Square-square Root And Cube-cube Root

14× 14 = 196

12× 12 = 144

13× 13 = 169

21× 21 = 441

25× 25 = 625

If a number ends with 1 or 9 its square ends with 1.

For example- 9× 9 = 81

If a number ends with 2 or 8 its square ends with 4.

For example- 12× 12 = 144

There are 2n non-perfect square numbers between the squares of the number n and (n + 1).

Here n = 5

⇒ Natural numbers lying between 5² and 6² = 2×5 = 10

No perfect square has 2, 3, 7 and 8 at its unit place.

The ones digit of any number with 3 at its unit place is 9.

i.e. 13² = 169

Area of a square board = side × side

⇒ side = √144

⇒ side = 12 units

∵ √25 = 5

So, C represents the location of it.

∵ (2m)² + (m² – 1)² = (m² + 1)²

∵ Every square can be expressed as a sum of successive odd natural numbers from 1.

⇒ The sum of first n odd natural numbers is n²

Here, n = 8 ⇒ 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 = 8× 8 = 64

∵ Every square can be expressed as a sum of successive odd natural numbers from 1.

⇒ The sum of first n odd natural numbers is n²

9× 9× 3 = 243

6× 6× 6 = 216

8× 7× 7 = 392

8× 8× 5× 3× 3× 3 = 8640

Using Pythagoras triplets: (2m)² + (m² – 1)² = (m² + 1)²

2m = 4x

⇒ m = 2

m² – 1 = (4-1)x = 3x

⇒ m² + 1 = (4 + 1)x = 5x

1× 1 = 1

2× 2 = 4

3× 3 = 9

4× 4 = 16

5× 5 = 25

6× 6 = 36

7× 7 = 49

If a number ends with 1 or 9 its square ends with 1.

For example- 9× 9 = 81

No perfect square has 2, 3, 7 and 8 at its unit place.

6× 6× 6 = 216

9× 9× 7 = 567

5× 5× 5 = 125

7× 7× 7 = 343

10× 10× 10 = 1000

Given n² = m

⇒ n = √m

A perfect square number having n digits where n is even will have square root with digits.

Given

⇒ n = m³

So,

Between 1 and 100, 81 = 9² is the highest perfect square.

Excluding 1 there are 8 perfect squares between them.

Between 1 and 1000, 729 = 9³ is the highest perfect cube.

Excluding 1 there are 8 perfect cubes between them.

If a number ends with 4 or 6 its square ends with 6.

500 × 500 = 250000

There are 2n non perfect square numbers between the squares of the number n and (n + 1).

Example: n = 5

⇒ Natural numbers lying between 5² and 6² = 2×5 = 10

A perfect square number having n digits where n is odd will have square root with digits.

Here n = 5, square root will have digits

5.5× 5.5 = 55× 55× 0.1× 0.1 = 3025× 0.01 = 30.25

100 × 100× 100 = 1000000

1 m = 100 cm

1 m² = 100× 100 cm² = 10000 cm²

1 m = 100 cm

1 m³ = 100× 100× 100 cm³ = 1000000 cm³

Ones digit in the cube of 38 is 2 because for any number which lasts with 8 has 2 at its end place in cube.

For example:

8× 8× 8 = 512

0.7× 0.7 = 7× 7× 0.1× 0.1 = 49× 0.01 = 0.49

∵ Every square can be expressed as a sum of successive odd natural numbers from 1.

⇒ The sum of first n odd natural numbers is n²

Here, n = 6 ⇒ 1 + 3 + 5 + 7 + 9 + 11 = 6× 6 = 36

If a number ends with 3 or 7 its square ends with 9.

Using Pythagoras triplets: (2m)² + (m² – 1)² = (m² + 1)²

m² + 1 = 17

⇒ m² = 16

⇒ m = 4

Now, m² – 1 = (16 - 1) = 15

And 2m = 2× 4 = 8

1.2× 1.2× 1.2 = 12× 12× 12× 0.1× 0.1× 0.1 = 1728× 0.001 = 1.728

For example,

3× 3× 3 = 27

Given

125 = 5×5× 5

So, 5 is the least number by which 125 be multiplied to make it a perfect square

72 = 2× 2×2× 3× 3

So, 3 is the least number by which 72 be multiplied to make it a perfect cube

72 = 2× 2×2× 3× 3

So, 9 is the least number by which 72 be multiplied to make it a perfect cube

If a number ends with 7 its cube ends with 3.

