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The minimum distance between the objects and the eye so that a clear image is formed on the retina is called the least distance for clear vision or near the point of the eye. This distance is 25 cm for the human eye.

The range of vision of human beings is 25 cm to infinity.

The minimum distance at which objects can be seen most distinctly without strain is called the least distance of distinct vision (or) distance of clear vision.

The eye lens forms a real and inverted image of an object on the retina.

The image formed on the retina in human eyes.

The human eye uses light and enables us to see the object. It has a lens in its structure.

Following are integers for given cases 

(i) 25 m below sea level = – 25 m 

(ii) 7 m above Earth = + 7 m 

(iii) On heating milk at 20°C temperature = + 20°C 

(iv) On freezing ice-cream at 5°C temperature = – 5°C

Writing the integers in ascending and descending order – 

(i) Ascending order = – 9, – 4, 0, 2, 7 

Descending order = 7, 2, 0, – 4, – 9 

(ii) Ascending order = – 5, – 3, 1, 3, 5, 6 

Descending order = 6, 5, 3, 1, – 3, – 5

Writing predecessor and successor, in the table –

Using appropriate symbols : 

1. Any number, smaller than 0 = – 1, – 2 

2. 50 meters below sea level = – 50 meter 

3. 10°C temperature below 0°C = – 10°C 

4. 15°C temperature above 0°C = + 15°C 

Writing the integers in ascending and descending order- 

(i) Ascending order = – 7, – 3, 3, 5 

Descending order = 5, 3, – 3, – 7 

(ii) Ascending order = – 2, – 1, 0, 3 

Descending order = 3, 0, – 1, – 2 

(iii) Ascending order = – 6, 1, 3 

Descending order = 3, 1, – 6 

(iv) Ascending order = – 5, – 1, 2, 4 

Descending order = 4, 2, – 1, – 5

Always positive.

Writing integers in ascending order between the pair of numbers given below – 

(i) 0 and – 4 or – 4 and 0 ⇒ – 4, – 3, – 2, – 1, 0 

(ii) – 3 and – 5 or – 5 and – 3 ⇒ – 5, – 4, – 3 

(iii) – 2 and 2 ⇒ – 2, – 1, 0, 1, 2 

(iv) – 10 and – 6 ⇒ – 10, – 9, – 8, – 7, – 6

(i) (- 7) + (+ 8) = – 7 + 8 = 8 – 7 = 1 

(ii) (-3) + 5 = -3 + 5 = 5- 3 = 2 

(iii) (- 3) + (- 2) = – 3 – 2 = – 5 

(iv) (+ 7) + (- 2) = 7 – 2 = 5

(i) F 

(ii) F 

(iii) F 

(iv) T

(i) Largest negative number is -1

Always negative.

Filling the blanks with sign 

(i) ( -2) + (-9) = -2 -9 = – 11 

and (-2) + (-4) = -2 -4 = – 6 

∴(-2) + (-9) < (-2) + (-4) 

(ii) (- 21) + (- 10) ……….. (- 10) + (- 21) 

(-21) + (-10) = – 21 – 10 = – 31 

and (-10) + (-21) = -10 -21 = -31 

∴ (-21) + (-10) = (-10) + (-21) 

(iii) 45- (-12) (- 12) + 45 

45 – (-12) = 45 + 12 = 57 

and (-12) + 45 = -12 + 45 = 33 

∴ 45- (-12) > (-12) + 45 

(iv) (-14) + (14) ……………….. (-7) + (1) 

(-14) + (14) = -14 + 14 = 0 

and (-7) + (1) = 7 + 1 = -6 

∴ (-14) + (14) > (-7) + (1)

(i) (-7) + (-4) + 11 = -7 -4 +11 = -11 + 11 = 0 

(ii) (-12) + (-3) – (-4) = -12 -3 + 4 = -15 + 4 = -11 

(iii) 14 – 8 – (-2) = 14 – 8 + 2 = 14 + 2- 8= 16 – 8 = 8 

(iv) (-24) + (-12) – (-8) = -24 -12 + 8 = -36 + 8 = -28

(i) -5 + 5 = 0 

(ii) 7 + (-7) = 0 

(iii) 11 + (-11) = 0 

(iv) (-3) + (-4) = – 7 

(v) 14 – (- 2) = 16 

(vi) (-4) +(-4) = – 8

(i) 137 + (- 354) + 125 = 137 – 354 + 125 = 137 + 125 – 354 = 262 – 354 = – 92 

(ii) (- 312) + 39 + 192 = – 312 + 39 + 192 = – 312 + 231 = – 81 

(iii) 37 + (- 3) + 24 + (- 8) = 37 – 3 + 24 – 8 = 37+ 24 – 3 – 8 = 61 – 11 = 50 

(iv) 102 + (- 24) + (24) + (- 11) = 102 – 24 + 24 – 11 = 102 + 24 – 24 – 11 = 102 – 11 = 91

