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

Solution:-

(b) 64

= 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15

= 64

Solution:-

(b) n²

The sum of first n odd natural numbers is given by n².

Solution:-

(b) 216

A perfect cube can always be expressed as the product of triplets of prime factors.

216 = 6 × 6 × 6 = 6³

Solution:-

We know that,

Pythagoras theorem = a = √(b² + c²)

Where ‘a’ is hypotenuse and b and c are the other two legs of triangle.

Then,

a = √ ((3x)² + (4x)²)

a = √(9x² + 16x²)

a = √(9x² + 16x²)

a = √(9x + 16x)

a = √25x

a = 5x

Hence the hypotenuse of the right triangle is 5x.

Solution:-

(b) 36, 49

The numbers given in the question are perfect square numbers.

1² = 1, 2² = 4, 3² = 9, 4² = 16, 5² = 25

So, 6 × 6 = 36 and 7 × 7 = 49

Solution:-

(d) 59²

Because, the square of a number having 1 or 9 at the units place ends in1.

Solution:-

(b) 8

Solution:-

(b) 567

Because, factors of 216 = 6 × 6 × 6

567 = 9 × 9 × 7

125 = 5 × 5 × 5

343 = 7 × 7 × 7

Solution:-

(a) 10

= ³√1000

= ³√(10 × 10 × 10)

= ³√(10³)

= 10

Solution:-

(d) √m

n² = m

Then,

= n = √m