG
∠B = ∠F = 55°
Now in ∆EFG,
∠F + ∠FEG + ∠FGE = 180°
(Angles of a triangle)
⇒ 55° + x + 55° = 180°
⇒ x+ 110°= 180°
∴ x= 180°- 110° = 70°
Hence x = 70, y = 55°

Question 9.
If the angles A, B and C of ∆ABC satisfy the relation B – A = C – B, then find the measure of ∠B.
Solution:
In ∆ABC,
∠A + ∠B + ∠C= 180° …(i)
and B – A = C – B

⇒ B + B = A + C ⇒ 2B = A + C
From (i),
B + 2B = 180° ⇒ 3B = 180°
∠B = \(\frac { { 180 }^{ \circ } }{ 3 }\) = 60°
Hence ∠B = 60°

Question 10.
In ∆ABC, if bisectors of ∠ABC and ∠ACB intersect at O at angle of 120°, then find the measure of ∠A.
Solution:
In ∆ABC, bisectors of ∠B and ∠C intersect at O and ∠BOC = 120°

But ∠BOC = 90°+ \(\frac { 1 }{ 2 }\)
90°+ \(\frac { 1 }{ 2 }\) ∠A= 120°
⇒ \(\frac { 1 }{ 2 }\) ∠A= 120°-90° = 30°
∴ ∠A = 2 x 30° = 60°

Question 11.
If the side BC of ∆ABC is produced on both sides, then write the difference between the sum of the exterior angles so formed and ∠A.
Solution:
In ∆ABC, side BC is produced on both sides forming exterior ∠ABE and ∠ACD
Ext. ∠ABE = ∠A + ∠ACB
and Ext. ∠ACD = ∠ABC + ∠A

Adding we get,
∠ABE + ∠ACD = ∠A + ∠ACB + ∠A + ∠ABC
⇒ ∠ABE + ∠ACD – ∠A = ∠A 4- ∠ACB + ∠A + ∠ABC – ∠A (Subtracting ∠A from both sides)
= ∠A + ∠ABC + ∠ACB = ∠A + ∠B + ∠C = 180° (Sum of angles of a triangle)

Question 12.
In a triangle ABC, if AB = AC and AB is produced to D such that BD = BC, find ∠ACD: ∠ADC.
Solution:
In ∆ABC, AB = AC
AB is produced to D such that BD = BC
DC are joined
In ∆ABC, AB = AC
∴ ∠ABC = ∠ACB
In ∆ BCD, BD = BC
∴ ∠BDC = ∠BCD
and Ext. ∠ABC = ∠BDC + ∠BCD = 2∠BDC (∵ ∠BDC = ∠BCD)
⇒ ∠ACB = 2∠BCD (∵ ∠ABC = ∠ACB)
Adding ∠BDC to both sides
⇒ ∠ACB + ∠BDC = 2∠BDC + ∠BDC
⇒ ∠ACB + ∠BCD = 3 ∠BDC (∵ ∠BDC = ∠BCD)
⇒ ∠ACB = 3∠BDC

Question 13.
In the figure, side BC of AABC is produced to point D such that bisectors of ∠ABC and ∠ACD meet at a point E. If ∠BAC = 68°, find ∠BEC.

Solution:
In the figure,

side BC of ∆ABC is produced to D such that bisectors of ∠ABC and ∠ACD meet at E
∠BAC = 68°
In ∆ABC,
Ext. ∠ACD = ∠A + ∠B
⇒ \(\frac { 1 }{ 2 }\) ∠ACD = \(\frac { 1 }{ 2 }\) ∠A + \(\frac { 1 }{ 2 }\) ∠B
⇒ ∠2= \(\frac { 1 }{ 2 }\) ∠A + ∠1 …(i)
But in ∆BCE,
Ext. ∠2 = ∠E + ∠l
⇒ ∠E + ∠l = ∠2 = \(\frac { 1 }{ 2 }\) ∠A + ∠l [From (i)]
⇒ ∠E = \(\frac { 1 }{ 2 }\) ∠A = \(\frac { { 68 }^{ \circ } }{ 2 }\)  =34°

Hope given RD Sharma Class 9 Solutions Chapter 11 Co-ordinate Geometry VSAQS 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.

