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RD Sharma Class 8 Solutions Chapter 4 Cubes and Cube Roots Ex 4.1

These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 4 Cubes and Cube Roots Ex 4.1

Other Exercises

• RD Sharma Class 8 Solutions Chapter 4 Cubes and Cube Roots Ex 4.1
• RD Sharma Class 8 Solutions Chapter 4 Cubes and Cube Roots Ex 4.2
• RD Sharma Class 8 Solutions Chapter 4 Cubes and Cube Roots Ex 4.3
• RD Sharma Class 8 Solutions Chapter 4 Cubes and Cube Roots Ex 4.4
• RD Sharma Class 8 Solutions Chapter 4 Cubes and Cube Roots Ex 4.5

Question 1.
Find the cubes of the following numbers:
(i) 7
(ii) 12
(iii) 16
(iv) 21
(v) 40
(vi) 55
(vii) 100
(viii) 302
(ix) 301
Solution:
(i) (7)³ = 7 x 7 x 7 = 343
(ii) (12)³ = 12 x-12 x 12 = 1728
(iii) (16)³ = 16 x 16 x 16 = 4096
(iv) (21)³ = 21 x 21 x 21 = 441 x 21 =9261
(v) (40)³ = 40 x 40 x 40 = 64000
(vi) (55)³ = 55 x 55 x 55 = 3025 x 55 = 166375
(vii) (100)³ = 100 x 100 x 100 =1000000
(viii)(302)³ = 302 x 302 x 302 = 91204 x 302 = 27543608
(ix) (301)³ = 301 x 301 x 301 = 90601 x 301 =27270901

Question 2.
Write the cubes of all natural numbers between 1 and 10 and verify the following statements :
(i) Cubes of all odd natural numbers are odd.
(ii) Cubes of all even natural numbers are even.
Solution:
Cubes of first 10 natural numbers :
(1)³ = 1 x 1 x 1 = 1
(2)³ = 2 x 2 x 2 = 8
(3)³ = 3 x 3 x 3 = 27
(4)³= 4 x 4 x 4 = 64
(5)³ = 5 x 5 x 5 = 125
(6)³ = 6 x 6 x 6 = 216
(7)³ = 7 x 7 x 7 = 343
(8)³ = 8 x 8 x 8 = 512
(9)³ = 9 x 9 x 9= 729
(10)³ = 10 x 10 x 10= 1000
We see that the cubes of odd numbers is also odd and cubes of even numbers is also even.

Question 3.
Observe the following pattern :

Write the next three rows and calculate the value of 1³ + 2³ + 3³ +…. + 9³ + 10³ by the above pattern.
Solution:
We see the pattern

Question 4.
Write the cubes of 5 natural numbers which are multiples of 3 and verify the followings :
The cube of a natural number which is a multiple of 3 is a multiple of 27′
Solution:
5 natural numbers which are multiples of 3
3,6,9,12,15.
(3)³ = 3 x 3 x 3 = 27
Which is multiple of 27
(6)³ = 6 x 6 x 6 = 216 ÷ 27 = 8
Which is multiple of 27
(9)³ = 9 x 9 x 9 = 729 + 27 = 27
Which is multiple of 27
(12)³= 12 x 12 x 12 = 1728 ÷ 27 = 64
Which is multiple of 27
(15)³ = 15 x 15 x 15 = 3375 ÷ 27 = 125
Which is multiple of 27
Hence, cube of multiple of 3 is a multiple of 27

