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(b) 51.4 cm

From the question it is given that, radius of semicircular disc 10 cm

We know that, perimeter of semicircular disc = circumference of semicircle + diameter

Circumference of semicircle = 2πr/2

= πr

= (22/7) × 10

= 31.4 cm

Then, total tape required = 31.4 + 10 + 10

= 51.4 cm

Correct option is B) 17.5

Let radius of the circle be r cm.

\(\therefore\) circumference of the circle is P = 2 π r

But given that circumference of the circle be 55 cm.

\(\therefore\) 2 π r = 55

⇒ 2 x 22/7 x r = 55

⇒ r = 55 x 7/22 x 1/2 = 35/4 cm

Now, the diameter of the circle is 2r = 35/2  = 17.5 cm

Correct option is D) 22

Radius of circle be r = 3.5 cm  = 7/2 cm

\(\therefore\) Circumference of the circle is P = 2 π r

= 2 x 22/7 x 7/2 = 22 cm

(c) 14 cm

Circumference of a circle = 2πr

88 = 2 × (22/7) × r

(88 × 7)/(2 × 22) = r

r = 14 cm

Correct option is D) 14

Let radius of the circle be r cm.

\(\therefore\) 2πr = 44 (\(\because\) circumference of the circle is 44 cm)

⇒ r = 44/2π = 22/π = 22 x 7/22 = 7 cm

\(\therefore\) Diameter of the circle is 2r = 2 x 7 = 14 cm

Correct option is D) 14

Let the height of parallelogram be x cm.

\(\therefore\) Base = 2x cm

Now, Area = Base x  Height

⇒ Area = 2x \(\times\) x = 2x² cm²

But the given that = 98 cm²

\(\therefore\) 2x² = 98

⇒ x² = 98/2 = 49 = 7²

⇒ x = 7 cm

\(\therefore\) Base = 2x = 2 x 7 = 14 cm

Area = side² 

Perimeter = 4 × side 

Diagonal = side√2

Area of the square = \(\frac{1}{2}\times Diagonal^2\)

= \(\frac{1}{2}\times 24\times 24\)

= 288 m²

Now, let the side of the square be x m.

Thus, we have:

Area = Side²

\(\Rightarrow\) 288 = x²

\(\Rightarrow\) x = \(12\sqrt{2}\)

\(\Rightarrow\) x = 16.92

Perimeter = 4 x Side

= 4 x 16.92

= 67.68m

Thus, the perimeter of thee square plot is 67.68m.

Correct option is B) 4.55

Correct option is A) 16

Let base of the triangle be x cm.

Given h = 5 cm and Area of triangle = 40cm²

⇒ 1/2 x base x height = 40

⇒ base = \(\frac{40\times2}{height}\) = 80/5 = 16 cm

Correct option is C) 4

We have area of parallelogram  = 36 cm² and base = 9 cm

\(\because\) Area of parallelogram = Base x  Height

\(\therefore\) Height = Area/Base = 36/9 = 4 cm

Area of a square = 128 cm² (given)

Let the side of square be ‘a’

Area of square = 1/2 × Diagonal²

128 = 1/2 × Diagonal²

Diagonal = 16 cm

Area of square = (side)²

128 = a²

⇨ a = 11.31 cm

Perimeter of square = 4a = 4 × 11.31 = 45.24

Perimeter of square is 45.24 cm.

Correct option is A) 10.5

Diagonals of rhombus are d₁ = 3 cm and d₂ = 7 cm

\(\therefore\) Area of rhombus = 1/2 d₁d₂

= 1/2 x 3 x 7 = 21/2 = 10.5 cm²

(a) Rs 5600.

Cost of painting both its surfaces 

= 2 × 70 × Area of one surface

= Rs \(\bigg(2\times70\times\frac12\times10\times8\bigg)\) = Rs 5600.

Answer is (A) 6 cm

Volume of I^st cube = (3)³ = 27 cm³

Volume of II^nd cube (4)³ = 64 cm³

Volume of III^rd cube = (5)³ = 125 cm³

Sum of the volumes of these three cubes 

= 27 + 64 + 125 = 216 cm³

Now cube is recasted with these three cubes.

∴ Volume of the cubes recasted

= Sum of the volumes of three cubes.

