Chapter 3: Squares and Square Roots & Cubes and Cube Roots

Solution:-

First we have to find out the factors by using prime factorisation method.

So, prime factors of 484 = 2 × 2 × 11 × 11

Now, grouping the prime factors = (2 × 2) × (11 × 11)

= 2² × 11²

There is no unpaired factor remains here.

∴484 is a perfect square.

(b) 11250

Solution:-

First we have to find out the factors by using prime factorisation method.

So, prime factors of 11250 = 2 × 3 × 3 × 5 × 5 × 5 × 5

Now, grouping the prime factors = 2 × (3 × 3) × (5 × 5) × (5 × 5)

= 2 × 3² × 5² × 5²

Factor 2 has no pair.

∴11250 is not a perfect square.

(c) 841

Solution:-

First we have to find out the factors by using prime factorisation method.

So, prime factors of 841 = 29 × 29

Now, grouping the prime factors = (29 × 29)

= 29²

There is no unpaired factor remains here.

∴841 is a perfect square.

(d) 729

Solution:-

First we have to find out the factors by using prime factorisation method.

So, prime factors of 729 = 3 × 3 × 3 × 3 × 3 × 3

Now, grouping the prime factors = (3 × 3) × (3 × 3) × (3 × 3)

= 3² × 3² × 3²

There is no unpaired factor remains here.

∴729 is a perfect square.

Solution:-

First we have to find out the factors by using prime factorisation method.

So, prime factors of 128 = 2 × 2 × 2 × 2 × 2 × 2 × 2

Now, grouping the prime factors = (2 × 2 × 2) × (2 × 2 × 2) × 2

= 2³ × 2³ × 2

Factor 2 remains unpaired after grouping in triplets.

∴128 is not a perfect cube.

(b) 343

Solution:-

First we have to find out the factors by using prime factorisation method.

So, prime factors of 343 = 7 × 7 × 7

Now, grouping the prime factors = (7 × 7 × 7)

= 7³

There is no unpaired factor remains here.

∴343 is a perfect cube.

(c) 729

Solution:-

First we have to find out the factors by using prime factorisation method.

So, prime factors of 729 = 3 × 3 × 3 × 3 × 3 × 3

Now, grouping the prime factors = (3 × 3 × 3) × (3 × 3 × 3)

= 3³ × 3³

There is no unpaired factor remains here.

∴729 is a perfect cube.

(d) 1331

Solution:-

First we have to find out the factors by using prime factorisation method.

So, prime factors of 1331 = 11 × 11 × 11

Now, grouping the prime factors = (11 × 11 × 11)

= 11³

There is no unpaired factor remains here.

∴1331 is a perfect cube.

Solution:-

By using distributive law,

We have,

101 = 100 + 1

So,

101² = (100 + 1)²

= (100 + 1) (100 + 1)

= 100 (100 + 1) + 1 (100 + 1)

= ((100 × 100) + (100 × 1)) + ((1 × 100) + (1 × 1))

= 10000 + 100 + 100 + 1

= 10201

∴The square of the given number i.e. 101² = 10201

(b) 72

Solution:-

By using distributive law,

We have,

72 = 70 + 2

So,

72² = (70 + 2)²

= (70 + 2) (70 + 2)

= 70 (70 + 2) + 2 (70 + 2)

= ((70 × 70) + (70 × 2)) + ((2 × 70) + (2 × 2))

= 4900 + 140 + 140 + 4

= 5184

∴The square of the given number i.e. 72² = 5184

Solution:-

From the question it is given that,

Sides of the right angle triangle are 6 cm, 10 cm and 8 cm

We know that, the square one side must be equal to the sum of square of other two sides of right angle triangle.

i.e. a² = b² + c²

Where, a = 10 cm, b = 8 cm, c = 6 cm

10² = 8² + 6²

100 = 64 + 36

100 = 100

∴Right angle triangle is formed by the given sides 6cm, 10cm and 8cm.

