Perimeter = 80 cm
No. of identical shapes = 9
Side of square = 4 cm
Possible shape:

\(\therefore\) P = 80cm

Shape	Equilateral triangle	Square	Regular pentagon	Regular hexagon	Regular octagon
Number of lines of symmetry	3	4	5	6	8
Order of rotational symmetry	3	4	5	6	8

(i) A regular polygon of 10 sides will have 10 lines of symmetry.
(ii) If a regular polygon has, 10 lines of symmetry, then its order ofrotational symmetry is 10
(iii) A regular polygon of 'n' sides has n, n lines of symmetry and the order of rotational symmetry is n.

(i) From the number line,
G is -3
K is -1 
- 1 > -3, -1 is greater.
\(\therefore\) k is greater.
(ii) From the number line, C represents the integer is -4.
(iii) From the number line 
G is -3
H is 4 
Integers between G and H are -2, -1, 0, 1, 2, 3
Ans: 6 integers.
(iv) 2 pairs
(i.e) (E,J) and (C,H)
(-5, 5) (-4,4)
(v) units to the left of D is - 6. So, False.

From the figure,
Complete squares = 9
half squares = \(\cfrac { 13 }{ 2 } \)
\(\therefore\) Approximate area = Complet esquares + half squares
= \(9+\cfrac { 13 }{ 2 } \)
= \(\cfrac { 18+13 }{ 2 } =\cfrac { 31 }{ 2 } \)
= 15.5 sq.units

Square of side = 10 m
Breadth of rectangle = 8 m
Square:
Area of the square = s² sq. units
= 102 = 100m²
P = 4s = 4(10) = 40 m
Rectangle:
Area of the rectangle = I\(\times\) b sq. units
b = 8 m
P = 40 m
2 (l+ b) = 40
2(l + 8) = 40
2l + 16 = 40
2l = 40 -16
\(l=\cfrac { 24 }{ 2 } =12\)
Area of the rectangle = (12 \(\times\) 8) m2
Difference = Area of the square - Area of the rectangle
= 100 - % = 4m²
Square plot has the greater area and 4 m²

To prove: Square B = 2 (Square A)
Let the side of the square B = 4 cm
P = 4s = 4 \(\times\) 4 = 1.6.cm
Next, The side of the square A = 2 cm
P = 4s = 4 \(\times\) 2 = 8 cm
\(\therefore\) Square B = 16 = 2 (8) = 2 square A
\(\therefore\) Perimeter of square B is twice that of square A.

l = 3b
P = 64 cm
2 (I + b) = 64
2(3b + b) = 64
2 (4b) = 64
\(8b=64\Rightarrow b=\cfrac { 64 }{ 8 } 8\)cm
I = 3 (8) = 24 crn
The sides of the rectangle I = 24 cm, b = 8 cm

Each side of shape = 3 cm
\(\therefore\) 20 equal sides perimeter = 20 
times 3cm
P = 20 \(\times\) 3 = 60 cm
\(\therefore\) P = 60 cm

(i) Given:
Length (1) = 5 cm
breadth (b) = 8 cm
Perimeter = 2 (I + b)
= 2 (5 + 8)
= 2 \(\times\)13 = 26 cm
Are a (A) = lb = 5 \(\times\) 8 = 40 cm²
(ii) Given :
Length (1) = 13 cm
Perimeter (P) = 54 cm
To find: b and A
\(b=\cfrac { P-2l }{ 2 } \)
= \(\cfrac { 54-2(13) }{ 2 } \)
= \(\cfrac { 54-26 }{ 2 } =\cfrac { 28 }{ 2 } =14\)
b = 14 cm
Area (A) = lb = 13 \(\times\) 14
A = 182 cm²
(iii) Given:
Breadth (b) = 15 cm
Perimeter (P) = 60 cm
To find: 1 and A
\(I=\cfrac { P-2b }{ 2 } \)
= \(\cfrac { 60-2(15) }{ 2 } =\cfrac { 60-30 }{ 2 } =\cfrac { 30 }{ 2 } \)
1 = 15 cm
A = lb = 15 \(\times\) 15 = 225
A = 225 cm²
(iv) Given :
Length (1) = 10m
Area (A) = 120 sq. m
To find: band P
\(b=\cfrac { A }{ l } \)
= \(\cfrac { 120 }{ 10 } =12\)
b = 12m
P = 2 (1 +  b)
= 2(10 + 12)
= 2 \(\times\) 22 = 44
P = 44 m
(v) Given:
Breadth (b) = 4 feet
Area (A) = 20 s. feet
To find: 1 and P
\(\therefore l=\cfrac { A }{ b } \)
= \(\cfrac { 20 }{ 4 } =5\)
1= 5 feet
P = 2(I + b)
= 2(5 + 4)
= 2(9) = 18
P = 18 feet

C3: The opposite to the second negative integer.
C5: The additive identity in whole numbers.
C6: The successor of the integer in C2.
C8: The predecessor of the integer in C7.
C9: The opposite to the integer in C5.