(core)³ = 216

core = \(\sqrt [ 3 ]{ 216 }\)

= \(\sqrt [ 3 ]{ 6\times 6\times 6 } \) = 6 cm

Given, the base diameter of cone = 14 m

∴ Radius (r) = \(\frac { 14 }{ 5 }\) = 7 m

Slant height (d) = 25 m

Total surface area of the cone = πr(l+r)

= \(\frac { 22 }{ 7 }\) × 7 × (25 + 7)

= \(\frac { 22 }{ 7 }\) × 7 × 32

= 22 × 32

= 704 m²

Hence, total surface area of cone 704 m²

Answer is (D)718.66 cm³

A Largest cone is cut off from a cube with its core 14 cm.

∴ Height of cone = 14 cm

And diameter of base = 14 cm

∴ Base radius of cone = \(\frac { 14 }{ 2 }\) = 7 cm

Volume of cone = \(\frac { 1 }{ 3 }\)πr²h

= \(\frac { 1 }{ 3 }\) × \(\frac { 22 }{ 7 }\) × (7)² × 14

= \(\frac { 1 }{ 3 }\) × \(\frac { 22 }{ 7 }\) × 7 × 7 × 14

= \(\frac { 1 }{ 3 }\) × 22 × 7 × 14

= \(\frac { 22\times 98 }{ 3 } \)

= \(\frac { 2156 }{ 3 }\)

= 718.66 cm³

Given :

Length of wall (l) = 8 m

Thickness of wall (b) = 35 cm 

= 0.35 m

Height of wall (h) = 4 m

Volume of wall = l × b × h

= 8 × 4 × 0.35

= 11.2 m³

Dimension of the gate in the wall 

= 2 m × 1 m × 0.35 ms [∵ Thickness of wall = 0.35 m]

∴ Volume of the gate in the wall 

= 2m × 1 m × 0.35 m = 0.7 m³

Measures of 2 windows in the wall 

= 1.20 m × 1 m × 0.35 m

Volume of 2 windows in the wall 

=2 × 1.2 × 1 × 0.35 = 0.84 m³

Volume of wall with out a gate and two windows

= 11.2 – (0.7 + 0.84)

= 9.66 m³

∵ Cost of building 1 m³ wall = Rs 1500

∴ Cost of building 9.66 m³ wall 

= 1500 × 9.66 = Rs 14490

∴ Hence cost of building the wall = Rs 14490

Base radius of conical vessel (R) = 10 cm

And the height (H) = 18 cm

Volume of conical vessel = \(\frac { 1 }{ 3 }\)πR²H

= \(\frac { 1 }{ 3 }\) × π × (10)² × 18

= π ×100 × 6

= 600 π cm³

Let the water lavel in the cylindrical vessel be h.

Radius (r) = 5 cm.

Now according to the problem

Volume of water in the cylindrical vessel = Volume of water in conical vessel

πr²h = 600 π

or r²h = 600

⇒ (5)²h = 600

⇒ 25 h = 600

h = \(\frac { 600 }{ 25 }\)

= 24 cm

Hence the water level in the cylindrical vessel = 24 cm.

Given,

Radius of conical vessel (R) = 10 cm

and its height = 18 cm

Volume of conical vessel = \(\frac { 1 }{ 3 }\)r²h

= \(\frac { 1 }{ 3 }\) × π × (10)² × 18

= \(\frac { 1 }{ 3 }\) × π × 100 × 18

π × 100 × 6

= 600 π cm³.

Let the height of water in cylindrical vessel be H and its radius = 5 cm.

Now, according to question.

Volume of cylindrical vessel = Volume of water in the conical vessel.

πR²H = 600 π

π(5)²H = 600 π

H = \(\frac { 600\pi }{ 25\pi } \) = 24 cm

Hence, height of water in cylindrical vessel = 24 cm

(i) rectangular

A polyhedron whose base is a polygon and outer surfaces are triangles of one vertex is called pyramid.

Answer is (B) 15 cm

Let the common radius of the cone and cylinder be r.

Let height of cone be h₁.

And the height of cylinder h₂ = 5 cm.

According to the problem

Volume of cone = Volume of cylinder

⇒ \(\frac { 1 }{ 3 }\)πr²h₁ = πr²h₂

⇒ \(\frac { { h }_{ 1 } }{ 3 }\) = h₂ = 5

⇒ h₁ = 3 × 5

h₁ = 15

∴ Height of cone = 15 cm

Finally, the last day of school arrived and the elf was free to go. As for homework, there was no more, so he quietly and slyly slipped out the back door in it. 

Correct option is C. help him do his homework.

