(a + b)² + (a - b)² = a²+ 2ab + b² + a²- 2ab + b²
= 2a² + 2b² = 2(a²+ b²)
(a + b)² + (a - b)²= 2(a² + b²)

(1.9) x (2.1) = (2 - 0.1) (2 + 0.1)
Substitute a = 2 and b = 0.1 in the identity
(a + b) (a - b) = a² - b²
(2 + 0.1) (2 - 0.1) = 2²-(0.1)²
= 4 - 0.01
(1.9) x (2.1) = 3.99 .

\(382.4 \div 1000 =\frac{382.4}{1000}=\frac{3824}{10} \times \frac{1}{1000} \)
\(=\frac{3824}{10000}=0.3824 \)

\(2\leftarrow ,1\downarrow \) and then \(1\leftarrow 2\downarrow \) or \(\\ 2\leftarrow 1\downarrow \) and then \(1\leftarrow ,2\downarrow \)

(i) yellow line 120 cm ⟶ ,210 cm ↓.
(ii) black line 270 cm\(\leftarrow \) , 330 cm \(\uparrow\)
(iii) blue line 150 cm ⟶

(i) Choosing from, vertex we can name as follows.
Image of \(\angle\)Lis \(\angle\)L', Image of \(\angle\)M of \(\angle\)M'
Image of \(\angle\)N of \(\angle\)N', Image \(\angle\)O of \(\angle\)O'
(ii) Corresponding sides are LM and L'M',
MN and M'N', No and N'O', OL and O'L'.

(i) Footprint A \(\rightarrow\) Footprint B
The transformation is Translation

(ii) Footprint A \(\rightarrow\) Footprint C
The transformation is Refleetion about horizontal line.

(iii) Footprint A \(\rightarrow\) Footprint D
The transformation is Reflection about vertical line.

(iv) Footprint A \(\rightarrow\) Footprint E
The transformation is rotation about the heel.

a) From the picture

7➝2↓
b) No, the fishing boat cannot make the same translation if it translate, it well be land on the Island
c)  Translating the fishing boat to reach D is 5➝3↓

(i)

(ii)

(i) First 10 prime numbers : 2,3, 5,7, 11, 13, 17 , 19, 23, 29
\(\text { Mean } =\frac{\text { Sum of Pr ime numbers }}{\text { Number of observation }} \)
\(=\frac{2+3+5+7+11+13+17+19+23+29}{10} \)
\(=\frac{129}{10}=12.9 \)
Mean = 12.9

(i) Mean
Total = 22 + 15 + 10 +10 + 24 + 21 = 102
\(The\ mean\ =\quad \cfrac { Total }{ number } =\cfrac { 102 }{ 6 } =17\) 
The mean = 17
(ii) Mode
Ascending order of given data
10, 10, 15, 21, 22, 24
Here, 10 occurs two times
mode = 10
(iii) Median
n = 6, even number
The median = \(\cfrac { 1 }{ 2 } \left( \cfrac { { n }^{ th } }{ 2 } term+\left( \cfrac { n }{ 2 } +1 \right) ^{ th }term \right) \) 
= \(\cfrac { 1 }{ 2 } \left( \cfrac { 6 }{ 2 } ^{ th }term+\left( \cfrac { 6 }{ 2 } +1 \right) ^{ th }term \right) \) 
= \(\cfrac { 1 }{ 2 } \left( { 3 }^{ rd }term+{ 4 }^{ th }term \right) \) 
= \(\cfrac { 1 }{ 2 } \left( 15+21 \right) =\cfrac { 1 }{ 2 } \times 36=18\) 
Hence Mean = 17, Mode = 10 and Median = 18

Ascending order 8, 10, 15, 16,25,30
n = 6, Even number
The median = \(\cfrac { 1 }{ 2 } \left( \cfrac { n }{ 2 } ^{ th }term+\left( \cfrac { n }{ 2 } +1 \right) ^{ th }term \right) \) 
= \(\cfrac { 1 }{ 2 } \left( \cfrac { { 6 }^{ th } }{ 2 } term+\left( \cfrac { 6 }{ 2 } +1 \right) ^{ th }term \right) \) 
= \(\cfrac { 1 }{ 2 } \left( { 6 }^{ th }term+{ 7 }^{ th }term \right) \) 
= \(\cfrac { 1 }{ 2 } \left( 15+16 \right) =\cfrac { 1 }{ 2 } \times 31=15.5\) 
The median = 15.5

Ascending order of ages of 11 players is
25, 25, 25, 26, 36, 36, 36, 38, 38, 39,40. Here 25 and 36 occur 3 times, the maximum number of times.
Mode = 25 and,36.

5.1,4.8,4.3,4.5,5.1,4.7,4.5,5.2,5.4,5.8,4.3,5.6, 5.2, 5.5
\(The\ mean\ of\ the\ data=\cfrac { Total }{ Number } \) 
= \(\cfrac { 5.1+4.8+4.3+4.5+5.1+4.7+4.5++5.2+5.4+5.8+4.3+5.6+5.2+5.5 }{ 14 } \) 
=  \(\cfrac { 70 }{ 14 } =5\) 
The mean of the data = 5

The mean height of 11 students =  150 cm
\(The\ mean=\cfrac { Total\ marks\ of\ students }{ Number\ of\ students } \) 
\(150=\cfrac { 154+145+y+y+y+4+160151+149+149+150+144+140 }{ 11 } \) 
150 x 11 = 1346+ 2y
1650 = 1346 + 2y
2y = 1650 - 1346 = 304
\(y=\cfrac { 304 }{ 2 } =152\) 
y = 15 Height of two students Y = 152 cm and
y + 4 = 152 + 4 = 156 cm

Let the side of the lawn be a m
Area = a x a = a²sq.m
The side of outer square = (a + 4 ) m
Outer square area = ( a + 4 )²sq.m
= a² + 8a + 16 [(a + b)² = a² + 2ab + b²]
Area of the path = Area of the outer square - Area of the inner square

= 8a + 16 = 136
8a = 136 - 16 = 120
a = 120/8 = 15 m
The side of the square a = 15 m
Area of the lawn = a² = 15 x 15 m² = 225 m²

Balaji, Principal P = Rs.5000
Time n = 2 years
Let rate of the interest r%
Simple interest I = \(\cfrac { Pnr }{ 100 } =\cfrac {5000\times 2\times r }{ 100 } =Rs.100\)
Charles, Principal P = Rs.3000
Time n = 4 years
Let rate of the interest r%
SImple interest I = \(\cfrac { Pnr }{ 100 } =\cfrac { 3000\times 2\times r }{ 100 } =Rs.120\)
Arun received simple interest from Balaji and Charles = 100r + 120r = 220r
Total interest Arun received = Rs.2200
220r = 2200

Rate of the interest r = 10%

The number of coloured squares = 58
Fraction = \(\cfrac { 58 }{ 100 } \)
Decimal = 0.58
Percentage = \(\cfrac { 58 }{ 100 } *100 =58\)%

Therefore 25.8 + 18.53 = 44.33

Underline the digit to be rounded 8.71
Since the digit next to the underlined in 7 which is greater than 5. Therefore we should add 1 to the underlined digit.
Hence we get 9
Hence the rounded value of 8.71 to the nearest whole number is 9.

Step 1 : Draw a rough diagram and mark the given measurements.
Step 2: Mark a point O on the paper.
Step 3: Extend the compass distance equal to the radius 4cm.
Step 4: At center 0, Hold the compass firmly and place the pointed end of the compass
Step 5: Slowly rotate the compass around to get the circle