(i)In rectangle, Perimeter = 14m
Length = 4m
Breadth =?
P = 2(I + b)
14 = 2(4 + b)
14 = 8 + 2b
14 - 8 = 2b
\(6=2b\Rightarrow b=\cfrac { 6 }{ 2 } =3\)
\(\therefore\) Breadth = 3m.
(ii) In isosceles triangle 2 sides are equal.
P = 21 cm, 3^rd side = 5 cm.
P = side 1 + side 2 + side 3
21 = (a + b + c)
21 = (a + a + 5)
\(21a=2a+5\Rightarrow 2a=21-5\)
2a = 16
\(2=\cfrac { 16 }{ 2 } =8\)
\(\therefore\) Equal sides = a = 3 cm, b = 8 cm.

7 units

Yes; Natural numbers are positive integers.

i) + 24 is read as 'plus 24' or 'positive 24'.
ii) - 13 is read as 'negative 13' or 'minus 13'.
iii) -9 is read as 'negative 9' or 'minus 9'.
iv) B is read as 'plus 8' or 'positive 8'.

Given: \(\cfrac { 3 }{ 5 } ,\cfrac { 9 }{ 10 } ,\cfrac { 11 }{ 15 } \)
LCM of 5; 10 and 15 is 30.
\(\cfrac { 3 }{ 5 } \cfrac { 6 }{ 6 } \times \cfrac { 18 }{ 30 } ,\cfrac { 9 }{ 10 } \times \cfrac { 3 }{ 3 } =\cfrac { 27 }{ 30 } ,\cfrac { 11 }{ 15 } \times \cfrac { 2 }{ 2 } =\cfrac { 22 }{ 30 } \)
\(\Rightarrow \cfrac { 18 }{ 30 } <\cfrac { 22 }{ 30 } <\cfrac { 27 }{ 30 } \)
\(\therefore\) Ascending order :\(\cfrac { 3 }{ 5 } ,\cfrac { 11 }{ 15 } ,\cfrac { 9 }{ 10 } \)

(i) \(\cfrac { 1 }{ 3 } \) is greater than \(\cfrac { 1 }{ 5 } \)
That is \(\cfrac { 1 }{ 3 } \ge \cfrac { 1 }{ 5 } \)
(ii) \(\cfrac { 2 }{ 5 } \) is smaller \(\cfrac { 5 }{ 8 } \)
That is \(\cfrac { 2 }{ 5 } \le \cfrac { 1 }{ 5 } \)

1. Fill the date of withdrawal.
2. Fill the account number is the proper place.
3. Fill the branch of bank
4. Fill the Name of the account holder.
5. Fill the amounf to be withdrawn in words.
6. Fill the amount to be withdrawn in numbers.
7. Put the signature ofthe account holder.

(i) By Euclid's game
Given numbers = 25 and 35
HCF\(\Rightarrow\)(25, 35) = HCF(25, 35 - 25)
= (25, 10)
= (25 - 10, 10) = (15, 10)
= (5, 10)
= (5, 10 - 5)
= (5, 5)
(ii) By Euclid's game
Given numbers = 36 and 12
HCF of 36 and and 12 \(\Rightarrow\)(36, 12)
\(\Rightarrow\) (36 -12, 12)
\(\Rightarrow\)(24, 12)
\(\Rightarrow\)(24 - 12, 12)
\(\Rightarrow\)(12, 12)
\(\therefore\) HCF = 12.
(iii) 
Given numbers are 15 and 29
By Euclid's game:
HCF of 15 and 29 is
\(\Rightarrow\)(15, 29)
\(\Rightarrow\)(15, 29 - 15)
\(\Rightarrow\)(15, 14)
\(\Rightarrow\)(15 - 14, 14)
\(\Rightarrow\)(1, 14)\(\Rightarrow\) (1, 14 - 1) \(\Rightarrow\)(1, 13)
\(\Rightarrow\)(1, 13 -1) \(\Rightarrow\)(1, 12) \(\Rightarrow\)(1, 12 -1), (1, 11) (1, 1)
\(\therefore\) HCF = 1

Fibonaxxi sequence 1,1,2,3,5,8,13,21,34,55

(i) 1 \(\times\) 1 = 1
11 \(\times\) 11 = 121
111 \(\times\) 111 = 12321
1111 \(\times\) 1111 = 1234321
11111 \(\times\) 11111 = 123454321
(ii)

The order of rotational symmetry of each face of a die are

(i) Hidden word isDECODE
(ii) Hidden word is KICK
(iii) Hidden word is BED
(iv) Hidden word WAY
(v) Hidden word MAT H
(vi) Hidden word TOM A T 0

A square of side = 2 cm
A rectangle of length = 15 cm
and
breadth= 10 cm
Perimeter of the combined shape = Perimeter of square + Perimeter of rectangle.