Given: square of 86

We know that, the unit’s digit of the square of a number having digits 4 or 6 at unit’s place is 6.

Hence, the given statement is true.

Given: two perfect squares

Let us take an example:

16 and 25 both are perfect square,

But 16 + 25 = 41 which is not a perfect square.

Hence, the given statement is false.

Given: two perfect squares

Let us take an example:

16 and 25 both are perfect square,

But 16 × 25 = 400 which is a perfect square.

Same is true for values too.

Hence, the given statement is true.

Given: number between 50 and 60

i.e. 51, 52, 53, 54, 55, 56, 57, 58, 59

we observe that there is no perfect square in between these.

Hence, the given statement is true.

Given: square root of 1521

As we know,

(31)² = 961

⇒ 1521 ≠ square root of 31

Hence, the given statement is false.

Given:

If a³ is the cube and m is one of the prime factor of a. Then, m appears three times in a³.

Hence, the given statement is True.

Given: 2.8

As we know,

(2.8)² = 7.84

and 7.84 ≠ 78.4

Hence, the given statement is false.

Given: 0.4

As we know,

(0.4)³ = 0.064

and 0.064 = 0.064

Hence, the given statement is True.

Given: square root of 0.9

As we know,

(0.3)² = 0.09

⇒ 0.9 ≠ square root of 0.3

Hence, the given statement is false.

Given: natural number

As we know, 1 is a natural number.

(1)² = 1 (which is not greater than 1)

⇒ (1)² >≠ 1

Hence, the given statement is false.

Given: cube root of 8000

As we know,

(200)³ = 8000000

⇒ 8000 ≠ cube root of 200

Hence, the given statement is false.

Given: perfect cubes between 1 and 100

i.e. 8, 27, 64, 125, 216, 343, 512 and 729.

we observe that there are 8 perfect cubes in between these.

Hence, the given statement is false.

Given: natural numbers between 100² and 101²

As we know, natural numbers between a and b = b-a-1

then natural numbers between 100² and 101² = 101² – 100² – 1

= (101² – 100²)– 1

= (101 + 100)(101-100)-1

= (201)(1)-1

= 201 -1 = 200

Hence, the given statement is true.

Given: first n odd natural numbers

Odd natural number will be = 2n-1

As we know, sum of odd natural numbers =

= n² + n-n

= n²

Hence, the given statement is true.

Given: 1000

Factors of 1000 = 2 ×2 × 2 × 2 × 5 × 5 × 5

= 2³× 5³

As we see, 1000 is a perfect cube not a perfect square.

Hence, the given statement is false.

Given: perfect square

As we see, A perfect square can never have 8 at its unit’s place.

Hence, the given statement is false.

Given: natural number m (2m-1, 2m²-2m,2m²-2m + 1)

For Pythagorean triplet square of one should be equal to sum of square of other two.

But it is not true for the given values.

(2m-1)²≠ (2m²-2m)² + (2m²-2m + 1)²

(2m²-2m)²≠(2m-1)² + (2m²-2m + 1)²

(2m²-2m + 1)²≠(2m²-2m)² + (2m-1)²

Hence, the given statement is false.

Given: Pythagorean triplet

For Pythagorean triplet square of one should be equal to sum of square of other two.

Pythagorean triplet as 5² = 4² + 3²

Here, we see 4 is not an odd number.

Hence, the given statement is false.

Given: integer a

For an integer (-1)

(-1)² = 1

And (-1)³ = -1

-1 < 1

Here we see (-1)³ not greater than (-1)²

Hence, the given statement is false.

Given: integer x and y

For an integer x and y as (-1) and (-2) respectively.

x²>y²⇒ (-2)² > (-1)² = 4 > 1 (true)

and x³>y³⇒ (-2)³ > (-1)³ = -8 > -1 (not true)

as -8 < -1

Hence, the given statement is false.

Given: x and y be natural numbers.

If x divides y ⇒ is a natural number.

⇒ is also a natural number.

⇒ is also a natural number.

⇒ x³ divides y³

Hence, the given statement is true.