Given (508)²

We have (a + b)² = (a² + 2ab + b²)(508)² can be written as 500+8 So here a=500 and b=8

Using the formula,

(500 + 8)² = (500² + 2 X 500 X 8 + 8²)

On simplifying we get

(500 + 8)² = (250000 + 8000 +64)

(508) ² = 258064

(i) We know that, 

(a – b)² = (a² – 2ab + b²) 

We have 

(196)² = (200 - 4)² 

= 2002 – 2 (200 × 4) + 4² 

= 40000 – 1600 + 16 

= 3814 

(ii) We know that, 

(a – b)² = (a² – 2ab + b²) 

We have 

(689)² = (700 - 11)² 

= 700² – 2 (700 × 11) + 11² 

= 490000 – 15400 + 121 

= 474721 

(iii) We know that, 

(a – b)² = (a² – 2ab + b²) 

We have 

(891)² = (900 - 9)² 

= 900² – 2 (900 × 9) + 9² 

= 810000 – 16200 + 81 

= 793881

Given (196)²

We have (a – b)² = (a² – 2ab + b²)(196)² can be written as 200-4 So here a=200 and b=4

Using the formula,

(200 – 4)² = (200² – 2 X 200 X 4 + 4²)

On simplifying we get

(200 – 4)² = (40000 – 1600 +16)

(196)² = 38416

Given (630)²

We have (a + b)² = (a² + 2ab + b²)(630)² can be written as 600+30 So here a=300 and b=10

Using the formula,

(600 + 30)² = (600² + 2 X 600 X 30 + 30²)

On simplifying we get,

(600 + 30)² = (360000 + 3600 +900)

(630)² = 396900

(i) We know that, 

(a + b)² = (a² + 2ab + b²) 

We have, 

310² = (300 + 10)² 

= [300² + 2 (300 × 10) + 10²] 

= 90000 + 6000 + 100 

= 96100 

(ii) We know that, 

(a + b)² = (a² + 2ab + b²) 

We have, 

508² = (500 + 8)² 

= [500² + 2 (500 × 8) + 8²] 

= 250000 + 8000 + 64 

= 258064 

(iii) We know that, 

(a + b)² = (a² + 2ab + b²) 

We have, 

630² = (600 + 30)² 

= [600² + 2 (600 × 30) + 30²] 

= 360000 + 36000 + 900 

= 396900

Given (689)²

We have (a – b)² = (a² – 2ab + b²)(689)² can be written as 700-11 So here a=700 and b=11

Using the formula,

(700 – 11)² = (700² – 2 X 700 X 11 + 11²)

On simplifying we get

(700 – 11)² = (490000 – 15400 +121)

(689)² = 474721

Given (891)²

We have (a – b)² = (a² – 2ab + b²)(891)² can be written as 900-9 So here a=900 and b=9

Using the formula,

(900 – 9)² = (900² – 2 X 900 X 9 + 9²)

On simplifying we get

(900 – 9)² = (810000 – 16200 +81)

(891)² = 793881

Given 69 X 71

We can write 69 as 70-1 and also 71 as 70+1 We know that (a + b) X (a – b) = a²– b²

Here a= 70 and b = 1

Applying the above formula we get

(70 + 1) X (70 – 1) 

= (70)² – (1)²

 = 4900 – 1

= 4899

Given 94 X 106 

We can write 94 as 100-6 and also 106 as 100+6 

We know that (a + b) X (a – b) = a²– b²

Here a= 100 and b = 6

Applying the above formula we get

(100 + 6) X (100 – 6) 

= (100)² – (6)²

= 10000 – 36

= 9964

(i) Here, figure (i) contains 7 faces, 10 vertices and 15 edges.
Using Euler’s formula, we see F + V – E = 2
Putting F = 7, V = 10 and E = 15,
F + V – E = 2
=> 7 + 10 – 5 = 2
=> 17 – 15 = 2
=> 2 = 2
=> L.H.S. = R.H.S.

(ii) Here, figure (ii) contains 9 faces, 9 vertices and 16 edges.
Using Euler’s formula, we see F + V – E = 2

F + V – E = 2
=> 9 + 9 – 16 = 2
=> 18 – 16 = 2
=> 2 = 2
=> L.H.S. = R.H.S.

