Draw a line ‘ l ’ and mark a point ‘P’ on it

Measure 5.5 cm using compass placing the pointer at ‘0’ and the pencil pointer at 5.5 cm.

Place the pointer of the compass at ‘ P ’ then draw a small arc on the line ‘l ’ with the pencil pointer. It cuts the line ‘ l ’ at a point and name that point as ‘Q’

Now, PQ is the required line segment of length 5.5 cm.

(i) and (iv) are complementary angles
(ii) and (iii) are non-complementary angles

Paralell line:
\( \overline{\mathrm{AB}} \ \text{and}\)  \( \overline{\mathrm{DC}},\)
\( \overline{\mathrm{AD}} \ \text{and}\) \(\overline{\mathrm{BC}},\)
\(\overline{\mathrm{DC}} \text { and } \overline{\mathrm{HG}}, \)
\(\overline{\mathrm{AD}} \text { and } \overline{\mathrm{EH}},\)
\( \overline{\mathrm{AE}} \text { and } \overline{\mathrm{DH}}, \)
\( \overline{\mathrm{DH}} \text { and } \overline{\mathrm{GG}}\)
Intersecting lines:
\(i) \ \overline{\mathrm{AB}}, \overline{\mathrm{AE}}, \overline{\mathrm{AD}}\\ ii) \ \overline{\mathrm{DA}}, \overline{\mathrm{DH}}, \overline{\mathrm{DC}}\\ iii) \ \overline{\mathrm{CB}}, \overline{\mathrm{CG}}, \overline{\mathrm{CD}}\\ iv)\ \overline{\mathrm{HD}}, \overline{\mathrm{HG}}, \overline{\mathrm{HE}}\\ v) \ \overline{\mathrm{EA}}, \overline{\mathrm{EH}}_{, . .}\)

(i) Total chairs in a stadium = 7,689
Chairs in one row = 90
\(\frac{7689}{90}=85\ \frac{39}{90}\)
There will be 85 rows
(ii) Yes, There are 39 chairs left over.

(i) Indian System :Two crore fifty nine lakh forty one thousand nine hundred
International System : Twenty five million nine hundred forty one thousand nine hundred
(ii) 5,50,500
(iii) Eighty six crore forty lakh seven hundred thirty
(iv) Ninteen million eight hundred eighty eight thousand eight hundred
(v) Indian System : 60,53,100 - Sixty lakh fifty three thousand one hundred
International System : 6,053,100 - Six million fifty three thousand one hundred

(i) There are 1706 tigers in 2011
(ii) No. of tigers in 2008 = 1400
No. of tigers in 1990 = 3500
There were 2100 lesser tigers
(iii) No. of tigers in 2014 = 2226
No. of tigers in 2011 = 1706
= 520
\(\therefore\) 2226 > 1706
\(\therefore\) The number of tigers increased from 2011 to 2014.
Yes, the number of tigers increased, 520 more tigers are there in 2014.

(a) green balls = 4, Yellow balls = 3, black balls =  5.
so, Black balls
(b) 5 parts of Black balls = 40 balls
\(\therefore \) 1 part of Black balls = \(\frac{40}{5}=8 \text { balls }\)
4part of Green balls = 4 \(\times\) 8 = 32
3 part of yellow balls = 3 \(\times\) 8 = 24
If there are 40 black balls, then total number of balls = 32 + 24 + 40 = 96 balls
(c) Number of green balls = 32
Number of Yellow balls = 24

(a) Given Ragi flour:raw rice = 4 : 1
If one cup of ragi flour used then the ratio of ragi flour : raw rice = \(\frac { 4 }{ 4 } :\frac { 1 }{ 4 } =1:\frac { 1 }{ 4 } \)
Raw rice required = \(\frac { 1 }{ 4 } \) cup
(b) Ratio of water : ragi flour = 8 : 4
If 16 cups of water used then ratio of water : ragi flour = 8\(\times\)2 : 4\(\times\)2 = 16 : 8.
So, 8 cups of ragi flour is used.
(c) We know that the quantities of same units can be compared. Here Ragi flour, Raw rice and water are in one unit, Sesame oil and salt are in different units, these different units cannot be compared and cannot be expressed as a ratio.

Population in the year 2011 is 46,81,087 \(\Rightarrow\) 46,81,000
Population in the year 2011 is 43,43,645 \(\Rightarrow\) 43,44,000
\(\therefore\) Increase in Population = 3,37,000.

In my purchase cost of 6 apples = Rs. 90
∴ cost of 1 apple = \(\frac { 90 }{ 6 } \) = Rs. 15
In my friend's purchase cost of 5 apples = Rs. 70
∴ Cost of apple =\(\frac { 70 }{ 5 } \) = Rs. 14
My friend buy for lower prize.
∴ My friend's purchase is a better buy.

The ratio of Golden Gate Bridge & Its model are = 6480 : 756 and 60 : 7
Product of extremes = 6480 \(\times\) 7 = 45,360
Product of means = 756 \(\times\) 60 = 45,360
Product of extremes = Product of means
∴ The model is in proportion to the ordinal bridge.

i) 5k = 8
k = 8 - 5 =
k = 3
ii) 6 + m = 7
m = 7 - 6 
m = 1
iii) 4 + 6 = n
n = 10
iv) 7 + 2 = a
a = 9
v) b + 3 = 9
= 9 - 3 
b = 6
vi) 6 + c = 10
c = 10 - 6 = 4
c = 4
vii) x +4 = 8
x = 8 - 4 
x = 4
viii) y = 3 + 6
y = 9

(i) Length of wire = '12 s' cm 
Sum of 3 sides of equilateral triangle = '12 s'
3 \(\times\) side = 12 s
The length of a side = \(\frac{12 \mathrm{~s}}{3}=\) '4 s' cm
(ii) Sum of 4 sides of square = '12 s'
The length of a side = \(\frac{12 \mathrm{~s}}{4}=\) '3 s' cm

\(\therefore\) Total people visited = 11,910.

i), iii) are obtuse angles

Total length 63 cm, on dividing 3 + 4 = 7 equal parts, we get the line segments with lengths
\(\frac{3}{7}\times63=27\ cm\)
\(\frac{4}{7}\times63=36\ cm\)
Length of each line segment is 27 cm & 36 cm

The value of ‘k’ is 15.

(1) An Acute angle:

(2) An Obtuse Angle: 
 
(3) A right Angle: 

(4) A straight Angle:
                                                                                

Given digits are 9, 4, 6 and 8.
Smallest number with these digits is 4689
Given that it is an even number.
\(\therefore\) It may be 4698
So password of ATM card is 4698.

789,213,456

2,30,51,980

NUMBER  : 50,000
Expanded form : 5 \(\times\) 10000
Read as  : Fifty Thousand
NUMBER : 676097 
Expanded form : 6 \(\times\) 100000 + 7 \(\times\) 10000 + 6 \(\times\) 1000 + 0 \(\times\) 100 + 9 \(\times\) 10 + 7 \(\times\) 1
Read as : Six Lakh Seventy Six Thousand Ninety Seven