Given: a

Let us take an example:

(35)² = 1225 (ends in 5)

And (35)³ = 42875 (doesn’t ends in 25)

Hence, the given statement is false.

Given: a

Let us take an example:

(7)² = 49 (ends in 9)

And (7)³ = 343 (doesn’t ends in 7)

Hence, the given statement is false.

Given: square root of a perfect square of n digit.

If perfect square of 3 digits then the square root has 2 digits. i.e.

If perfect square of 5 digits then the square root has 3 digits. i.e.

Which is true.

Hence, the given statement is true.

Given: Square root of a number x.

It is denoted as

Hence, the given statement is true.

Given: 7 at unit place.

(7)² = 49

(17)² = 289

(27)² = 729

And so on.

49, 289, 729 don’t have 3 at the units place.

Hence, the given statement is false.

Given: 7 at unit place.

(7)³ = 343

(17)³ = 4913

(27)³ = 19683

And so on.

343, 4913, 19683 all have 3 at the units place.

Hence, the given statement is true.

Given: one-digit number

(3)³ = 27

Which is a two digit number.

Hence, the given statement is false.

Given: even number

Cube of an even number = (2)³ = 8

Which is not an odd number.

Hence, the given statement is false.

Given: odd number

Cube of an odd number = (3)³ = 27

Which is not an even number.

Hence, the given statement is false.

Given: even number

Cube of an even number = (2)³ = 8

Which is an even number.

Hence, the given statement is true.

Given: odd number

Cube of an odd number = (3)³ = 27

Which is an odd number.

Hence, the given statement is true.

Given: 999

Factors of 999 = 3 ×3 × 3 × 37

= 3³×37

As we see, 999 is not a perfect cube.

Hence, the given statement is false.

Given: 363×81

Factors of 363×81 = 3 ×11 × 11 × 3 × 3 ×3 × 3

= 3³ ×11 × 11 × 3 × 3

As we see, 363×81 is not a perfect cube.

Hence, the given statement is false.

Given: Cube roots of 8

Cube roots of 8 = 2 only

Hence, the given statement is false.

Given:

..(1)

..(2)

From (1) and (2)

Hence, the given statement is false.

Given: Cube roots of a negative integer.

Cube roots of -8 = -2 only

Hence, the given statement is false.

Given: a number

(-2)² = 4

(-2)³ = -8

-8 which is not a positive.

Hence, the given statement is false.

to find: first five square numbers

first five square numbers are:

(1)², (2)², (3)², (4)² and (5)² = 1, 4, 9, 16 and 25.

to find: cube of first three multiple of 3.

cube of first three multiple of 3 are:

(3)³, (6)³ and (9)³ = 27, 216 and 729.

Given: 500

Factors of 500 = 2 ×2 × 5 × 5 × 5

= 2² × 5³

As we see, 500 is not a perfect square.

Hence proved.

given: 81

As we know,

(81) = (9)² = 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 = sum of first nine consecutive odd numbers.

(a) Given: 484

Factors of 484 = 2 ×2 × 11 × 11

= 2² × 11²

As we see, 484 is a perfect square.

(b) Given: 11250

Factors of 11250 = 2 × 3 × 3 × 5× 5 × 5 × 5

= 2¹ × 3²× 5²× 5²

As we see, 2 has no pair.

So, 11250 is not a perfect square.

(c) Given: 841

Factors of 841 = 29 ×29

= 29²

As we see, 841 is a perfect square.

(d) Given: 729

Factors of 729 = 3 × 3 × 3 × 3 × 3 × 3

= 3²× 3²× 3²

So, 729 is a perfect square.

(a) Given: 128

Factors of 128 = 2 ×2 × 2×2×2×2×2

= 2³ × 2³×2

As we see, 128 is not a perfect cube.

(b) Given: 343

Factors of 343 = 7 × 7 × 7

= 7³

So, 343 is a perfect cube.

(c) Given: 729

Factors of 729 = 3 × 3 × 3 × 3 × 3 × 3

= 3³× 3³

So, 729 is a perfect cube.

(d) Given: 1331

Factors of 1331 = 11 × 11 × 11

= 11³

So, 1331 is a perfect cube.

