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Given: For the frustum shaped tub, 

height (h) = 21 cm,

radii (r₁ ) = 20 cm, and (r₂ ) = 15 cm 

To find: Capacity (volume) of the tub. 

Volume of frustum = 1/3 πh (r₁² + r₂² + r₁ × r₂ )

= 1/3 x 22/7 x 21 (20² + 15² + 20 x 15)

= 22(400 + 225 + 300)

= 22 x 925

= 20350 cm³

= (20350/1000) liters    ....[1liter = 1000 cm³]

= 20.35 liters

∴ The capacity of the tub is 20.35 liters.

8cm, 10cm, 10cm; 8cm, 8cm, 12cm

Let the edge of a cube be a.

If edge of the cube became tripled i.e. a = 3 x a = 3a

._.. Volume of the cube = a³

.^.. Volume of the cube with edge tripled = (3a)³ = 27 a³

Hence, volume is 27 times of the original volume.

Surface area of a cube = 6 a², where a is side of a cube.

.^.. Side of cube = 5 cm

.^.. Surface area of the cube = 6 x (5)² 

= 6 x 25

= 150 cm²

Now, surface area of the cube with side 1 cm = 6 x (1)²= 6 cm²

.^.. Surface area of 5 cubes with side 1 cm = 5 x 6 = 30 cm²

Ratio of the surface area of the original cube to that of the sum of the surface area of the smaller cubes

= 30/150

= 3/15

= 1:5

Volume of a cube = (Side)³

Side of cube = 0.5 cm

Volume of the cube = (0.5)² = 0.125 cm³

The number of cubes required to make volume of 8 cm³ cube

= 8/0.125

= 8000/125

= 64 cubes

(A) – (iii), (B) – (iii), (C) – (ii), (D) – (i)

To find the cost of a frame of a picture, we need to find the perimeter of the picture.

True

Length of the boundary of the remaining park is

(B) 400 m.

To find the cost of painting a wall we need to find the perimeter of the wall.

False

False.

To find the cost of painting a wall we need to find the area of the wall.

(A) 2 times

As we know that, perimeter of square = side × 4

= 10 × 4

= 40 cm

Given, side of square = 10 cm

The side of the square is doubled = 10 + 10

= 20 cm

Therefore, the new perimeter become 2 times the side of the square is doubled.

False.

An engineer who plans to build a compound wall on all sides of a house must find the perimeter of the compound.

False

We know that, area of square = side × side

As per the condition given in the question, side of the square is doubled = 2 × side

Then, new area = 2side × 2side

= 4 side²

Therefore, side of the square is doubled then the area of the square obtained is 4 times the old area.

True.

A farmer who wants to fence his field, must find the perimeter of the field.

Area of a square is doubled if the side of the square is doubled.

False

A farmer who wants to fence his field, must find the perimeter of the field. 

True

False.

Perimeter of regular octagon = number of sides × length of reach sides

= 8 × 6

= 48 cm

Perimeter of a regular octagon of side 6 cm is 36 cm.

False

Given: For the sphere, 

radius (r) = 7 cm 

To find: Surface area of the sphere.  

Surface area of sphere = Aπr² 

= 4 × 22/7 × (7) 

= 88 × 7 

= 616 cm² 

∴ The surface area of the sphere is 616 cm

(A) 15 cm

Volume of cone = volume of cylinder 

∴ 1/3πr²h = πr²H

∴ h/3 = 5

∴ h = 15 cm

Inner area of rectangle = length× breadth

Inner area of rectangle = 112× 78 = 8736 m²

Width of path = 2.5 m

Length of outer rectangle = 112 +2.5 + 2.5 = 117 m

Breadth of outer rectangle = 78 +2.5 + 2.5 = 83 m

Outer area of rectangle = length× breadth = 117 × 83 = 9711 m²

Area of path = outer area of rectangle – inner area of rectangle

Area of path = 9711 – 8736 = 975 m²

Cost of construction for 1 m² = 4.50 Rs

Cost of construction for 975 m² = 975(4.50) = 4387.5 Rs

Length of rectangular park (l) = 175 cm

Breadth of rectangular park (b) = 125 cm

Length of wire required for fencing the park =

Perimeter of the park

= 2 × (l + b)

= 2 × (175 + 125)

= 2 × 300 = 600 m

Cost for fencing 1 m of the park = Rs 12.

Cost for fencing 600 m of the square park = 600 × 12

= Rs. 7200

Correct option is: C) 25 cm

We have l = 9 cm, b = 12 cm & h = 20 cm.

\(\therefore\) The length of the diagonal of cuboid =  \(\sqrt {l^2 + b^2 + h^2}\)

= \(\sqrt {9^2 + 12^2 + 20^2} = \sqrt {81 + 144 + 400} = \sqrt {625} = 25 cm\)

Hence, the diagonal of the cuboid is 25 cm.