The distance from school to Library via Bus Stop.
= \(\left[ \cfrac { 3 }{ 4 } +3\cfrac { 1 }{ 2 } \right] \)
= \(\cfrac { 3 }{ 4 } +\cfrac { 7 }{ 2 } \)
= \(\cfrac { 3+(7\times 2) }{ 4 } =\cfrac { 3+14 }{ 4 } =\cfrac { 17 }{ 4 } \)
= \(4\frac { 1 }{ 4 } \) km
The distance between school and library via Hospital
= \(\left( 4\frac { 1 }{ 2 } +1\frac { 1 }{ 4 } \right) \)
= \(\cfrac { 9 }{ 2 } +\cfrac { 5 }{ 4 } \)
= \(\cfrac { (9\times 2) }{ 4 } =\cfrac { 18+5 }{ 4 } =\cfrac { 23 }{ 4 } \)
= \(5\frac { 3 }{ 4 } \)
(iii) The shortest between (i) and (ii)
via Bus stop = \(4\frac { 1 }{ 4 } \)
via Hospital = \(5\frac { 3 }{ 4 } \)
Ans: via Bus stop.
(iv) Distance between School and Hospital = \(4\frac { 1 }{ 2 }\ km \)
Distance between School and Bus stop = \(\frac{3}{4}\ km\)
Required  Answer = \(4 \frac{1}{2} \div \frac{3}{4}\)
\(=\frac{9}{2} \div \frac{3}{4}=\frac{9}{2} \times \frac{4}{3}=6\)
The distance between school and Hospital is 6 times the distance between school and Bus stop.

Given:
A painter painted = \(\cfrac { 3 }{ 8 } \) of the wall
Yellow colour painted = \(\cfrac { 1 }{ 3 } \times \cfrac { 3 }{ 8 } =\cfrac { 1 }{ 8 } \)
\(\therefore \cfrac { 1 }{ 8 } \) is the yellow colour of the entire wall

(i) \(\cfrac { 3 }{ 7 } \div 4\)
= \(\cfrac { 3 }{ 7 } \times \cfrac { 1 }{ 4 } =\cfrac { 3\times 1 }{ 7\times 4 } =\cfrac { 3 }{ 28 } \)
(ii) \(\cfrac { 4 }{ 3 } \div \cfrac { 5 }{ 9 } \)
= \(\cfrac { 4 }{ 3 } \times \cfrac { 9 }{ 5 } \)
= \(\cfrac { 4\times 3 }{ 5 } =\cfrac { 12 }{ 5 } =2\cfrac { 2 }{ 5 } \)
(iii) \(4\cfrac { 1 }{ 5 } \div 3\cfrac { 3 }{ 4 } \)
= \(\cfrac { 21 }{ 5 } \div \cfrac { 15 }{ 4 } \)

= \(\cfrac { 21 }{ 5 } \times \cfrac { 4 }{ 15 } =\cfrac { 7\times 4 }{ 5\times 5 } =\cfrac { 28 }{ 25 } =1\cfrac { 3 }{ 25 } \)

(iv) \(9\cfrac { 2 }{ 3 } \div 1\cfrac { 2 }{ 3 } \)
= \(\cfrac { 29 }{ 3 } \div \cfrac { 5 }{ 3 } \)
= \(\cfrac { 29 }{ 3 } \times \cfrac { 3 }{ 5 } =\cfrac { 29 }{ 5 } 5\cfrac { 4 }{ 5 } \) 

(i) \(3\cfrac { 7 }{ 8 } \)
Improper fraction = \(\cfrac { \left( 3\times 18 \right) +7 }{ 18 } \)
= \(\cfrac { 54+7 }{ 18 } =\cfrac { 61 }{ 18 } \)
(ii) \(\cfrac { 99 }{ 7 } \)
\(Mixed\ fraction=Qutient+\frac { Remainder }{ Divisor } \)
= \(14+\cfrac { 1 }{ 7 } \)
= \(14\cfrac { 1 }{ 7 } \)
(iii) \(\cfrac { 47 }{ 6 } \)
Mixed fraction = \(7\cfrac { 5 }{ 6 } \)
(iv) \(12\cfrac { 1 }{ 9 } \)
Mixed fraction = \(\cfrac { \left( 12\times 9 \right) +1 }{ 9 } \)
= \(\cfrac { 108+1 }{ 9 } =\cfrac { 109 }{ 9 }\)