RD Sharma Class 9 Solutions Chapter 6 Factorisation of Polynomials Ex 6.4

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 6.4

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

In each of the following, use factor Theorem to find whether polynomial g(x) is a factor of polynomial f(x) or, not: (1-7)
Question 1.
f(x) = x³ – 6x² + 11x – 6; g(x) = x – 3
Solution:
We know that if g(x) is a factor of p(x),
then the remainder will be zero. Now,
f(x) = x³ – 6x² + 11x – 6; g(x) = x -3
Let x – 3 = 0, then x = 3
∴ Remainder = f(3)
= (3)³ – 6(3)² +11 x 3 – 6
= 27-54 + 33 -6
= 60 – 60 – 0
∵  Remainder is zero,
∴ x – 3 is a factor of f(x)

Question 2.
f(x) = 3X⁴ + 17x³ + 9x² – 7x – 10; g(x) = x + 5
Solution:
f(x) = 3x⁴ + 17X³ + 9x² – 7x – 10; g(x) = x + 5
Let x + 5 = 0, then x = -5
∴  Remainder = f(-5) = 3(-5)⁴ + 17(-5)³ + 9(-5)² – 7(-5) – 10
= 3 x 625 + 17 x (-125) + 9 x (25) – 7 x (-5) – 10
= 1875 -2125 + 225 + 35 – 10
= 2135 – 2135 = 0
∵  Remainder = 0
∴ (x + 5) is a factor of f(x)

Question 3.
f(x) = x⁵ + 3x⁴ – x³ – 3x² + 5x + 15, g(x) = x + 3
Solution:
f(x) = x⁵ + 3X⁴ – x³ – 3x² + 5x + 15, g(x) = x + 3
Let x + 3 = 0, then x = -3
∴ Remainder = f(-3)
= (-3)⁵ + 3(-3)⁴ – (-3)³ – 3(-3)² + 5(-3) + 15
= -243 + 3 x 81 -(-27)-3 x 9 + 5(-3) + 15
= -243 +243 + 27-27- 15 + 15
= 285 – 285 = 0
∵  Remainder = 0
∴  (x + 3) is a factor of f(x)

Question 4.
f(x) = x³ – 6x² – 19x + 84, g(x) = x – 7
Solution:
f(x) = x³ – 6x² – 19x + 84, g(x) = x – 7
Let x – 7 = 0, then x = 7
∴  Remainder = f(7)
= (7)³ – 6(7)² – 19 x 7 + 84
= 343 – 294 – 133 + 84
= 343 + 84 – 294 – 133
= 427 – 427 = 0
∴  Remainder = 0
∴ (x – 7) is a factor of f(x)

Question 5.
f(x) = 3x³  + x² – 20x + 12, g(x) = 3x – 2
Solution:

Question 6.
f(x) = 2x³ – 9x² + x + 12, g(x) = 3 – 2x
Solution:
f(x) = 2x³ – 9x² + x + 12, g(x) = 3 – 2x

Question 7.
f(x) = x³ – 6x² + 11x – 6, g(x) = x² – 3x + 2
Solution:
g(x) = x² – 3x + 2
= x² – x – 2x + 2
= x(x – 1) – 2(x – 1)
= (x – 1) (x – 2)
If x – 1 = 0, then x = 1
‍∴ f(1) = (1)³ – 6(1)² + 11(1) – 6
= 1-6+11-6= 12- 12 = 0
‍∴ Remainder is zero
‍∴ x – 1 is a factor of f(x)
and if x – 2 = 0, then x = 2
∴ f(2) = (2)³ – 6(2)² + 11(2)-6
= 8 – 24 + 22 – 6 = 30 – 30 = 0
‍∴ Remainder = 0
‍∴ x – 2 is also a factor of f(x)

Question 8.
Show that (x – 2), (x + 3) and (x – 4) are factors of x³ – 3x² – 10x + 24.
Solution:
f(x) = x³ – 3x² – 10x + 24
Let x – 2 = 0, then x = 2
Now f(2) = (2)³ – 3(2)² – 10 x 2 + 24
= 8 – 12 – 20 + 24 = 32 – 32 = 0
‍∴ Remainder = 0
‍∴ (x – 2) is the factor of f(x)
If x + 3 = 0, then x = -3
Now, f(-3) = (-3)³ – 3(-3)² – 10 (-3) + 24
= -27 -27 + 30 + 24
= -54 + 54 = 0
∴ Remainder = 0
∴ (x + 3) is a factor of f(x)
If x – 4 = 0, then x = 4
Now f(4) = (4)³ – 3(4)² – 10 x 4 + 24 = 64-48 -40 + 24
= 88 – 88 = 0
∴ Remainder = 0
∴ (x – 4) is a factor of (x)
Hence (x – 2), (x + 3) and (x – 4) are the factors of f(x)