(a) Given: 101

As we know,

(101)² is to find using distributive law:

(101)² = 101 × 101

= 101 (100 + 1)

= 10100 + 101

= 10201

(b) Given: 72

As we know,

(72)² is to find using distributive law:

(72)² = 72 × 72

= 72 (70 + 2)

= 5040 + 144

= 5184

Given: sides of triangle 6cm, 10cm and 8cm

As we know,

For right angle triangle, square of one side must equal to sum of square of other two sides.

i.e.

(10)² = (6)² + (8)²

100 = 36 + 64

⇒ 100 = 100

Hence, 6cm, 10cm and 8cm are sides of triangle.

Given:

We know that for any natural number greater than 1, (2m, m²-1, m² + 1) is a the Pythagorean triplet.

So if one number is 2m then other two number are m²-1 and m² + 1.

i.e. 2m = 4 (given)

⇒ m = 2

Then m²-1 = 2²-1 = 4-1 = 3

And m² + 1 = 2² + 1 = 4 + 1 = 5

Hence, Pythagorean triplet is 3, 4 and 5

(a) Given: 11025

Factors of 11025 = 3 × 3× 5× 5× 7× 7

Then square roots of 11025 = 3 × 5 × 7

= 105

(b) Given: 4761

Factors of 4761 = 3 × 3× 23× 23

Then square roots of 4761 = 3 × 23

= 69

(a) Given: 512

Factors of 512 = 2 × 2× 2× 2× 2× 2× 2× 2× 2

Then cube roots of 512 = 2× 2× 2

= 8

(b) Given: 2197

Factors of 2197 = 13 × 13 × 13

Then cube roots of 2197 = 13

Given: 176

Factors of 176 = 2 × 2 × 2 × 2 × 11

As we see, 11 do not have a pair.

Hence 176 should be multiply with 11 to get perfect square.

⇒ 176 × 11 = 2 × 2 × 2 × 2 × 11 × 11

⇒ 1936 = 2²× 2²× 11²

⇒ 1936 = 44²

Which is perfect square of 44.

Given: 9720

Factors of 9720 = 2 × 2 × 2 × 3 × 3 × 3 × 3 × 3 × 5

As we see, 3 and 5 do not have a pair to make a cube.

Hence 9720 should be divided with 3 × 3 × 5 to get perfect cube.

⇒ 9720 should divide with 45.

Now, after division we get 216 = 6³

Now, 216 is a perfect cube.

Given:

We know that for any natural number greater than 1, (2m, m²-1, m² + 1) is a the Pythagorean triplet.

So if one number is m² + 1 then other two number are m²-1 and 2m.

i.e. m² + 1 = 5 (given)

⇒ m² = 4

⇒ m = 2

Then m²-1 = 2²-1 = 4-1 = 3

And 2m = 2(2) = 4

Hence, Pythagorean triplet is 3, 4 and 5

Similarly another triplet is:

13² = 12² + 5²

Hence, 3,4,5 and 5,12,13 are two Pythagorean triplet.

Given: 216

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

As we see, 2 and 3 do not have pair.

Hence 216 should be divided with 2 × 3 to get perfect square.

⇒ 216 should divide with 6.

Now, after division we get 36 = 6²

Now, 36 is a perfect square and 6 is its square root.

Given: 3600

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

As we see, 2,3 and 5 do not have a pair to make a cube.

Hence 3600 should be multipled with 2 × 2 × 3 × 5 to get perfect cube.

⇒ 3600 should multiply with 60.

Now, 3600× 60 = 216000

Now, 216000 is a perfect cube and the cube root is 2 × 2 × 3 × 5 i.e. 60.

(a) Given: 1369

Hence, = 37

(b) Given: 5625

Hence, = 75

(a) Given: 27.04

Hence, = 5.2

(b) Given: 1.44

Hence, = 1.2

Given: 1385

Hence, the least number is 16, which should be subtracted from 1385 to get a perfect square and the required perfect square number will be = 1385 – 16 = 1369

Then, = 37

Given: 6200

Hence, (78)² = 6084 < 6200

Next perfect sqaure is (79)² = 6241

Hence, the least number is (6241-6200 = 41), which should be added to 6200 to get a perfect square and the required perfect square number will be 6241

Then, = 41

Given: least number of four digit.

i.e. 1000

Hence, the least number of four digits that is a perfect square = 1000 + 24 = 1024

Given: least number of three digit.