Correct option is B. elf

C. about the details of God’s creation.

We have,

Dimension of rectangular strip = 25 cm × 7 cm

When this strip is rotated about its longer side,

Height of cylinder becomes = 25 cm

Radius = 7 cm

∴ Total surface area of cylinder = 2πr (h+r)

= 2 × 22/7 × 7 (25 + 7)

= 2 × 22/7 × 7 × 32

= 1408 cm²

We have,

Dimensions of rectangular sheet of paper = 44cm × 20cm

When this sheet of paper is rolled along its length,

Circumference of base becomes = 44 cm

By using the formula,

Circumference of base = 2πr

So, 2πr = 44

2 × 22/7 × r = 44

r = 44×7 / 44

= 7cm

Radius = 7cm

Height = 20 cm

∴ Total surface area of cylinder = 2πr (h+r)

= 2 × 22/7 × 7 (20 + 7)

= 2 × 22/7 × 7 × 27

= 1188 cm²

The molecules of all gases do not have cohesive force amongst themselves. During cubical expansion molecules of all types of gases are separate from each other under similar conditions.

Hence, all gases show equal expansion on being heated for same range of temperature at a constant pressure.

The relation between angle of deviation and angle of incidence, angle of emergence and angle of prism is given by 

Angle of deviation = i₁ + i₂ – A

For maximum deviation, Angle of incidence (i₁) = 90°

Angle of emergence (i₂) = 90°

We know, μ = 1/ sin C . 1/ sin A/2 [Where C = Critical angle, A = Angle of prism]

∴ A = 2C

∴ Maximum deviation = i₁ + i₂ – A = 90 + 90 – 2C = 180 – 2C = n – 2C.

Velocity of light would be minimum in medium 'A'.

• Almost every discipline, under natural and social sciences is linked with geography.
• Geomorphology, Climatology, Oceanography and Biogeography are the branches of physical geography.
• Geomorphology studies landforms, rock types, processes of formation of rocks, landforms, etc. Therefore, it is related to geology.
• Climatology studies elements of atmosphere such as temperature, winds, rainfall, humidity, natural disasters like cyclones, anticyclones, storms, etc. Therefore, it is related to meteorology.
• Oceanography studies oceans and seas on the surface of the earth, ocean currents, ocean routes, etc. Therefore, it is related to Hydrology.
• Biogeography studies the distribution of plants and animals, their species, ecosystems, etc. Therefore, it is related to biology.
• Knowledge of mathematics is important for cartographic techniques, such as drawing of maps and diagrams. Similarly, knowledge of statistics is useful to do data analysis since various statistical techniques and hypotheses testing are used in data analysis.
• Thus, physical geography is related to various branches of sciences.

• Geography is the study of the earth as a home of man, and various phenomena related to it.
• Therefore, geography is the study of the physical environment in relation to man. The physical environment has direct effect on cultural and social environments.
• The earth is dynamic in nature. Hence, we find variations in its physical and cultural/social environments.
• In geography we study the relation between the physical environment and production, distributions and their patterns and variations.
• Geographers study the location, geographical phenomena, whether physical or human, which are highly dynamic and its causes.
• Since geography is the study of space and time it makes geography dynamic in nature.
• In geographical study, the geographer tries to answer questions like what, why, where and when.

• There are two contrasting approaches to study the subject of geography. They are possibilism and environmental determinism. This contrast in approaches is called as dualism in geography.
• Some geographers are of the view that nature is more dominant than man. It is called environmental determinism. According to them, when we study geography, we study the earth. We study how natural resources have influence on economic activities, as well as food habits of people.
• For example, in coastal areas fishing activity is more developed and fish is the main food of the people.
• Some geographers are of opinion that man dominates the nature. It is called possibilism. According to them man can make changes in nature due to his intelligence.
• For example, there are polyhouses even in polar areas, in which temperature is controlled artificially and vegetables are grown.
• There are many other thinkers who have different approaches in geography.

Thus, the study of geography is dualistic in nature.

• Human geography is the branch of geography dealing with human activities and their influence on culture, communities and economies.
• In human geography every social science studies separately has interface with branch of human geography; because of their spatial attributes.
• Social sciences like sociology, political science, economics, history and demography are very closely related with branches of human geography, such as social geography, political geography, economic geography and historical geography, respectively.
• Since, we study varied branches of geography in human geography, it is said to be multidisciplinary in nature.

(d) Both A and R are correct but R is not correct explanation A.