Perimeter of the combined shape = 10 cm + 15 cm + 10 cm + 2 cm + 2 cm + 2 cm + 13 cm
\(\therefore\)  P = 54 cm.

Perimeter = 40 cm
Two sides are 13 cm and 15 cm
To find: 3^rd side (c)
a + b + C = 40 \(\Rightarrow\) 13 + 15 + c = 40
\(\Rightarrow\) 28 +c = 40
c = 40 - 28 = 12cm
\(\therefore\) Third side = 12 cm

Area of a rectangle = 820 sq. cm
Width (b) = 20 cm
Length (l) =?
Perimeter =?
\(I=\cfrac { A }{ b } \)
= \(\cfrac { 820 }{ 20 } =41\\ \)
= =2(41+20)
= 2 (61) = 122 cm
\(\therefore\) I = 41 cm , P = 122 cm

Side of the Square (s) = 8 cm
P = 4s
P = 4 \(\times\) 8 = 32 cm
Area = S²
= 82 = 64 cm²

Given :
G is 3 and C is - 1

To find: The numbers are A and K. From the number line.
A = - 3
K = 7

The depth of 10 feet from the ground level and its opposite

Opposite number of - 10 feet is 10 feet.

Length of the staircase = \(5\cfrac { 1 }{ 2 } \) m
To one step = \(\cfrac { 1 }{ 4 } \) m
No. of the steps will be there = \(\cfrac { 5\frac { 1 }{ 2 } }{ \frac { 1 }{ 4 } } \)
= \(\cfrac { \frac { 11 }{ 2 } }{ \frac { 1 }{ 4 } } \Rightarrow \cfrac { 11 }{ 2 } \times \cfrac { 4 }{ 1 } =22\)
22 steps will be there in the staircase.

Mangai bought = \(6\cfrac { 3 }{ 4 } kg\) of apple 
Kalai bought = \(1\cfrac { 1 }{ 2 } \) times as Mangai bought
= \(\left[ 1\cfrac { 1 }{ 2 } \times 6\cfrac { 3 }{ 4 } \right] kg\) of apples

= \(\cfrac { 3 }{ 2 } \times \cfrac { 27 }{ 4 } \)
= \(\cfrac { 81 }{ 8 } \)
\(\therefore \) Kalai bought = \(10\cfrac { 1 }{ 8 } \) kg of apples.

Difference between \(2\cfrac { 1 }{ 2 }\ and\ 3\cfrac { 1 }{ 2 } \)
Difference = \(3\cfrac { 2 }{ 3 } -2\cfrac { 1 }{ 2 } \)
= \(\cfrac { 11 }{ 3 } -\cfrac { 5 }{ 2 } \)
= \(\cfrac { \left( 11\times 2 \right) -\left( 5\times 3 \right) }{ 6 } =\cfrac { 22-15 }{ 6 } \)
= \(\cfrac { 7 }{ 6 } \) ...(1)
Also given
Sum of \(1\cfrac { 1 }{ 2 }\ and\ 2\cfrac { 1 }{ 4 } \)
Sum =\(1\cfrac { 1 }{ 2 } +2\cfrac { 1 }{ 4 } \)
= \(\cfrac { 3 }{ 2 } +\cfrac { 9 }{ 4 } \)
= \(\cfrac { \left( 3\times 2 \right) +9 }{ 4 } =\cfrac { 6+9 }{ 4 } =\cfrac { 15 }{ 4 } \)
From (1) and (2),
Difference = \(\cfrac { 7 }{ 6 } \), Sum = \(\cfrac { 15 }{ 4 } \)
\(\cfrac { 7 }{ 6 } ,\cfrac { 15 }{ 4 } \)
\(\Rightarrow\) Lcm of 6 and 4 is 12
\(\cfrac { 7 }{ 6 } \times \cfrac { 2 }{ 2 } =\cfrac { 14 }{ 12 } ,\cfrac { 15 }{ 4 } \times \cfrac { 3 }{ 3 } =\cfrac { 45 }{ 12 } \)
\(\cfrac { 14 }{ 12 } <\cfrac { 45 }{ 12 } \Rightarrow \cfrac { 7 }{ 6 } <\cfrac { 15 }{ 4 } \)
\(\therefore \) The difference between \(2\cfrac { 1 }{ 2 }\ and\ 3\cfrac { 2 }{ 3 } \)

Total distance = \(5\cfrac { 3 }{ 4 } km=\cfrac { 23 }{ 4 } km\)
He had walked = \(2\cfrac { 1 }{ 2 } km=\cfrac { 5 }{ 2 } km\)
Difference = \(\left( \cfrac { 23 }{ 4 } -\cfrac { 5 }{ 2 } \right) km\)
= \(\cfrac { 23-\left( 5\times 2 \right) }{ 4 } =\cfrac { 23-10 }{ 4 } =\cfrac { 13 }{ 4 } km\)
\(\therefore 3\cfrac { 1 }{ 4 } km\) distance still he has to walk to reach his house