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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 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.

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 7.3

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 7.3

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.
Draw the graph of each of the following linear equations in two variables.

Solution:
(i)x + y = 4
x = 4 – y
If y = 0, then x = 4
If y = 4, then x = 0
Now plot the points (4, 0) and (0, 4) on the graph and join them ro get the graph of the given equation

(ii)x – y = 2
x = 2 +y
If y = 0, then x = 2 and if y = 1,
Then x = 2 + 1 = 3

Now plot the points (2, 0) and (3, 1) on the graph and join them to get the graph of the equation.

(iii) -x+y = 6 ⇒  y = 6+x

If x = 0, then y = 6 + 0 = 6
If x = -1, then y = 6 – 1 = 5

Now plot the points (0, 6) and (-1, 5) on the graph and join them to get a graph of the line.
(iv) y = 2x
If x = 0, then y =  2 x 0 = 0
If x = 1, then y = 2 x 1 = 2


Now plot the points (0, 0) and (1, 2) on the graph and join them to get the graph of the line.


Now plot the points (5, 0) and (0, 3) on the graph and join them to get the graph of the line.

Now plot the points (4, 0) and (2, -3) on the graph and join them to get the graph of the line.


Now plot the points (-1, 2) and (2, 3) on the graph and join then to get the graph of the line.


Now plot the points (1, 0) and (-1, 1) on the graph and join then to get the graph of the line.

Question 2.
Give the equations of two lines passing through (3, 12). How many more such lines are there, and why ?
Solution:
∵  Points (3, 12) lies on the lines passing through the points
∴ Solutions is x = 3,y- 12
∴  Possible equation can be
x + y = 15
-x+y = 9
4x-y = 0
3x – y + 3 = 0

Question 3.
A three-wheeler scooter charges ₹15 for first kilometer and ₹8 each for every subsequent kilometer. For a distance of x km, an amount of ₹y is paid. Write the linear equation representing the above information.
Solution:
Charges for the first kilometer = ₹15
Charges for next 1 km = ₹8
Distance = x km
and total amount = ₹y
∴ Linear equation will be,
15 + (x- 1) x 8 =y
⇒  15 + 8x – 8 = y
⇒   7 + 8x = y
∴  y = 8x + 7

Question 4.
Plot the points (3, 5) and (-1, 3) on a graph paper and verify that the straight line passing through these points also passes through the point (1, 4).
Solution:

Points (3, 5) and (-1, 3) have been plotted on the graph and joined to get a line. We see that die point (1,4) also lies out.

Question 5.
From the choices given below, choose the equation whose graph is given in figure.
(i) y = x                   
(ii) x + y = 0
(iii) y = 2x                   
(iv) 2 + 3y = 7x
Solution:

From the graph, we see that Points (-1, 1) and (1, -1) be on the graph of the line these will satisfy the equation of the line
∴  -x = y ⇒ x+ y = 0
i.e, required equation
∵ x + y = 0 is the graph of the equation.

Question 6.
From the choices given below, choose the equation whose graph is given in figure.
(i) y = x + 2              
(ii) y = x – 2
(ii) y = -x + 2           
(iv) x + 2y = 6
Solution:
From the graph
Points (-1,3) and (2, 0) lie on the graph of the line
Now there points, by observation, satisfy the equation y= -x+2
∴ Required equation is y = -x + 2 whose graph is given.

Question 7.
If the point (2, -2) lies on the graph of the linear equation 5x + ky =4, find the value of k.
Solution:
∵  Point (2, -2) lies on the graph of the linear equation 5x + ky = 4
∴  It will satisfy it
∴ Now substituting the values of x and y 5 x 2 + k (-2) = 4
⇒ 10 – 2k = 4 ⇒  -2k = 4 – 10 = -6 -6
⇒ k= \(\frac { -6 }{ -2 }\) =3
Hence k = 3

Question 8.
Draw the graph of the equation 2x + 3p = 12. From the graph find the co-ordinates of the point.
(i) whose y -coordinates is 3
(ii) whose x-coordinates is -3
Solution:


Plot the points (6, 0) and (0, 4) on the graph and join them to get the graph if the line.
(i) If y = 3, then draw perpendicular from y = 3 to the line, which get meets it at P then x-coordinate of p will be
∴ coordinates of P are ( \(\frac { 3 }{ 2 }\) ,3)
(ii) If x = -3, draw perpendicular from x = -3 to the line, which meets it Q.
The y coordinates of Q will be y = 6
∴ co-ordinates of Q are (-3, 6)

Question 9.
Draw the graph of each of the equations given below. Also, find the coordinates of the points where the graph cuts the coordinates axes:
(i) 6x – 3y = 12        
(ii) -x + 4y = 8
(iii) 2x + y = 6          
(iv) 3x + 2y + 6 = 0
Solution:


Now plot the points of each equation and join then we get four lines as shown on the graphs.
Equation (i) cuts the axes at (2, 0) and (0, -4)
Equation (ii) cuts the axes at (-8, 0) and (0, 2)
Equation (iii) cuts the axes at (3, 0), (0, 6) and
Equation (iv), cuts the axes at (-2, 0) and (0,-3)

Question 10.
A lending library has a fixed charges for the first three days and an additional charge for each day thereafter. Aarushi paid ₹27 for a book kept for seven days. If fixed charges are ₹x and per day charges are ₹y. Write the linear equation representing the above information.
Solution:
Let fixed charges for first 3 days = ₹x
and additional charges for each day = ₹y
Total period = 7 days
and amount charges = ₹27
∴ x + (7 – 3) x y  = 27
⇒  x + 4y = 27
Hence x + 4y = 27

Question 11.
A number is 27 more than the number obtained by reversing its digits. If its unit’s and ten’s digit are x and y respectively, write the linear equation representing the above statement.
Solution:
Let unit’s digit = x
and tens digit = y
∴  Number = x + 10y
By reversing the digits, units digit = y
and ten’s digit = x
∴  number = y + 10x
Now difference of these two numbers = 27 (x + 10y) – (y +10x) = 27
x + 10y – y – 10x = 27
⇒  -9x + 9y – 27 = 0
⇒ x-_y + 3 = 0                   (Dividing by -9)
Hence equation is x – y + 3 = 0

Question 12.
The sum of a two digit number and the number obtained by reversing the order of its digits is 121. If units and ten’s digit of the number are x and y respectively, then write the linear equation representing the above statement.
Solution:
Let unit digit = x
and tens digit = y
∴ Number = x + 10y
By reversing the digits,
units digit = y
and tens digit = x
∴ Number =y+ 10x
Now sum of these two numbers = 121
∴ x + 10y + y + 10x = 121
⇒  1 lx + 11y = 121
⇒  x + y = 11                        (Dividing by 11)
∴  x + y = 11

Question 13.
Draw the graph of the equation 2x + y = 6. Shade the region bounded by the graph and the coordinate axes. Also find the area of the shaded region.
Solution:
2x + y = 6
⇒  y = 6 – 2x
If x = 0, then y = 6- 2 x 0 = 6 – 0 = 6
If x = 2, then y = 6- 2 x 2 = 6- 4 = 2

Now plot the points (0, 6) and (2, 2) on the graph and join them to get a line which intersects x-axis at (3, 0) and y-axis at (0,6)
Now co-ordinates if vertices of the shaded portion are (6, 0) (0, 0) and (3, 0) Now area of the shaded region.

Question 14.
Draw the graph of the equation  \(\frac { x }{ 3 }\) +  \(\frac { y }{ 4 }\)  = 1 Also find the area of the triangle formed by the line and the co-ordinate axes.
Solution:

Now plot the points (3, 0) and (0, 4) and join them to get a line which interest x-axis at A (3, 0) and y-axis at B (0, 4)

Question 15.
Draw the graph of y = | x |
Solution:
y = | x |

⇒   y = x       [∵ | x |=x]
∴  Now taking z points.


Now plot the points (1, 1) (2, 2) and (3, 3) and join them to get a graph of a line.

Question 16.
Draw the graph of y = | x | + 2
Solution:
y – | x | + 2
⇒  y = x + 2         [| x | = x]
If x = 0, then y = 0 + 2 = 2
If x = 1, then y = 1+2 = 3
If x = 2, then y = 2 + 2 = 4

Now plot the points (0, 2), (1, 3) and (2, 4) on the graph and join them to get a line.

Question 17.
Draw the graphs of the following linear equation on the same graph paper.
2x + 3y = 12, x -y = 1
Find the co-ordinates of the vertices of the triangle formed by the two straight lines and the y-axis. Also find the area of the triangle.
Solution:

Now plot the points (6, 0) (0, 4) on the graph to get a line.

Now plot the points (1, 0) and (2, 1) on the graph to get another line.
Area of the triangle FEB so formed,
= \(\frac { 1 }{ 2 }\) FB x FL = \(\frac { 1 }{ 2 }\) x 5 x 3
= \(\frac { 15 }{ 2 }\)
= 7.5 sq. units
co-ordinates of E, F, B are E (3, 2), (0, -1) and (0, 4)

Question 18.
 Draw the graphs of the linear equations 4x – 3y + 4 = 0 and 4x + 3y – 20 = 0. Find the area bounded by these lines and x-axis.
Solution:



Now plot the points (5, 0) and (2, 4)and join them to get a line we see that the ΔABC is formed by bounding there line with x-axis.