Question 5.
Write the cubes of 5 natural numbers which are of the form 3n+ 1 (e.g.,4, 7, 10, …………) and verify the following :
‘The cube of a natural number of the form 3n + 1 is a natural number of the same form i.e. when divided by 3 it leaves the remainder 1’.
Solution:
3n + 1
Let n = 1, 2, 3, 4, 5, then
If n = 1, then 3n +1= 3 x 1+1= 3+1= 4
If n = 2, then 3n +1=3 x 2+1=6+1=7
If n = 3, then 3n + 1= 3 x 3 + 1= 9 + 1 = 10
If n = 4, then 3n + 1= 3 x 4+1 = 12 + 1= 13
If n = 5, then 3n +1=3 x 5 + 1 = 15 +1 = 16
Now
(4)³ = 4 x 4 x 4 = 64
Which is \(\frac {64 }{ 3 }\)=21, Remainder = 1
(7)³ = 7 x 7 x 7 = 343
Which is \(\frac {343 }{ 3 }\) =114, Remainder = 1
(10)³ = 10 x 10 x 10 = 1000 ÷ 3 = 333, Remainder = 1
(13)³ = 13 x 13 x 13 = 2197 ÷ 3 = 732, Remainder = 1
(16)³ = 16 x 16 x 16 = 4096 ÷ 3 = 1365, Remainder = 1
Hence cube of natural number of the form, 3n + 1, is a natural of the form 3n + 1

Question 6.
Write the cubes of 5 natural numbers of the form 3n + 2 (i.e. 5, 8, 11,……… ) and verify the following :
‘The cube of a natural number of the form 3n + 2 is a natural number of the same form i.e. when it is dividend by 3 the remainder is 2’.
Solution:
Natural numbers of the form 3n + 2, when n
is a natural number i.e. 1, 2, 3, 4, 5,………….
If n = 1, then 3n + 2 = 3 x 1+2 = 3+ 2 = 5
If n = 2, then 3n + 2 = 3 x 2 + 2 = 6 + 2 = 8
If n = 3, then 3n + 2 = 3 x 3 + 2 = 9 + 2 = 11
If n = 4, then 3n + 2 = 3 x 4 + 2 = 12 + 2 = 14
and if n = 5, then 3n + 2 = 3 x 5 + 2 = 15 + 2= 17
Now (5)³ = 5 x 5 x 5 = 125
125 + 3 = 41, Remainder = 2
(8)² = 8 x 8 x 8 = 512 512 -s- 3 = 170, Remainder = 2
(11)³ = 11 x 11 x 11 = 1331
1331 + 3 = 443, Remainder = 2
(14)³ = 14 x 14 x 14 = 2744
2744 + 3 = 914, Remainder = 2
(17)³ = 17 x 17 x 17 = 4913
4913 = 3 = 1637, Remainder = 2
We see the cube of the natural number of the
form 3n + 2 is also a natural number of the
form 3n + 2.

Question 7.
Write the cubes of 5 natural numbers of which are multiples of 7 and verify the following :
‘The cube of a multiple of 7 is a multiple of 7^3′.
Solution:
5 natural numbers which are multiple of 7,are 7, 14, 21, 28, 35
(7)³ = (7)³ which is multiple of 7³
(14)³ = (2 x 7)3 = 2³ x 7³, which is multiple of 7³
(21)³ = (3 x 7)3 = 33 x 7³_, which is multiple of 7³
(28)³ = (4 x 7)³ = 4³ x 7³, which is multiple of 7³ (35)³ = (5 x 7)³ = 5³ x 73 which is multiple of 7³
Hence proved.

Question 8.
Which of the following are perfect cubes?
(i) 64
(ii) 216
(iii) 243
(iv) 1000
(v) 1728
(vi) 3087
(vii) 4608
(viii) 106480
(ix) 166375
(x) 456533
Solution:
(i) 64 = 2 x 2 x 2 x 2 x 2 x 2

Grouping the factors in triplets of equal factors, we see that no factor is left
∴ 64 is a perfect cube
(ii) 216 = 2 x 2 x 2 x 3 x 3 x 3

Grouping the factors in triplets of equal factors, we see that no factor is left
216 is a perfect cube.
(iii) 243 = 3 x 3 x 3 x 3 x 3

