(i) Total parts = 8
Shaded paris of portion = 3
Fraction = \(\cfrac { 3 }{ 8 } \)
(ii) Shaded portion:
Total no. parts = 15
Shaded portion = 5
\(\therefore \) Fraction \(\cfrac { 5 }{ 15 } =\cfrac { 1 }{ 3 } \)
(iii) Shaded portion = 3
\(\therefore \) Fraction = \(\cfrac { 3 }{ 9 } =\cfrac { 1 }{ 3 } \)
(iv) Total no. of parts = 9
Shaded portion = 5
\(\therefore\) Fraction = \(\cfrac { 5 }{ 9 } \)

Now the HCF of 16, 28
16 = 2 \(\times\) 2 \(\times\)2 \(\times\) 2
28 = 2 \(\times\)2 \(\times\) 7
HCF of (16, 28) = 2 \(\times\)2 = 4
Now the HCF of (16 , 28 -16)
16 = 2 \(\times\) 2 \(\times\)2 \(\times\) 2
12 = 2 \(\times\) 2 \(\times\) 3
HCF of (16, 12) = 2 \(\times\) 2 = 4
Therefore HCF of (16, 28) = HCF of (16, 28 - 16).
Hence, HCF of two numbers a and b, a > b, is same as the HCF of a and a – b.

Base, b = 18 cm
Height, h = 12 cm
Area, \(A={1\over2}(b x h)\ sq\ units\)
\(={1\over2}(18\times12)\)
= 108 sq. cm. (or) 108 cm²

Thendral
Perimeter of the rectangle, P = 12 cm
2 ( l + b ) = 12
l + b = \({12\over6}=6cm\)

The possible pairs of measures whose sum is 6 are (5, 1) and (4, 2).
Hence, Thendral can make a rectangle in 2 ways. She can make a rectangle of length 5 cm and breadth 1 cm and another one with length 4 cm and breadth 2 cm.
Tharani

Perimeter of the square, P = 12 cm
4 \(\times\) s = 12
\(s={12\over4}=3 \ cm\)
Hence, Tharani can make only one square of side 3 cm.
Thanam
Perimeter of the triangle, P = 12 cm
a + b + c = 12 cm
The possible triplets of measures whose sum is 12 and also satisfying the triangle inequality are (2, 5, 5); (3, 4, 5); (4, 4, 4).
Hence, Thanam can make 3 triangles of sides 2 cm, 5 cm & 5 cm; 3 cm, 4 cm & 5 cm and 4 cm, 4 cm & 4 cm.

Draw number line and plot the numbers −14 and −11 as follows

Fixing −11, we find −14 is to the left of −11. So, −14 is smaller than −11. That is, −14 < −11.

\({17\over3}=Quotient{Remainder\over Divisor}=5{2\over3}\)

(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

Perimeter (P) = 40 m
Side (8) =?
Area (A) =?
4S = P
\(4S=40\Rightarrow S=\cfrac { 40 }{ 4 } =10m\)
A = S² = 10² = 100m²
\(\therefore\) Side (S) = 10 m, Area (A) = 100 m²

i) Wrong, Integers are not continuously marked
ii) Correct, Integers are correctly marked.
iii) Wrong, Integer −2 is marked wrongly.
iv) Correct, Integers are marked at equal distance.
v) Wrong, negative integers marked wrongly.

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 \(2\cfrac { 1 }{ 2 }\ and\ 3\cfrac { 2 }{ 3 } \)

(i) 4 square tiles horizontally and 4 square tiles vertically.
\(\Rightarrow\) 4 \(\times\) 4 = 16 sq. units.
No. of tiles tilled = 7
No. of tiles required = 16 -7 = 9
(ii) 4 square tiles vertically and 3 square tiles horizontally .
Area \(\Rightarrow\) 4 \(\times\) 3 = 12 sq. units.
No. of tiles filled = 6
(iii) 4 square tiles vertically and 3 square files horizontally.
Area = 4 \(\times\) 3 = 12 sq. units.
No. of tiles filled = 6
Required tiles = 12 - 6 = 6
(iv) 4 square tiles horizontally and 4 square tiles vertically.
Area = 4 \(\times\) 4 = 16 sq. units.
No. of tiles filled = 8
Required tiles = 16 - 8 = 8 

Euclid's game
(28, 35, 41)
\(\Rightarrow\) (28, (35 - 28), (42 - 28))
\(\Rightarrow\) (28, 7, 14)
\(\Rightarrow\) (28 -7), 7, 14 - 7)
\(\Rightarrow\) (21, 7, 7) \(\Rightarrow\)((21 -7), 7, 7)
\(\Rightarrow\) (14, 7, 7) 
\(\Rightarrow\) (14 - 7, 7, 7)
\(\Rightarrow\)(7, 7, 7)
\(\therefore\) HCF of (28, 35, 42) is 7

(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.

(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