(i) name a pair of complementary angles:
∠FAE;∠EAD
(ii) name a pair of supplementary angles:
∠FAD;∠DAC
∠BAC;∠CAE
∠FAB;∠BAC
∠FAB;∠FAE
∠EAD;∠DAB

Construction:
(i) Placed the centre of the protractor at the vertex L. Lined up the ray \(\overline { LM } \) with the 0° line. Then draw and labelled a point N at 38° mark on the
(a) inner scale (anticlockwise) and
(b) outer scale (clockwise).
(ii) Removed the protractor and draw \(\overline { LN } \) to complete the angle.

Construction:
(i) Placed the centre of the protractor at the vertex A. Line up the ray \(\overline { AB } \) with the 0° line. Then draw and labelled a point C at the 65° mark on the (a) inner scale (anti clockwise) (b) outer scale (clockwise)
(ii)Removed the protractor and draw \(\overline { AC } \) to complete the angle.
Now

   
1. Draw the base ray PQ
2. Placed the center of the protractor at the vertex P.  lined up the ray \(\overline { PQ } \) with the \({ 0 }^{ 0 }\) line. Then drawn and labeled a pointed.

A, B, C, D are four triangles.
Combining A & D, B & D and D & C do not form any triangles.
Combining A, B & D; A, C & D and B, D & C also do not form any triangles.
Combining A, B, C & D = 1 triangle
\(\therefore\)Total number of triangles = 4 +1 = 5

Pictograph of Number of computers sold

Using place value chart
The number should be the biggest odd

The password is 87543.

Using distributive property of multiplication over addition.
50 \(\times\) 102 = 50 \(\times\) 10 + 50 \(\times\) 2
= 5000 + 100
= 5100
50 \(\times\)102 = 5100.

Number of times Antony brushes a day = 2
∴ Number of times Antony brushes a week = 2\(\times\)7 = 14
Number of times Shabeen brushes a day = 1
∴ Number of times Shabeen brushes a week = 1\(\times\)7 = 7
Number of times Antony brushes : Number of times Antony brushes =14 : 7 = 2 : 1
∴ The required ratio = 2 :1

The ratio of Students : teachers : administrators = 1500 : 50 : 5.
The new ratio of Students : teachers : administrators = 1800 : new teachers : new administrators
Given the ratios are maintained.
i.e., they are proportional.
∴  \(\frac { 1500 }{ 1800 } =\frac { 50 }{ New\quad teachers } =\frac { 5 }{ New\quad admins } \)
From the first two ratios, taking, cross product
\(\frac { 1500 }{ 1800 } =\frac { 50 }{ New\quad teachers } \)
New Teachers \(\times\) 1500 = 50 \(\times\) 1800
New Teacher = \(\frac { 50\times 1800 }{ 1500 } \)
New Teacher = 60
Also \(\frac { 1500 }{ 1800 } =\frac { 5 }{ New(admin) } \)
New admins \(\times\) 1500 = 5 \(\times\) 1800
New admms = \(\frac { 5\times 1800 }{ 1500 } \) = 6
∴ For 1800 students 60 teachers and 6 administrators are needed.

Correct answers given by Kaarmugilan and Kavitha are in the ratio 10:11
Total of 84 points will be divided into 10 + 11 = 21 equal parts
Points scored by Kavitha = \(\frac { 11 }{ 21 } \) \(\times\) 84 = 44 points
∴ Kavitha scored 44 points in the quiz.

\(4:5=\frac{4}{5}\times\frac{3}{3}=\frac{12}{15}\)
\(8:15=\frac{8}{15}\)
Here \(\frac{12}{15}\) larger than \(\frac{8}{15}\) and 4: 5 is larger than 8: 15

(i) parallel lines:
\(\overline{CD}\) and \( \overline{EF}\), \(\overline{CD} \) and \(\overline{IJ}\), \(\overline{EF} \ and\ \overline{IJ}\)
(ii) intersecting lines:
\(\overline{AB}\) and \( \overline{CD},\) \(\overline{AB}\) and \( \overline{EF},\) \(\overline{AB}\) and \( \overline{IJ},\) \(\overline{GH}\) and \( \overline{IJ},\) \(\overline{AB}\) and \( \overline{GH}\)
(iii) points of intersection:
P, Q and R

Age of Athiyan = ‘p’ years
Age of Mugilan = ‘p + 6’ years (algebraic statement)
If p = 20, then Mugilan’s age is = 20 + 6 = 26 years.

QR = 10 cm

Construction:
Step 1: Drawn a line 'l' and marked the point Q on it
Step 2: Measured 10 cm using compass by placing pointer at 'O' and pencil pointer at 10 cm.
Step 3: Placing pointer at Q drawn the arc on 'l' with pencil pointer and named the point R
Step 4: QR is the required line segments of 10 cm.

AB = 7.5 cm

Construction:
Step 1: Drawn a line I and marked the point A
Step 2: Measured 7.5 cm using compass placing the pointer at 'O' and pencil pointer at 7.5 cm
Step 3: Placing compass pointer at A, drawn an arc on l with the pencil pointer. It cut 'l' at B
Step 4: AB is the required segment of length 7.5 cm

One hundred three million four hundred fifty six thousand seven hundred eighty nine

75,32,105

Seventy five lakh thirty two thousand one hundred five

7345671 = Seven million three hundred forty five thousand six hundred seventy

Age ofArivazhagan's father = n 
Arivazhagan’s age = ‘n - 30’

Total number of students = 50.
Number of boys = 30.
Then number of girls = 50 - 30 = 20.
(i) Ratio of number of boys to no. of girls = \(\frac { 30 }{ 20 } \)
\(=\frac { 30\div 10 }{ 20\div 10 } =\frac { 3 }{ 2 } =3:2\)
(ii) \(\frac { Number\ of\ girls }{ Total\ number\ of\ students } =\frac { 20 }{ 50 } =\frac { 20\div 10 }{ 50\div 10 } =\frac { 2 }{ 5 } =2:5\)
(iii) \(\frac { Number\ of\ boys }{ Total\ number\ of\ students } =\frac { 30 }{ 50 } =\frac { 30\div 10 }{ 50\div 10 } =\frac { 3 }{ 5 } =3\ :\ 5\)

Distance covered by Akilan in 1 hour = 10 km
Distance covered by Selvi in 1 hour = 6 km
\(\frac { Distance\ covered\ by\ Akilan }{ Distance\ covered\ by\ Selvi } =\frac { 10\ km }{ 6\ km } \)
\(=\frac { 10\div 2 }{ 6\div 2 } \)
\(=\frac { 5 }{ 3 } \)
Distance covered by Akilan : Selvi = 5 : 3.