By Euler’s formula, we have

F + V − E = 2

(i) F + 6 − 12 = 2 F − 6 = 2

F = 8

(ii) 5 + V − 9 = 2 V − 4 = 2

V = 6

(iii) 20 + 12 − E = 2

32 − E = 2 E = 30

Thus, the table can be completed as

No, it can be a cuboid also.

(i) Number of faces = F = 7 

Number of vertices = V = 10 

Number of edges = E = 15

We have, F + V − E = 7 + 10 − 15 = 17 − 15 = 2 

Hence, Euler’s formula is verified.

(ii) Number of faces = F = 9

Number of vertices = V = 9

Number of edges = E = 16

F + V − E = 9 + 9 − 16 = 18 − 16 = 2

Hence, Euler’s formula is verified.

We have,

24a³b³ / -8ab

By using the formula aⁿ / a^m = a^n-m

24/-8 a^3-1 b^3-1

-3a²b²

A square prism has a square as its base. However, its height is not necessarily same as the side of the square. Thus, a square prism can also be a cuboid.

(i) A cylinder can be thought of as a circular prism i.e., a prism that has a circle as its base.

(ii) A cone can be thought of as a circular pyramid i.e., a pyramid that has a circle as its base.

(i) It is not a polyhedron as it has a curved surface. Therefore, it will not be a prism also.

(ii) It is a prism.

(iii) It is not a prism. It is a pyramid.

(iv) It is a prism.

Given:

∠1 = 60° and ∠2 = ( 2/3 )rd of a right angle.

To prove: parallel to m

Proof ∠1 = 60°

∠1 = 2/3 x 90^o = 60^o

∠1 = ∠2 = 60^o

Parallel to m as pair of corresponding angles are equal.

According to the question,

We have from the figure ∠1 = 60° and ∠6 = 120°

Since, ∠1 = 60° and ∠6 = 120°

Here, ∠1 = ∠3 [since they are vertically opposite angles]

∠3 = ∠1 = 60°

Now, ∠3 + ∠6 = 60° + 120°

⇒ ∠3 + ∠6 = 180°

We know that,

If the sum of two interior angles on same side of l is 180°, then the lines are parallel.

Therefore, m || n

(c) (108–b)^o

From the question it is given that, ∠1 = (2a + b)^o and ∠6 = (3a – b)^o

Since ∠5 and ∠6 these two forms a linear pair

Then,

∠5 = (180-3a + b)^o … [equation 1]

∠5 = ∠1 = (180-3a + b)^o [Because Corresponding angles] …equation (2)

On equating equation (1) and equation (2) we get,

2a + b = 180-3a + b

 5a = 180

 a = 36⁰

Since ∠1 and ∠2 forms a linear pair so

∠2 = 180⁰– 2a-b

Corresponding angles are equal so,

∠1 = ∠3 = 85°

By using co - interior angle property,

∠2 + ∠3 = 180°

∠2 + 85° = 180°

∠2 = 95°

Correct option is B) Intersecting

Given,

∠1 = 60°

∠2 = \(\frac{2}{3}\) of right angle

To prove:

l ‖ m

Proof:

∠1 = 60°

∠2 = \(\frac{2}{3}\) x 90° = 60°

Since,

∠1 = ∠2 = 60°

Therefore

 l ‖ m as pair of corresponding angles are equal.

Let AB and CD be perpendicular to MN

∠ABD = 90°(AB perpendicular to MN) (i)

∠CON = 90°(CD perpendicular to MN) (ii)

Now,

∠ABD = ∠CON = 90°

Since,

AB ‖ CP

Therefore, corresponding angles are equal.

Dodo and cheetah.

Trisomy of 21st chromosome.

Let, ∠A = 2x and

∠B = 3x

Now,

∠AHB = 180°(Co. interior angles are supplementary)

2x + 3x = 180°

5x = 180°

x = 36°

∠A = 2x = 72°

∠B = 3x = 108°

Now,

∠A = ∠C = 72°(Opposite sides angles of a parallelogram are equal)

∠B = ∠D = 108°(Opposite sides angles of a parallelogram are equal)

Given that,

l ‖ m and n cuts them

Let,

∠1 = 65°

∠2 = x

∠3 = 40°

∠1 = ∠4 = 65°(Alternate angle) (i)

∠3 + ∠4 + ∠5 = 180°(Angle sum property)

40° + 65° + ∠5 = 180°

∠5 = 75°

Now,

∠2 + ∠5 = 180°(Linear pair)

x + 75° = 180°

x = 105°