Question 9.
Show that (x + 4), (x – 3) and (x – 7) are factors of x³ – 6x² – 19x + 84.
Solution:
Let f(x) = x³ – 6x² – 19x + 84
If x + 4 = 0, then x = -4
Now, f(-4) = (-4)³ – 6(-4)² – 19(-4) + 84
= -64 – 96 + 76 + 84
= 160 – 160 = 0
∴ Remainder = 0
∴ (x + 4) is a factor of f(x)
If x – 3 = 0, then x = 3
Now, f(3) = (3)³ – 6(3)² – 19 x 3 + 84
= 27 – 54 – 57 + 84
= 111 -111=0
∴ Remainder = 0
∴ (x – 3) is a factor of f(x)
and if x – 7 = 0, then x = 7
Now, f(7) = (7)³ – 6(7)² – 19 x 7 + 84
= 343 – 294 – 133 + 84
= 427 – 427 = 0
∴ Remainder = 0
∴ (x – 7) is also a factor of f(x)
Hence (x + 4), (x – 3), (x – 7) are the factors of f(x)

Question 10.
For what value of a (x – 5) is a factor of x³ – 3x² + ax – 10?
Solution:
f(x) = x³ – 3x² + ax – 10
Let x – 5 = 0, then x = 5
Now, f(5) = (5)³ – 3(5)² + a x 5 – 10
= 125 – 75 + 5a – 10
= 125 – 85 + 5a = 40 + 5a
∴ (x – 5) is a factor of fix)
∴ Remainder = 0
⇒  40 + 5a = 0 ⇒  5a = -40
⇒ a = \(\frac { -40 }{ 5 }\)= -8
Hence a = -8

Question 11.
Find the value of a such that (x – 4) is a factor of 5x³ – 7x² – ax – 28.
Solution:
Let f(x) =  5x³ – 7x² – ax – 28
and Let x – 4 = 0, then x = 4
Now, f(4) = 5(4)³ – 7(4)² – a x 4 – 28
= 5 x 64 – 7 x 16 – 4a – 28
= 320 – 112 – 4a – 28
= 320 – 140 – 4a
= 180 – 4a
∴ (x – 4) is a factor of f(x)
∴ Remainder = 0
⇒  180 -4a = 0
⇒  4a = 180
⇒  a = \(\frac { 180 }{ 4 }\) =  45
∴  a = 45

Question 12.
Find the value of a, if x + 2 is a factor of 4x⁴ + 2x³ – 3x² + 8x + 5a.
Solution:
Let f(x) = 4x⁴ + 2x³ – 3x² + 8x + 5a
and Let x + 2 = 0, then x = -2
Now, f(-2) = 4(-2)⁴ + 2(-2)³ – 3(-2)² + 8 x ( 2) + 5a
= 4 x 16 + 2(-8) – 3(4) + 8 (-2) + 5a
= 64- 16- 12- 16 +5a
= 64 – 44 + 5a
= 20 + 5a
∴  (x + 2) is a factor of fix)
∴  Remainder = 0
⇒  20 + 5a = 0 ⇒  5a = -20
⇒  a =\(\frac { -20 }{ 5 }\)  = -4
∴ a = -4

Question 13.
Find the value of k if x – 3 is a factor of k²x³ – kx² + 3kx – k.
Solution:
Let f(x) = k²x³ – kx² + 3kx – k
and Let x – 3 = 0, then x = 3
Now,f(3) = k²(3)³ – k(3)² + 3k(3) – k
= 27k² – 9k + 9k-k
= 27k²-k
∴ x – 3 is a factor of fix)
∴ Remainder = 0
∴ 27k² – k = 0
⇒ k(27k – 1) = 0 Either k = 0
or 21k – 1 = 0
⇒ 21k = 1
∴  k= \(\frac { 1 }{ 27 }\)
∴  k = 0,\(\frac { 1 }{ 27 }\)