Now, if we subtract 38 from 999 we will get a perfect square

Hence, the three-digit number is 999-38 = 961

And = 31

Given:

the least square number which is exactly divisible by 3, 4, 5, 6, and 8 is equal to L.C.M. 0f 3,4,5,6 and 8.

Hence, their L.C.M. is 2× 2× 2× 3× 5 = 120

As we see, (2× 2)× 2× 3× 5

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

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

As, 120× 30 = 3600

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

Given: length of diagonal = 10cm

i.e. let side of square (AB) = xcm

length of diagonal (AD)² = (AB)² + (BD)²

(10)² = (x)² + (x)²

⇒ 100 = 2x²

⇒ x² = 50

⇒

Side of square =

Given:

Let the number is y

Acc. To condition:

y × y = 51.84

⇒ y² = 51.84

Hence, required number is 7.2

Given:

Let the number is y

Acc. To condition:

y × y = 84.64

⇒ y² = 84.64

Hence, required number is 9.2

Given: square field of side = 150m

Area of square = side × side

= 150 × 150

= 22500m²

The area farmer have to plough = 22500m²

Given:

Let the number is y.

Then the number of unit square = y × y = y²

But total number of unit squares = 256

⇒ y² = 256

Then y = = 16

Hence, the number of unit squares = 16

Given: cube of side = 15m

volume of cube = (side)³

= (15)³

= 3375m³

Hence, volume of cube = 3375m³

Given: dimensions of a rectangular field

i.e. let length (AB) = 80m

and breadth (BD) = 18m

length of diagonal (AD) =

Given: perimeter = 96 m

Let length of square = a

Then 4a = 96m

⇒ a = 24m

Area of square = side × side

= 24 × 24

= 576m²

Given: cube of volume = 512cm³

volume of cube = (side)³

512 = (x)³

⇒ x³ = 8× 8× 8 = 8³

Hence, side of cube (x) = 8cm

Given:

Let the three numbers are 1x, 2x and 3x

Acc. To given condition:

i.e (1x)³ + (2x)³ + (3x)³ = 4500

⇒ (x)³ + 8(x)³ + 27 (x)³ = 4500

⇒36(x)³ = 4500

⇒ (x)³ = 125 = 5³

⇒ x = 5

Hence, numbers are 1(5) = 5

2(5) = 10

3(5) = 15

Given:

Let length of square = 6.5m

Area of square = side × side

= 6.5 × 6.5

= 42.25m²

Given:

Length of rectangle (l) = 6.4m

Breadth of rectangle (b) = 2.5m

Area of rectangle = l× b

= 6.4 × 2.5

= 16m²

Then area of square = 16m² (equal to rectangle)

As we know,

Area of square = side × side

16 = (side)²

4² = (side)²

⇒ side = 4m

Given: Difference of two perfect cubes = 189.

The cube root of the smaller of the two numbers = 3

The cube of the smaller of the two numbers = 3³ = 27

Let the other number = x

Acc. To given condition:

i.e. x³ – 27 = 186

⇒ x³ = 186 + 27

⇒ x³ = 216

⇒ x³ = 6³

⇒ x = 6

Then, the cube root of the greater of the two numbers = 6

Given:

Let the number of plants in each row = y

number of plants in a row = the number of rows = y

Hence, total plants = y × y

Acc. To given condition:

Y² = 1024

⇒ y² = 32 × 32

⇒ y² = 32²

⇒ y = 32

Hence, the number of plants in each row = 32

Given:

Let the number of seats in each row = y

number of seats in a row = the number of rows = y

Hence, total seats = y × y

Acc. To given condition:

Y² = 2704

⇒ y² = 52 × 52

⇒ y² = 52²

⇒ y = 52

Hence, the number of seats in each row = 52

Given: 7500 soldier

He is arranging in a form of sqaue. 7500 must complete the square.

If not then the remaining will be left out soldier.

Hence, the number of soldiers were left out = 104

Given:

Let the number of students in each row = y

number of students in a row = the number of rows = y

Hence, total students = y × y

Acc. To given condition:

Y² = 8649

⇒ y² = 93 × 93

⇒ y² = 93²

⇒ y = 93

Hence, the number of students in each row = 93

Given:

AC = ?