Grouping the factors in triplets, we see that two factors 3 x 3 are left
∴ 243 is not a perfect cube.
(iv) 1000 = 2 x 2 x 2 x 5 x 5 x 5

Grouping the factors in triplets of equal factors, we see that no factor is left
∴ 1000 is a perfect cube.
(v) 1728 = 2 x 2 x 2 x 2 x 2 x 2 x 3 x 3 x 3

Grouping the factors in triplets of the equal factors, we see that no factor is left
∴ 1728 is a perfect cube,
(vi) 3087 = 3 x 3 x 7 x 7 x 7

Grouping the factors in triplets of the equal factors, we see that two factor 3×3 are left
∴ 3087 is not a perfect cube.
(vii) 4608 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 3 x 3

Grouping the factors in triplets of equal factors, we see that two factors 3, 3 are left
∴ 4609 is not a perfect cube.
(viii) 106480 = 2 x 2 x 2 x 2 x 5 x 11 x 11 x 11

Grouping the factors in triplets of equal factors, we see that factors 2, 5 are left
∴ 106480 is not a perfect cube.
(ix) 166375 = 5 x 5 x 5 x 11 x 11 x 11

Grouping the factors in triplets of equal factors, we see that no factor is left
∴ 166375 is a perfect cube.
(x) 456533 = 7 x 7 x 7 x 11 x 11 x 11

Grouping the factors in triplets of equal factors, we see that no factor is left
∴ 456533 is a perfect cube.

Question 9.
Which of the following are cubes of even natural numbers ?
216, 512, 729,1000, 3375, 13824
Solution:
We know that the cube of an even natural number is also an even natural number
∴ 216, 512, 1000, 13824 are even natural numbers.
∴ These can be the cubes of even natural number.

Question 10.
Which of the following are cubes of odd natural numbers ?
125, 343, 1728, 4096, 32768, 6859
Solution:
We know that the cube of an odd natural number is also an odd natural number,
∴ 125, 343, 6859 are the odd natural numbers
∴ These can be the cubes of odd natural numbers.

Question 11.
What is the smallest number by which the following numbers must be multiplied, so that the products are perfect cubes ?
(i) 675
(ii) 1323
(iii) 2560
(iv) 7803
(v) 107311
(vi) 35721
Solution:
(i) 675 = 3 x 3 x 3 x 5 X 5

Grouping the factors in triplet of equal factors, 5 x 5 are left without triplet
So, by multiplying by 5, the triplet will be completed.
∴ Least number to be multiplied = 5
(ii) 1323 = 3 x 3 x 3 x 7 x 7

Grouping the factors in triplet of equal factors. We find that 7 x 7 has been left
So, multiplying by 7, we get a triplet
∴ The least number to be multiplied = 7
(iii) 2560 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 5

Grouping the factors in triplet of equal factors, 5 is left.
∴ To complete a triplet 5 x 5 is to multiplied
∴ Least number to be multiplied = 5 x 5 = 25
(iv) 7803 = 3 x 3 x 3 x 17 x 17

Grouping the factors in triplet of equal factors, we find the 17 x 17 are left
So, to complete the triplet, we have to multiply by 17
∴ Least number to be multiplied = 17
(v) 107811 = 3 x 3 x 3 x 3 x 11 x 11 x 11

Grouping the factors in triplet of equal factors, factor 3 is left
So, to complete the triplet 3 x 3 is to be multiplied
∴ Least number to be multiplied = 3 x 3 = 9
(vi) 35721 = 3 x 3 x 3 x 3 x 3 x 3 x 7 x 7

Grouping the factors in triplet of equal factors, we find that 7 x 7 is left
So, in order to complete the triplets, we have to multiplied by 7
∴ Least number to be multiplied = 7

Question 12.
By which smallest number must the following numbers be divided so that the quotient is a perfect cube ?
(i) 675
(ii) 8640
(iii) 1600
(iv) 8788
(v) 7803
(vi) 107811
(vii) 35721
(viii) 243000
Solution:
(i) 675 = 3 x 3 x 3 x 5 x 5