Question 14.
Find the values of a and b, if x² – 4 is a factor of ax⁴ + 2x³ – 3x² + bx – 4.
Solution:
f(x) = ax⁴ + 2x³ – 3x² + bx – 4
Factors of x² – 4 = (x)² – (2)²
= (x + 2) (x – 2)
If x + 2 = 0, then x = -2
Now, f(-2) = a(-2)⁴ + 2(-2)³ – 3(-2)² + b(-2) – 4
16a- 16 – 12-26-4
= 16a -2b-32
∵ x + 2 is a factor of f(x)
∴ Remainder = 0
⇒  16a – 2b – 32 = 0
⇒ 8a – b – 16 = 0
⇒ 8a – b = 16         …(i)
Again x – 2 = 0, then x = 2
Now f(2) = a x (2)⁴ + 2(2)³ – 3(2)² + b x 2-4
= 16a + 16- 12 + 26-4
= 16a + 2b
∵  x – 2 is a factor of f(x)
∴ Remainder = 0
⇒  16a + 2b = 0
⇒ 8a + b= 0                             …(ii)
Adding (i) and (ii),
⇒ 16a = 16
⇒ a = \(\frac { 16 }{ 16 }\) = 1
From (ii) 8 x 1 + b = 0
⇒ 8 + b = 0
⇒  b = – 8
∴ a = 1, b = -8

Question 15.
Find α and β, if x + 1 and x + 2 are factors of x³ + 3x² – 2αx +β.
Solution:
Let f(x) = x³ + 3x² – 2αx + β
and Let x + 1 = 0 then x = -1
Now,f(-1) = (1)³ + 3(-1)² – 2α (-1) +β
= -1 + 3 + 2α + β
= 2 + 2α + β
∵  x + 1 is a factor of f(x)
∴  Remainder = 0
∴ 2 + 2α + β = 0
⇒  2α + β = -2                    …(i)
Again, let x + 2 = 0, then x = -2
Now, f(-2) = (-2)³ + 3(-2)² – 2α(-2) + β
= -8 + 12 + 4α+ β
= 4 + 4α+ β
∵ x + 2 is a factor of(x)
∴ Remainder = 0
∴ 4+ 4α + β = 0
⇒  4α + β = -4 …(ii)
Subtracting (i) from (ii),
2α = -2
⇒  α = \(\frac { -2 }{ 2 }\) = -1
From (ii), 4(-1) + β = -4
-4 + β= -4
⇒  β =-4+ 4 = 0
∴  α = -1, β = 0

Question 16.
If x – 2 is a factor of each of the following two polynomials, find the values of a in each case:
(i) x³ – 2ax² + ax – 1
(ii) x⁵ – 3x⁴ – ax³ + 3ax² + 2ax + 4
Solution:
(i) Let f(x) = x³ – 2ax² + ax – 1 and g(x) = x – 2
and let x – 2 = 0, then x = 2
∴ x – 2 is its factor
∴ Remainder = 0
f(2) = (2)³ – 2a x (2)² + a x 2 – 1
= 8-8a+ 2a-1 = 7-6a
∴ 7 – 6a = 0
⇒  6a = 7
⇒ a = \(\frac { 7 }{ 6 }\)
∴ a =  \(\frac { 7 }{ 6 }\)
(ii) Let f(x) = x⁵ – 3x⁴ – ax³ + 3 ax² + 2ax + 4 and g(x) = x – 2
Let x – 2 = 0, then x=2
∴ f(2) = (2)⁵ – 3(2)⁴ – a(2³) + 3a (2)² + 2a x 2 + 4
= 32 – 48 – 8a + 12a + 4a + 4
= -12 + 8a
∴ Remainder = 0
∴ -12 + 8a = 0
⇒ 8a= 12
⇒ a = \(\frac { 12 }{ 8 }\) = \(\frac { 3 }{ 2 }\)
∴ Hence a = \(\frac { 3 }{ 2 }\)

Question 17.
In each of the following two polynomials, find the values of a, if x – a is a factor:
(i) x⁶ – ax⁵ + x⁴-ax³ + 3x-a + 2
(ii) x⁵ – a²x³ + 2x + a + 1
Solution:
(i) Let f(x) ⁼ x⁶ – ax⁵+x⁴-ax³ + 3x-a + 2 and g(x) = x – a
∴ x – a is a factor
∴ x – a = 0
⇒ x = a
Now f(a) = a⁶-a x a⁵ + a⁴-a x a³ + 3a – a + 2
= a⁶-a⁶ + a⁴-a⁴ + 2a + 2
= 2a + 2
∴ x + a is a factor of p(x)
∴ Remainder = 0