(AC) =

Hence, diagonal distance = 37m

Given:

Let distance between the wall and the foot of the ladder = (AB) = xm

(BC)² = (AC)² + (AB)²

⇒ (5.5)² = (4.4)² + (x)²

⇒ (x)² = (5.5)² - (4.4)²

⇒ (x)² = 30.35 – 19.36

⇒ (x)² = 10.89 = (3.3)²

⇒ (x) = 3.3

Hence, distance between the wall and the foot of the ladder = (AB) = 3.3m

Given:

Acc. To given condition:

The total amount he will get at the end = 1 + 3 + 5 + …

As we see it an odd natural number series,

And the the number of terms (n) = 30

Sum of odd natural numbers = n² = 30² = 900

He will get 900 coins in that month.

Given:

Let the three numbers are 2x, 3x and 5x

Acc. To given condition:

i.e (2x)² + (3x)² + (5x)² = 608

⇒ 4(x)² + 9(x)² + 25 (x)² = 608

⇒38(x)² = 608

⇒ (x)² = 16

⇒ x = 4

Hence, numbers are 2(4) = 8

3(4) = 12

5(4) = 20

Given:

the least square number which is exactly divisible by 8, 9 and 10 is equal to L.C.M. 0f 8, 9 and 10.

Hence, their L.C.M. is 2× 2× 2× 3× 3× 5 = 360

As we see, (2× 2)× 2× (3× 3) × 5

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

Hence, to make it perfect square, it must be multiplied with 2 × 5 = 10

As, 360× 10 = 3600

Hence 3600 is least square number which is exactly divisible by 8, 9 and 10.

Given:

area of square plot = (equal to rectangle)

As we know,

Area of square = side × side

⇒ length of one side of the plot = m

Given: 324

Now, we subtract successive odd numbers starting from 1 as:

324 – 1 = 323

323 – 3 = 320

320 – 5 = 315

315 – 7 = 308

308 – 9 = 299

299 – 11 = 288

288 – 13 = 275

275 – 15 = 260

260 – 17 = 243

243 – 19 = 224

224 – 21 = 203

203 – 23 = 180

180 – 25 = 155

155 – 27 = 128

128 – 29 = 99

99 – 31 = 68

68 – 33 = 35

35 – 35 = 0

Here we see 324 reduces to 0 after subtracting 18 odd numbers.

So 324 is perfect square of 18.

Given:

Let the three numbers are 2x, 3x and 4x

Acc. To given condition:

i.e (2x)³ + (3x)³ + (4x)³ = 0.334125

⇒ 8(x)³ + 27(x)³ + 64(x)³ = 0.334125

⇒99(x)³ = 0.334125

⇒ (x)³ = 0.003375

⇒ x = 0.015

Hence, numbers are 2(0.015) = 0.03

3(0.015) = 0.045

4(0.015) = 0.06

Given:

= 3 + 0.2 + 0.4

= 3.6

Given:

= {37}³

= 37 × 37 × 37

= 50653

Given:

= {44}³

= 44 × 44 × 44

= 85184

Given:

Let the perfect square is abcd.

Acc. To condition:

a = even number

b = even number

c = odd number

d = even number

then one of the number will be 2234.

Given:

Numbers in circles can be:

6 + 19 = 25 (perfect square)

19 + 30 = 49 (perfect square)

30 + 6 = 36 (perfect square)

Given: perimeter of one square = 40 m

Let length of square = a₁

Then 4a₁ = 40m

⇒ a₁ = 10m

Area of square = a₁ × a₁

= 10 × 10

= 100m²

perimeter of second square = 96 m

Let length of square = a₂

Then 4a₂ = 96m

⇒ a₂ = 24m

Area of square = a₂ × a₂

= 24 × 24

= 576m²

Because, perimeter of another square equal in area to the sum of the first two squares.

Hence,

area of another square = 100 + 576 = 676m

Area of square = a₃ × a₃

676 = (a₃)²

⇒ a₃ = 26m

Hence, perimeter = 4a₃

= 4 × 26 = 104m

Given:

Three-digit perfect squares are 196 and 961, which looks same when seen upside down.

Given:

(1) one such pair is 12 and 21

(12)² = 144 and (21)² = 441

(1) another such pair is 102 and 201

(102)² = 10404 and (201)² = 40401