Question 19.
The path of a train A is given by the equation 3x + 4y – 12 = 0 and the path of another train B is given by the equation 6jc + 8y – 48 = 0. Represent this situation graphically.
Solution:
Path of the train A = 3x + 4y – 12 = 0
Path of the train B = 6x + 8y – 48 = 0
Now, 3x + 4y – 12 = 0


Now plot the points (4, 0) and (0, 3) on the graph and join them to get a line, and 6x + 8y – 48 = 0
⇒  6x = 48 – 8y

Now plot the points (0, 6) and (4, 3) on the graph and join them to get another line.

Question 20.
Ravish tells his daughter Aarushi, “Seven years ago, I was seven times as old as you were then. Also, three years from now, I shall be three times as old as you will be”. If present ages of Aarushi and Ravish are x and y years respectively, represent this situation algebraically as well as graphically.
Solution:
Present age of Aarushi = x years
and age of Ravish = y years
7 years ago,
age of Aarushi = x – 7
years and age of Ravish =y-7 years
∴ y- 7 = 7 (X – 7)
⇒  y – 7 = 7x – 49
⇒  7x – y = -7 + 49  = 42
7x – y = 42
⇒  y = 7x – 42
If x = 6, then
y = 7 x 6 – 42 = 42 – 42 = 0,
If x = 7, then
y = 7 x 7 – 42 – 49 – 42 = -7

Plot the points (6, 0) (7, -7) on the graph and join them.
After 3 years,
age of Aarushi = x + 3
and age of Ravish = y + 3
⇒  y + 3 = 3(x + 3)
⇒ y + 3 = 3x + 9
⇒ y = 3x+ 9-3
⇒ y = 3x + 6
If x = -2, then y = 3 x (-2) + 6 =6-6=0
If x = 1, then y = 3 x (1) + 6 =3+6=9

Plot the points (1, 9), (-2, 0) on the graph Arundeep’s Mathematics (R.D.) 9th and join them to get another line.

Question 21.
Aarushi was driving a car with uniform speed of 60 km/h. Draw distance-time graph. From the graph, find the distance travelled by Aarushi in.
(i) 2\(\frac { 1 }{ 2 }\) Hours             
(ii) \(\frac { 1 }{ 2 }\) Hour
Solution:
Speed of car = 60 km / h.

Now plot the points (60, 1), (120, 2) are the graph and join then to get the graph of line.
From the graph, we see that

Hope given RD Sharma Class 9 Solutions Chapter 7 Introduction to Euclid’s Geometry Ex 7.3 are helpful to complete your math homework.

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RD Sharma Class 8 Solutions Chapter 3 Squares and Square Roots Ex 3.8

These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 3 Squares and Square Roots Ex 3.8

Other Exercises

• RD Sharma Class 8 Solutions Chapter 3 Squares and Square Roots Ex 3.1
• RD Sharma Class 8 Solutions Chapter 3 Squares and Square Roots Ex 3.2
• RD Sharma Class 8 Solutions Chapter 3 Squares and Square Roots Ex 3.3
• RD Sharma Class 8 Solutions Chapter 3 Squares and Square Roots Ex 3.4
• RD Sharma Class 8 Solutions Chapter 3 Squares and Square Roots Ex 3.5
• RD Sharma Class 8 Solutions Chapter 3 Squares and Square Roots Ex 3.6
• RD Sharma Class 8 Solutions Chapter 3 Squares and Square Roots Ex 3.7
• RD Sharma Class 8 Solutions Chapter 3 Squares and Square Roots Ex 3.8
• RD Sharma Class 8 Solutions Chapter 3 Squares and Square Roots Ex 3.9

Question 1.
Find the square root of each of the following correct to three places of decimal :
(i) 5
(ii) 7
(iii) 17
(iv) 20
(v) 66
(vi) 427
(vii) 1.7
(vii’) 23.1
(ix) 2.5
(x) 237.615
(xi) 15.3215
(xii) 0.9
(xiii) 0.1
(xiv) 0.016
(xv) 0.00064
(xvi) 0.019
(xvii) \(\frac {7 }{ 8 }\)
(xviii) \(\frac {5 }{ 12 }\)
(xix) 2 \(\frac {1 }{ 2 }\)
(xx) 287 \(\frac {8 }{ 8 }\)
Solution:

Question 2.
Find the square root of 12.0068 correct to four decimal places.
Solution:

Question 3.
Find the square root of 11 correct to five decimal places.
Solution:

Question 4.
Given that √2, = 1-414, √3 = 1.732, √5 = 2.236 and √7 = 2.646. Evaluate each of the following :

Solution:

Question 5.
Given that √2 = 1-414, √3 = 1-732, √5= 2.236 and √7= 2.646, find the square roots of the following :

Solution:

Hope given RD Sharma Class 8 Solutions Chapter 3 Squares and Square Roots Ex 3.8 are helpful to complete your math homework.

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