Grouping the factors in triplet of equal factors, 5 x 5 is left
5 x 5 is to be divided so that the quotient will be a perfect cube.
∴ The least number to be divided = 5 x 5 = 25
(ii) 8640 = 2 x 2 x 2 x 2 x 2 x 2 x 3 x 3 x 3

Grouping the factors in triplets of equal factors, 5 is left
∴ In order to get a perfect cube, 5 is to divided
∴ Least number to be divided = 5
(iii) 1600 = 2 x 2 x 2 x 2 x 2 x 2 x 5 x 5

Grouping the factors in triplets of equal factors, we find that 5 x 5 is left
∴ In order to get a perfect cube 5 x 5 = 25 is to be divided.
∴ Least number to be divide = 25
(iv) 8788 = 2 x 2 x 13 x 13 x 13

Grouping the factors in triplets of equal factors, we find that 2 x 2 has been left
∴ In order to get a perfect cube, 2 x 2 is to be divided
∴ Least number to be divided = 4
(v) 7803 = 3 x 3 x 3 x 17 x 17

Grouping the factors in triplets of equal factors, we see that 17 x 17 has been left.
So, in order to get a perfect cube, 17 x 17 is be divided
∴ Least number to be divided = 17 x 17 = 289
(vi) 107811 = 3 x 3 x 3 x 3 x 11 x 11 x 11

Grouping the factors in triplets of equal factors, 3 is left
∴ In order to get a perfect cube, 3 is to be divided
∴ Least number to be divided = 3
(vii) 35721 = 3 x 3 x 3 x 3 x 3 x 3 x 7 x 7

Grouping the factors in triplets of equal factors, we see that 7 x 7 is left
So, in order to get a perfect cube, 7 x 7 = 49 is to be divided
∴ Least number to be divided = 49
(viii) 243000 = 2 x 2 x 2 x 3 x 3 x 3 x 3 x 3 x 5 x 5 x 5

Grouping the factors in triplets of equal factors, 3 x 3 is left
∴ By dividing 3 x 3, we get a perfect cube
∴ Least number to be divided = 3 x 3=9

Question 13.
Prove that if a number is trebled then its cube is 27 times the cube of the given number.
Solution:
Let x be the number, then trebled number of x = 3x
Cubing, we get:
(3x)³ = (3)³ x³ = 27x³
27x³ is 27 times the cube of x i.e., of x³

Question 14.
What happenes to the cube of a number if the number is multiplied by
(i) 3 ?
(ii) 4 ?
(iii) 5 ?
Solution:
number (x)³ = x³
(i) If x is multiplied by 3, then the cube of
∴ (3x)³ = (3)³ x x³ = 27x³
∴ The cube of the resulting number is 27 times of cube of the given number
(ii) If x is multiplied by 4, then the cube of
(4x)³ = (4)³ x x³ = 64x³
∴ The cube of the resulting number is 64 times of the cube of the given number
(ii) If x is multiplied by 5, then the cube of
(5x)³ = (5)³ x x³ = 125x³
∴ The cube of the resulting number is 125 times of the cube of the given number

Question 15.
Find the volume of a cube, one face of which has an area of 64 m².
Solution:
Area of one face of a cube = 64 m²
∴ Side (edge) of cube = √64
= √64 = 8 m
∴ Volume of the cube = (side)³ = (8 m)³ = 512 m³

Question 16.
Find the volume of a cube whose surface area is 384 m².
Solution:
Surface area of a cube = 384 m² Let side = a
Then 6a² = 384 ⇒ a² = \(\frac {384 }{ 6 }\)= 64 = (8)²
∴ a = 8 m
Now volume = a³ = (8)³ m³ = 512 m³

Question 17.
Evaluate the following :

Solution:

Question 18.
Write the units digit of the cube of each of the following numbers :
31,109,388,833,4276,5922,77774,44447, 125125125.
Solution:
We know that if unit digit of a number n is
= 1, then units digit of its cube = 1
= 2, then units digit of its cube = 8
= 3, then units digit of its cube = 7
= 4, then units digit of its cube = 4
= 5, then units digit of its cube = 5
= 6, then units digit of its cube = 6
= 7, then the units digit of its cube = 3
= 8, then units digit of its cube = 2
= 9, then units digit of its cube = 9
= 0, then units digit of its cube = 0
Now units digit of the cube of 31 = 1
Units digit of the cube of 109 = 9
Units digits of the cube of 388 = 2
Units digits of the cube of 833 = 7
Units digits of the cube of 4276 = 6
Units digit of the cube of 5922 = 8
Units digit of the cube of 77774 = 4
Units digit of tl. cube of 44447 = 3
Units digit of the cube of 125125125 = 5

Question 19.
Find the cubes of the following numbers by column method :
(i) 35
(ii) 56
(iii) 72
Solution:

Question 20.
Which of the following numbers are not perfect cubes ?
(i) 64
(ii) 216
(iii) 243
(iv) 1728
Solution:
(i) 64 = 2 x 2 x 2 X 2 x 2 x 2

Grouping the factors in triplets, of equal factors, we see that no factor is left
∴ 64 is a perfect cube.
(ii) 216 = 2 x 2 x 2 x 3 x 3 x 3

Grouping the factors in triplets, of equal factors, we see that no factor is left
∴ 216 is a perfect cube.
(iii) 243 = 3 x 3 x 3 x 3 x 3

Grouping the factors in triplets, of equal factors, we see that 3 x 3 are left
∴ 243 is not a perfect cube.
(iv) 1728 = 2 x 2 x 2 x 2 x 2 x 2 x 3 x 3 x 3

Grouping the factors m triplets, of equal factors, we see that no factor is left.
∴ 1728 is a perfect cube.

Question 21.
For each of the non-perfect cubes, in Q. 20, find the smallest number by which it must be
(a) multiplied so that the product is a perfect cube.
(b) divided so that the quotient is a perfect cube.
Solution:
In qustion 20, 243 is not a perfect cube and 243 = 3 x 3 x 3 x 3 x 3
Grouping the factors in triplets, of equal factors, we see that 3 x 3 is left.
(a) In order to make it a perfect cube, 3 is to be multiplied which makes a triplet.
(b) In order to make it a perfect cube, 3 x 3 or 9 is to be divided.

Question 22.
By taking three different values of n verify the truth of the following statements :
(i) If n is even, then n³ is also even.
(ii) If n is odd, then n³ is also odd.
(iii) If n leaves remainder 1 when divided by 3, then it3 also leaves 1 as remainder when divided by 3.
(iv) If a natural number n is of the form 3p + 2 then n³ also a number of the same type.
Solution:
(i) n is even number.
Let n = 2, 4, 6 then
(a) n³ = (2)³ = 2 x 2 x 2 = 8, which is an even number.
(b) (n)³= (4)³ = 4 x 4 x 4 = 64, which is an even number.
(c) (n)³ = (6)³ = 6 x 6 x 6 = 216, which is an even number.

(ii) n is odd number.
Letx = 3, 5, 7
(a) (n)³ = (3)³ = 3 x 3 x 3 = 27, which is an odd number.
(b) (n)³ = (5)³ = 5 x 5 x 5 = 125, which is an odd number.
(c) (n)³ = (7)³ = 7 x 7 x 7 = 343, which is an odd number.