Solution:-

Given, one of the number in the Pythagorean triplet i.e. 4.

Then,

2m = 4

m = 4/2

m = 2

So,

= m² + 1

= 2² + 1

= 4 + 1

= 5

And,

= m² – 1

= 2² – 1

= 4 – 1

= 3

∴The Pythagorean triplet is 3, 4 and 5.

Solution:-

First we have to find out the factors by using prime factorisation method.

So, prime factors of 11025 = 3 × 3 × 5 × 5 × 7 × 7

Now, grouping the prime factors = (3 × 3) × (5 × 5) × (7 × 7)

Then, Square root of 11025 = √11025

= √((3 × 3) × (5 × 5) × (7 × 7))

= √(3² × 5² × 7²)

= 3 × 5 × 7

= 105

∴The Square root of 11025 is 105.

(b) 4761

Solution:-

First we have to find out the factors by using prime factorisation method.

So, prime factors of 4761 = 3 × 3 × 23 × 23

Now, grouping the prime factors = (3 × 3) × (23 × 23)

Then, Square root of 4761 = √4761

= √((3 × 3) × (23 × 23))

= √(3² × 23²)

= 3 × 23

= 69

∴The Square root of 4761 is 69.

Solution:-

First we have to find out the factors by using prime factorisation method.

So, prime factors of 512 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2

Now, grouping the prime factors = (2 × 2 × 2) × (2 × 2 × 2) × (2 × 2 × 2)

Then, cube root of 512 = ³√512

= ³√((2 × 2 × 2) × (2 × 2 × 2) × (2 × 2 × 2))

= ³√(2³ × 2³ × 2³)

= 2 × 2 × 2

= 8

∴The cube root of 512 is 8.

(b) 2197

Solution:-

First we have to find out the factors by using prime factorisation method.

So, prime factors of 2197 = 13 × 13 × 13

Now, grouping the prime factors = (13 × 13 × 13)

Then, cube root of 2197 = ³√2197

= ³√(13 × 13 × 13)

= ³√(13³)

= 13

∴The cube root of 2197 is 13.

Solution:-

First we have to find out the factors by using prime factorisation method.

So, prime factors of 176 = 2 × 2 × 2 × 2 × 11

Now, grouping the prime factors = (2 × 2) × (2 × 2) × (11)

= 2² × 2² × 11

Factor 11 has no pair.

∴176 is not a perfect square.

The smallest number it should be multiplied to get a perfect square is 11.

Then,

176 × 11 = 1936

Factors of 1936 = 2 × 2 × 2 × 2 × 11 × 11

= √1936

= √(2² × 2² × 11²)

= 2 × 2 × 11

= 44

∴Square root of 1936 is 44.

Solution:-

First we have to find out the factors by using prime factorisation method.

So, prime factors of 9720 = 2 × 2 × 2 × 3 × 3 × 3 × 3 × 3 × 5

Now, grouping the prime factors = (2 × 2 × 2) × (3 × 3 × 3) × 3 × 3 × 5

= 2³ × 3³ × 3 × 3 × 5

Factors 3 and 4 has no pair.

∴9720 is not a perfect cube.

The smallest number it should be divided to get a perfect cube is 3 × 3 × 5 = 45.

Then,

= 9720 ÷ 45

= 216

Factors of 216 = (2 × 2 × 2) × (3 × 3 × 3)

³√216 = ³√((2 × 2 × 2) × (3 × 3 × 3))

= ³√(2³ × 3²)

= 6

Solution:-

Given, one of the number in the Pythagorean triplet i.e. 5.

= m² + 1 = 5

= m² = 4

= m = 2

Then,

2m = 2 × 2

m = 4

So,

And,

= m² – 1

= 2² – 1

= 4 – 1

= 3

∴The Pythagorean triplet is 3, 4 and 5.

Similarly,

The other Pythagorean triplet is 5, 12 and 13.