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

Solution:-

Factors of 216 = 2 × 2 × 2 × 3 × 3 × 3

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

Factors 2 and 3 has no pair.

∴216 is not a perfect square.

The smallest number it should be divided to get a perfect square is 3 × 2 = 6.

Then,

= 216 ÷ 6

= 36

Factors of 36 = 6 × 6

√36 = √(6× 6)

= √(6)

= 6

∴36 is a perfect square and 6 is the square root of 36.

Solution:-

Factors of 3600 = 2 × 2 × 2 × 2 × 3 × 3 × 5 × 5

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

Factors 2, 3 and 5 has no pair.

∴3600 is not a perfect cube.

The smallest number it should be multiplied to get a perfect cube is 2 × 2 × 3 × 5 = 60

Then,

Product = 3600 × 60 = 216000

Factors of 21600 = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3 × 5 × 5 × 5

= ³√216000

= ³√(2³ × 2³ × 3³ × 5³)

= 2 × 2 × 3 × 5

= 60

Solution:-

We have,

∴ √1369 = 37

(b) 5625

Solution:-

We have,

∴ √5625 = 75

Solution:-

We have,

∴ √27.04 = 5.2

(b) 1.44

Solution:-

We have,

∴ √1.44 = 1.2

Solution:-

From the long division method, the least number that should be subtracted from 1385 to get a perfect square is 16.

Now,

= 1385 – 16

= 1369

So,

√1369 = √(37 × 37)

= √37²

= 37

∴The square root of a perfect square number 1369 is 37.

Solution:-

So, 78² = 6084

By comparing 6084 and 6200, 6084 < 6200

Then, next perfect square is 79² = 6241

∴The least number is (6241 – 6200) = 41, which should be added to 6200 to get a perfect square.

The required perfect square number is 6241

So,

√6241 = √(41 × 41)

= √41²

= 41

Solution:-

We know that, the least number of four digit is 1000

By using long division method we can able to find a perfect square,

∴The least number of four digits that is a perfect square = 1000 + 24

= 1024

Solution:-

Solution:-

We know that, the greatest number of three digit is 999.

By using long division method we can able to find a perfect square,

∴The greatest number of three digits that is a perfect square = 999 – 38

= 961

Solution:-

First we have to find out the LCM of 3, 4, 5, 6 and 8

So, LCM of 3, 4, 5, 6 and 8 = 2 × 2 × 2 × 3 × 5

= 120

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

i.e. 2,3 and 5 is not able to make their pair.

So, to make it perfect square, 120 must be multiplied with 2× 3× 5 = 30

Then,

= 120× 30 = 3600

∴ 3600 is least square number which is exactly divisible by 3, 4, 5, 6, and 8.

Solution:-

From the question it is given that,

Length of Diagonal = 10 cm

Let us assume square ABCD.

Sides AB = y, BC = y, CD = y, DA = y

Diagonal AC = 10 cm

Now, consider triangle ABC

By the rule of Pythagoras theorem,

AC² = AB² + BC²

10² = y² + y²

100 = 2y²

y² = 100/2

y² = 50

y = √50 cm

y = 5√2 cm

∴The length of the side of square is √50 cm or 5√2.