Correct option is: C) 25 cm

Correct option is: A) 1 : 2

Height of hemisphere = Radius of hemisphere = r.

\(\because\) Heights of cone and hemisphere are equal.

\(\therefore\) Height of cone = r = Radius of base of cone.

\(\therefore\) \(\frac {V_1}{V_2} \) =  \(\frac {Volume \,of \,cone}{Volume \, of\, hemisphere} = \frac {\frac 13 \pi r^2 h}{\frac 23 \pi r^3}\)

= \(\frac h{2r} = \frac r {2r} = \frac 12 \) (\(\because\) h = r)

= 1 : 2

Hence the ratio of their volumes is 1 : 2

Correct option is: A) 1 : 2

Correct option is: D) 4

Let the side of the cube be a.

Then the diagonal of the cube = \(\sqrt {a^2+a^2 + a^2} = \sqrt3 \,a\) 

Given that diagonal of the cube is \(4\sqrt3\).

\(\therefore\) \(\sqrt3 \, a\) = \(4\sqrt3\)

= a = 4

Hence, the side of the cube is 4.

Correct option is: D) 4

Dimensions of cuboidal tin box are 20 cm x 16 cm x 14 cm,

.^..Area of metal sheet for 1 box = Surface area of cuboid

= 2(lb + bh+hl)

= 2(20 x 16+16 x 14+ 14 x 20)

= 2(320 + 224 + 280)

= 2(824)

= 1648 cm²

.^.. Area of metal sheet required to make 10 such boxes = 10 x 1648 = 16460 cm²

l x b x h = 6 m x 400 cm x 1.5 m 

l = 6m, 

b = 4 m, 

h = 1.5 m 

∴ Total surface area of the cuboid = Outer surface area 

= 2(lb + bh + hl) = 2((6 x 4) + (4 x 1.5) + (1.5 x 6)) 

= 2(24 + 6 + 9) = 2(39) m²

Cost of painting 1 m² = Rs 22 

Cost of painting 78 m² = 78 x 22 = Rs 1716

Given,

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

When this sheet of paper is rolled along its length:

So,

Circumference of base thus form = 44 cm

= 2πr = 44

= r = \(\frac{44}{2π}\) = \(\frac{44\times 7}{44}\) = 7 cm

Radius = 7 cm

Height = 20 cm

∴ Total surface area of cylinder = 2πr(h + r) = 2 x \(\frac{22}{7}\) x 7 x 27 = 1188 cm²

Given,

Circumference of base of cylinder = 88 cm

Height of cylinder = 15 cm

So,

= 2πr = 88 Given

= r = \(\frac{88}{2π}\) = \(\frac{88\times 7}{2\times 22}\) = 14 cm

Radius of cylinder = 14 cm

∴ Curved surface area of cylinder = 2πrh = 2 x \(\frac{22}{7}\) x 14 x 15 = 1320 cm²

∴ Total surface area of cylinder = 2πr(h + r) = 2 x \(\frac{22}{7}\) x 14 x 29 = 2552 cm²

Given,

Dimensions of rectangular sheet = 44 cm × 20 cm

When it rolled along its length:

Radius of base = \(\frac{length}{2π}\) = \(\frac{44\times 7}{2\times 22}\) = 7 cm

Height of cylinder = 20 cm

∴ Volume of cylinder = πr²h = \(\frac{22}{7}\)x 7 x 7 x 20 = 3080 cm³

Given,

Dimensions of rectangular srip = 25 cm × 7 cm

When it rotated about longer side:

Radius of base = 7 cm

Height of cylinder = 25 cm

∴ Volume of cylinder = πr²h = \(\frac{22}{7}\) x 7 x 7 x 25 = 3850 cm³

Given,

Diameter of roller = 84 cm

Radius of roller = \(\frac{84}{2}\) = 42 cm

Length of roller = 120 cm

So,

Curved surface area of roller = 2πrh = 2 x \(\frac{22}{7}\) x 42 x 120 = 31680 cm²

No. of revolution it takes to level the ground = 500 given

Hence,

Area of play ground = 500 × 31680 = 15840000 cm² = 1584 m²

Perimeter of rhombus = 4 × side

Side of rhombus = \(\frac{perimeter}{4}\) = \(\frac{40}{4}\) = 10 m

Rhombus is a type of parallelogram, Hence area of rhombus = area of parallelogram

Therefore area of parallelogram = base× altitude

Base× altitude = area of parallelogram

10 × altitude = 84

Altitude of rhombus = 8.4 m

Area of trapezium = \(\frac{1}{2}\)(sum of lengths of parallel sides) x altitude

Length of bases of trapezium = 15 cm and 9 cm

Similarly length of altitude in m = 8 cm

Area of trapezium = \(\frac{1}{2}\)(15 + 9) x 8

Area of trapezium = \(\frac{1}{2}\) x 24 x 8 m² = 96 m²