Question 18.
In each of the following, two polynomials, find the value of a, if x + a is a factor.
(i)  x³ + ax² – 2x + a + 4
(ii) x⁴ – a²r + 3x – a
Solution:

Question 19.
Find the values of p and q so that x⁴ + px³ + 2x² – 3x + q is divisible by (x² – 1).
Solution:

Question 20.
Find the values of a and b so that (x + 1) and (x – 1) are factors of x⁴ + ax³ 3x² + 2x + b.
Solution:

Question 21.
If x³ + ax² – bx + 10 is divisible by x² – 3x + 2, find the values of a and b.
Solution:

Question 22.
If both x + 1 and x – 1 are factors of ax³ + x² – 2x + b, find the values of a and b.
Solution:

Question 23.
What must be added to x³ – 3x² – 12x + 19 so that the result is exactly divisibly by x² + x – 6?
Solution:

Question 24.
What must be subtracted from x³ – 6x² – 15x + 80-so that the result is exactly divisible by x² + x – 12?
Solution:

Question 25.
What must be added to 3x³ + x² – 22x + 9 so that the result is exactly divisible by 3x² + 7x – 6?
Solution:

Hope given RD Sharma Class 9 Solutions Chapter 6 Factorisation of Polynomials Ex 6.4 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.

RD Sharma Class 9 Solutions Chapter 7 Introduction to Euclid’s Geometry Ex 7.4

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 7.4

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.
Give the geometric representations of the following equations.
(a) on the number line
(b) on the cartesian plane.
(i) x – 2
(ii) y + 3 = 0
(iii) y = 3
(iv) 2x + 9 = 0
(v) 3x – 5 = 0
Solution:
(i) x = 2
(i) on the number line

(ii) x = 2 is a line parallel to 7-axis at a distance of 2 units to right of y-axis.

(ii) y = -3 is a line parallel to x-axis at a distance of 3 units below x-axis.

(iii) y = 3
(i) y = 3

(ii) y = 3 is a line parallel to x-axis at a distance of 3 units above x-axis.


x = -4.5 is a line parallel to 7-axis at a distance of 4.5 units to left of y-axis.


(ii) x = 1\(\frac { 2 }{ 3 }\) is a line parallel to y-axis at a  distance of 1\(\frac { 2 }{ 3 }\) unit to right side of y-axis.

Question 2.
Give the geometrical representation of 2x + 13 = 0 as an equation in
(i) One variable
(ii) Two variables
Solution:
(i) In one variable,
2x + 13 = 0
⇒ 2x = – 13
⇒ x = \(\frac { -13 }{ 2 }\)

is a line parallel to y-axis at a distance of -6 \(\frac { 1 }{ 2 }\) units on the left side of y-axis.

Question 3.
Solve the equation 3x + 2 = x -8, and represent on
(i) the number line
(ii) the Cartesian plane.
Solution:
3x + 2 = x – 8
⇒  3x – x = -8 – 2
⇒  2x = -10
⇒  x = \(\frac { -10 }{ 2 }\) = -5
(i) on the number line s = -5

(ii) x = -5 is a line parallel to  y-axis at a distance of 5 knot’s left of y-axis.

Question 4.
Write the equal of the line that is parallel to x-axis and passing through the points.
(i) (0, 3)                     
(ii)  (0, -4)
(iii) (2, -5)                     
(iv)    (3, 4)
Solution:
∵  A line parallel to x-axis will be of the type y = a
∴ (i) y = 3
(ii) y = -4
(iii) y = -5 and y = 4 are equations of the lines parallel to x-axis

Question 5.
Write the equation of the line that is parallel to y-axis and passing through the points.
(i) (4, 0)                      
(ii) (-2, 0)
(iii) (3, 5)                    
(iv) (-4, -3)
Solution:
∵  A line parallel to y-axis will be of the type x = a
∴  (i) x = 4, (ii)  x = -2, x = 3 and x = -4 are the equations of the lines parallel to y-axis.

Hope given RD Sharma Class 9 Solutions Chapter 7 Introduction to Euclid’s Geometry Ex 7.4 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.