Correct option: (a) GIS

Correct option: (d) River

a) Water cycle: The process of water evaporating from the seas, forming clouds in the sky, coming down as rain and flowing down the slopes on land in the form of rivers and finally joining the sea is called the water cycle. 

b) Evaporation: The change of water into vapour is known as evaporation. The process in which water vapour changes into water is called condensation. Clouds are tiny droplets of water hanging in the air above.

No, age cannot be in minus because we all start growing up from the moment we are born. ‘Minus 87’ means the man is 87 years back to his actual age.

Let normals at A and B meet at P. 

As mirrors are perpendicular to each other, therefore, BP || OA and AP || OB. 

So, BP ⊥ PA, i.e., ∠ BPA = 90° 

Therefore, ∠ 3 + ∠ 2 = 90° (Angle sum property) (1) 

Also, ∠1 = ∠2 and ∠4 = ∠3 (Angle of incidence = Angle of reflection) 

Therefore, ∠1 + ∠4 = 90° [From (1)] (2) Adding (1) and (2), we have 

∠1 + ∠2 + ∠3 + ∠4 = 180° 

i.e., ∠CAB + ∠DBA = 180° 

Hence, CA || BD

∠A+ ∠B + ∠C = 180° 

(Sum of the angles of a triangle is 180° ) 

50° +∠B + ∠B = 180° 

∠B = ∠C Base angles of an isosceles triangle 

50 + 2∠B= 180° 

2∠B = 180 – 50 

2∠B = 130° 

∠B = \(\frac{130^o}{2}\)

∠B = 65° 

∠B = ∠C = 65°

In Δ ABD, AB = AD 

∴∠ABD = ∠ APB = 70° 

∠APB +∠ ADC = 180° [Linear pair] 

70° +∠ADC = 180° 

∠ADC = 180° – 70° 

∠ADC = 110° 

In Δ ADC, 

∠DAC = 180° – ( 110° + 40°) 

= 180°-150 °

∠DAC = 30° 

In Δ ADC, 

∠ACD ∠DAC 

AD > CD 

but AB = AD 

∴ AB > CD

In ∆ADC and ∆AEB 

AC = AB [data] 

∠ADC = ∠AEB[BE & CD altitudes] 

∠DAC =∠EAB [Common angle] 

∆ ADC = ∆ AEB[AS∆postulate] 

AD=AE [Corresponding sides]

Let c be hypotenuse and the other two sides be b and a

According to the Pythagoras theorem, we have

c² = a² + b²

2.52 = 1.52 + b²

b² = 6.25 −2.25 = 4

b = 2 cm

Hence, the length of the other side is 2 cm.

∠A + ∠B = 115° + 65°

= 180°

Therefore,

AB ‖ BC, as sum of co interior angles are supplementary.

∠B + ∠C = 65° + 115°

= 180°

Therefore,

AB ‖ CD, as sum of co interior angles are supplementary.

Given,

∠4 = 110°,

∠7 = 65°

To find: m ‖ n

Here,

∠7 = ∠5 = 65°(Vertically opposite angle)

Now,

∠4 + ∠5 = 110° + 65°

= 175°

Therefore,

m is parallel to n as the pair of co interior angles is not supplementary.

Let us Assume the length of the third side be x.

Then,

(5 + 9) > x

∴the length of its third side is less than 14 cm

(i) Given 5, 7, 9

Yes, these numbers can be the lengths of the sides of a triangle because the sum of any two sides of a triangle is always greater than the third side.

Here, 5 + 7 > 9, 5 + 9 > 7, 9 + 7 > 5

(ii) Given 2, 10, 15

No, these numbers cannot be the lengths of the sides of a triangle because the sum of any two sides of a triangle is always greater than the third side, which is not true in this case.

(iii) Given 3, 4, 5

Yes, these numbers can be the lengths of the sides of a triangle because the sum of any two sides of triangle is always greater than the third side.

Here, 3 + 4 > 5, 3 + 5 > 4, 4 + 5 > 3

(iv) Given 2, 5, 7

No, these numbers cannot be the lengths of the sides of a triangle because the sum of any two sides of a triangle is always greater than the third side, which is not true in this case.

Here, 2 + 5 = 7

(v) Given 5, 8, 20

No, these numbers cannot be the lengths of the sides of a triangle because the sum of any two sides of a triangle is always greater than the third side, which is not true in this case.

Here, 5 + 8 < 20