(iii) If n leaves remainder 1 when divided by 3, then n³ is also leaves 1 as remainder,
Let n = 4, 7, 10 If n = 4,
then «³ = (4)³ = 4 x 4 x 4 = 64
= 64 + 3 = 21, remainder = 1
If n = 7, then
n³ = (7)³ = 7 x 7 x 7 = 343
343 + 3 = 114, remainder = 1
If n – 10, then
(n)³ = (10)³ = 10 x 10 x 10 = 1000
1000 + 3 = 333, remainder = 1

(iv) If the natural number is of the form 3p + 2, then n³ is also of the same type
Let p =’1, 2, 3, then
(a) If p = 1, then
n = 3p + 2 = 3 x 1+2=3+2=5
∴ n³ = (5)³ = 5 x 5 x 5 = 125
125 = 3 x 41 + 2 = 3p +2

(b) If p = 2, then
n = 3p + 2 = 3 x 2 + 2 = 6 + 2 = 8
∴ n³ = (8)³ = 8 x 8 x 8 = 512
∴ 512 = 3 x 170 + 2 = 3p + 2

(c) If p = 3, then
n = 3p + 2 = 3 x 3 + 2 = 9 + 2 = 11
∴ n³ = (11)³ = 11 x 11 x 11 = 1331
and 1331 =3 x 443 + 2 = 3p + 2
Hence proved.

Question 23.
Write true (T) or false (F) for the following statements :
(i) 0 392 is a perfect cube.
(ii) 8640 is not a perfect cube.
(iii) No cube can end with exactly two zeros.
(iv) There is no perfect cube which ends in 4.
(v) For an integer a, a³ is always greater than a² > b²
(vi) If a and b are integers such that a² > b² , then a³ > b³.
(x) If a² ends in an even number of zeros, then a ends in an odd number of zeros.
Solution:

∵ 5 is left.
(iii)True : A number ending three zeros can be a perfect cube.
(iv) False : v (4)³ = 4 x 4 x 4 = 64, which ends with 4.
(v) False : If n is a proper fraction, it is not possible.
(vi) False : It is not true if a and b are proper fraction.
(vii) True.
(viii) False : as a² ends in 9, then a³ does not necessarily ends in 7. It ends in 3 also.
(ix) False : it is not necessarily that a³ ends in 25 it can end also in 75.
(x) False : If a² ends with even zeros, then a³ will ends with odd zeros but of multiple of 3.

Hope given RD Sharma Class 8 Solutions Chapter 4 Cubes and Cube Roots Ex 4.1 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.

RD Sharma Class 9 Solutions Chapter 9 Triangle and its Angles VSAQS

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 VSAQS

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.
How many least number of distinct points determine a unique line?
Solution:
At least two distinct points determine a unique line.

Question 2.
How many lines can be drawn through both of the given points?
Solution:
Through two given points, one line can be drawn.

Question 3.
How many lines can be drawn through a given point?
Solution:
Through a given point, infinitely many lines can be drawn.

Question 4.
in how many points two distinct lines can intersect?
Solution:
Two distinct lines can intersect at the most one point.

Question 5.
In how many points a line, not in a plane, can intersect the plane?
Solution:
A line not in a plane, can intersect the plane at one point.

Question 6.
In how many points two distinct planes can intersect?
Solution:
Two distinct planes can intersect each other at infinite number of points.

Question 7.
In how many lines two distinct planes can intersect?
Solution:
Two distinct planes intersect each other in one line.

Question 8.
How many least number of distinct points determine a unique plane?
Solution:
Three non-collinear points can determine a unique plane.

Question 9.
Given three distinct points in a plane, how many lines can be drawn by joining them?
Solution:
Through three given points, one line can be drawn of they are collinear and three if they are non-collinear.

Question 10.
How many planes can be made to pass through a line and a point not on the line?
Solution:
Only one plane can be made to pass through a line and a point not on the line.

Question 11.
How many planes can be made to pass through two points?
Solution:
Infinite number of planes can be made to pass through two points.

Question 12.
How many planes can be made to pass through three distinct points?
Solution:
Infinite number of planes can be made of they are collinear and only one plane, if they are non-collinear.

Hope given RD Sharma Class 9 Solutions Chapter 9 Triangle and its Angles 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 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.

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