i) Legs of the table, railway track, edges of the scale
ii) Adjacent sides of a Board, Cross bars of windows, Adjacent sides of the textbook
iii) Cross bars of windows, Ladder, blades of a scissor.

Construction:
(i) Placed the center of the protractor at the vertex X. Lined up the ray \(\overline { XY } \)with the 0° Line. Then draw and labelled a point Z at 120° mark on the (a) inner scale
(anti clockwise) and (b) outer scale (clockwise).
(ii) Removed the protractor and draw \(\overline { XY } \) to complete the angle.
Now, LX = LZXY = LYXZ = 120°.

The number of dots given are 1, 3, 6, 10, 15, ...

1^st number = 1
2^nd number = 1 + 2 = 3
3^rd number = 3 + 3 = 6
4^th number = 6 + 4 = 10
5^th number = 10 + 5 = 15
6^th number will be 15 + 6 = 21
7^th number will be 21 + 7 = 28
8^th number will be 28 + 8 = 36
9^th number will be 36 + 9 = 45
10^th number will be 45 + 10 = 55
Number of dots in the 10^th figure = 55.

Taking smaller triangles we have AFB, BFG, BGC, CGH, HCD, DHI, DIE, EIJ, EJA and AJF \(\Rightarrow \)10 triangles.
Taking the combination of 2.
CGD, CID, DJE, DHE, EIA, EFA, AGB, AJB, BHC and BFC are 10 triangles.
Combining 3 at a time we have CED, DAE, EBA, ACB, BDC, CEF, DBJ, GDA, EBH, ACI => 10 triangles.
Other triangles are BCE, CAD, DBE, ADB and ACE are 5 triangles.
\(\therefore \) Total 10 + 10 + 10 + 5 = 35 triangles.

A, B, C, D, & E are five triangles.
No combination of these triangles taken 2 at a time, form any triangle.
Only the combinations of A, F & C; A, F & D; B, F & E; B, F & D and C, F & E form triangles when taken 3 triangles at a time.
Hence, the total number of triangles = 5 + 5 = 10.

Number of bags of apples = 25
Number of apples in each bag = 9
\(\therefore\) Total apples   = 25 \(\times\) 9 = 225
Number of friends = 6
Number of apples each friend get = 225 \(\div\) 6 = 37
Number of apples each get = 37
Remaining apples = 3.

Using distributivity of multiplication over subtraction 500 \(\times\) 689 - 500 \(\times\) 89
 = 500 \(\times\) (689 - 89)
 = 500 \(\times\) 600
 = 300000
500 \(\times\) 689 - 500 \(\times\) 89 = 3,00,000.

Matches won by Team A = 26 out of 52
\(\text { i.e: } \frac{26}{52}=1 / 2 \)
Matches won by team B = \(\frac { 3 }{ 4 } \times 52\) = 39 out of 52
∴ Team B won 39 out of 52 and therefore team B has the better record.

Let x be the part owned by C then A : B : C =  2(2x) : 2x : x
= 4x : 2x : x
∴ A : B = 4x : 2x
= 2 : 1
B : C = 2x : x
= 2 : 1
 ∴ A : B :: B : C i.e., They are in proportion

1 hour = 60 minutes
Distance covered by cholan in 60 minutes = 6 km
Distance covered in one minutes = \(\frac{6}{60}\)
Distance covered in 20 minutes = \(\frac{6}{60} \times 20\)
= 2 km

20 in the ratio 3 : 2
Dividing 20 into 3 + 2 = 5 equal parts, then
\(20\times\frac{3}{5}=12\)
\(20\times\frac{2}{5}=8\)
\(\therefore\) The required answer is 12, 8

Step (i): Preparing place value chart with 8 digits 6 in ten lakh place and 9 in Ten thousand place
Step (ii): Fill the other places with any of the numbers

\(\therefore\) The number may be 56897432. Similarly we can write many numbers

11 added to 10 times x

Divide the whole money into 2 + 3 = 5 equal parts then, Vimala gets 2 parts out of 5 parts and Yazhini gets 3 parts out of 5 parts.
Amount Vimala gets =Rs. 600 ×\(2\over 5\) = Rs. 240
Amount Yazhini gets =Rs. 600 ×\(3\over 5\) =Rs. 360
Vimala received Rs. 240 and Yazhini gets Rs. 360, which is Rs. 120 more than that of Vimala.

(i) \(\overline{EF}and\overline{GH},\overline{CD}and\overline{AB}\)
(ii) \(\overline{CD}and\overline{EF},\overline{AB}and\overline{EF},\overline{CD}and\overline{GH},\overline{AB}and\overline{GH}\)
(iii) P, Q, R and S

CD = 3.6 cm

Construction:
Step 1: Drawn a line l and marked the point C on it
Step 2: Measured 3.6 cm using compass by placing pointer at 0 and pencil at 3.6 cm
Step 3: Place the pointer of the compass at C 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 D.
Step 4: Now, CD is the required line of length 3.6 cm.

Eight million three hundred forty three thousand seven hundred ten

The Largest six digit number is 999999

634,567,105 = Six hundred thirty four million five hundred sixty seven thousand One hundred five.

\((i) \ \overline{XY}=2.4 \ cm\)
\((ii) \ \overline{ AB}=3.4 \ cm\)
\((iii) \ \overline{EF}=4 \ cm\)
\((iv) \ \overline{ PQ}=3 \ cm\)

(i) The next even number = u + 2
(ii) The previous even number = u – 2

5p