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Let the length of one parallel side of trapezium = 3x meter

Length of other parallel side of trapezium = 5x meter

Area of trapezium = 384 cm²

Distance between parallel sides = 12 cm

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

384 = \(\frac{1}{2}\)(3x + 5x) x 12

4x = \(\frac{384}{12}\)

4x = 32

x = \(\frac{32}{4}\) = 8

Therefore length of one parallel side of trapezium = 3x = 3× 8 = 24 cm

Length of other parallel side of trapezium = 5x = 5× 8 = 40 cm

Given,

Diameter of well = 7 m

Radius of well = \(\frac{7}{2}\) = 3.5 m

Depth of well = 10 m

Volume of well = πr²h = \(\frac{22}{7}\) x 3.5 x 3.5 x 10 m³

volume of well = Area of spread out × height of embankment

= \(\frac{22}{7}\) x 3.5 x 3.5 x 10 = 30 x 20 - \(\frac{22}{7}\) x \(\frac{7}{2}\)  x \(\frac{7}{2}\) x h

= \(\frac{22}{7}\) x 3.5 x 3.5 x 10 = \(\frac{1123}{2}\) x h

= h = \(\frac{22\times 35}{1123}\) = 68.56 cm

Height of embankment = 68.6 cm

(a) perimeter a square = 4 × side of the square

= 30 = 4 × side of the square

∴ Side of the square = 30/4 = 7.5 cm

Side of the square = 7.5 cm.

(b) perimeter an equilateral triangle = 3 × side of the ∆

30 = 3 × side of the ∆

Side of the triangle = 30/3 = 10 cm

Side of the equilateral ∆ = 10 cm.

(c) perimeter of a regular hexagon = 6 × side of the regular hexagon

30 = 6 × side

Side = 30/6 = 5 cm

Side of the regular hexagon = 5 cm.

Height of the label = 20 cm − 2 cm − 2 cm = 16 cm

Radius of the label = (14/2) cm = 7cm

Label is in the form of a cylinder having its radius and height as 7 cm and 16 cm.

Area of the label = 2π (Radius) (Height)

= { 2 x (22/7) x 7 x16 } cm² = 704 cm²

We find area when a region covered by a boundary, such as outer and inner surface area of a cylinder, a cone, a sphere and surface of wall or floor. 

When the amount of space occupied by an object such as water, milk, coffee, tea, etc., then we have to find out volume of the object. 

(a) Volume (b) Surface are (c) Volume

(a) Total area of the region = 100 × 144 = 14400 cm²

Area of one tile = 12 × 5 = 60 cm²

Number of tiles required = 14400/60 = 240

Therefore, 240 tiles are required.

(b) Total area of the region = 70 × 36 = 2520 cm²

Area of one tile = 60 cm²

Number of tiles required  = 2520/60 = 42

(D) 50cm

From the question it is given that, perimeter of each hexagon = 30 cm

We know hat, hexagon has 6 sides.

So, length of each side of hexagon = 30/6

= 5 cm

Now, consider two hexagons are joined together.

Then, perimeter of new figure = number of side × length of each side

= 10 × 6

= 50 cm

(A) Perimeter remains same but area changes.

We know that, area of big rectangle = length × breadth

= 10 × 20

= 200 cm²

Area of small rectangle = 5 × 2

= 10 cm²

Perimeter of rectangle = 2(length + breadth)

= 2(20 + 10)

= 2 × 30

= 60 cm

Then, perimeter of new figure = 20 + 8 + 5 + 2 + 15 + 10

= 60

Area of new figure = Area of big rectangle – Area of new figure

= 200 – 10

= 190 cm²

By comparing all the results, Perimeter remains same but area changes.

False

Three cubes having side x are joined face-to-face, then the cuboid so formed has the same height and breadth as the cubes but its length will be thrice that of the cubes.

Hence, the length, breadth and height of the cuboid so formed are 3x, x and x, respectively. Then, its surface area = 2 (lb + bh + hl)

= 2(3x x x + x x x + x x 3 x )

= 2(3x² + x² + 3x²)

= 2 x 7x² = 14x²

Now, the surface area of the cube of side x = 6 (Side)² = 6x²

Hence, the statement is false.

False

A cuboid has rectangular faces with different lengths and breadths. Only opposite faces of cuboid have the same length and breadth.

Therefore, areas of only opposite faces of a cuboid are equal.

True

Since, all the faces of a cube are squares of same side length, therefore the areas of any two faces of a cube are equal.

Same

Explanation:

For cylinder A, h= 10 cm and r = 10/π

Thus, CSA of cylinder A = 2πrh = 200

For cylinder B, h= 12 cm and r = 5/π

Thus, CSA of cylinder B = 2πrh = 200

Lateral

Explanation:

We know that, the rooms are in the cuboid shape. The walls are considered as the lateral faces of the cuboid shaped room.

• Perimeter-Length of the boundary of a simple closed figure
• Area-Lateral Surface area and Total surface area
• Volumes-Amount of space occupied by a solid
• Different types of shapes and solids such as rectangle, square, triangle, parallelogram, circle, trapezium, rhombus, cube, cuboid, cylinder.

Volume

Explanation:

The space occupied by any solids or three dimensional shaped are always measured in terms of volume.

(a) False – Boundry wall is around her house, so she must know the perimeter of the land and not the area. 

(b) True – Track is prepared along the boundary of the sports ground.

Given,

Volume of cylinder 1 = volume of cylinder 2

Ratio of their height = 1 : 2

= \(\frac{h_1}{h_2}\) = \(\frac{1}{2}\)

We have,

= V₁ = V₂

= πr¹₂h₁ = πr²₂h₂

= \(\frac{r^2_1}{r^2_2}\) = \(\frac{2}{1}\)

= \(\frac{r_1}{r_2}\) = \(\sqrt{\frac{2}{1}}\) = \({\frac{\sqrt2}{1}}\)

Volume of the cylinder is given by:

\(V = \pi r^2h\)

Where r is the radius and h is the height of the cylinder.

We know that:

\(V_1 = V_2 \space\text{and}\space h_1 = 2h_2\)

Therefore:

\(\pi r_1^22h_2 = \pi r_2^2h_2\implies \frac{r_2}{r_1}=\sqrt2\)

Given,

Height of cylinder = 10.5 m

= 3( A + A) = 2 curved surface area (A = circular area of box)

= 3 x 2A = 2(2πrh)

= 6πr² = 4 πrh

= r = \(\frac{2h}{3}\) = \(\frac{2\times 10.5}{3}\) = 7 m

Volume of cylinder = πr²h = \(\frac{22}{7}\)x 7 x 7 x 10.5 = 1617 cm³

\(\text{Volume of the cylinder is given by:}\\\text{V= area of circular face x height}\equiv A\cdot h\\\text{We know that A }= \pi r^2\text{ and }B\equiv\text{area of curved surface }=2\pi rh\\\text{Therefore:}\\\text{If }3\cdot(2A) = 2B\\\text{then}\\6\pi r^2 = 4\pi rh\implies r=\frac23h\\\text{It follows that:}\\V=\pi\frac49h^3=\pi\frac492^3=\frac{32}9\pi\)

Given,

Height of cylinder = 21 m

Diameter of well = 6 m

radius of well = \(\frac{6}{2}\) = 3 m

so,

Amount of earth can be dug out from this well = πr²h = \(\frac{22}{7}\) x 3 x 3 x 21 = 594 m³

The correct answer is option (a) 12 cm

Explanation:

The area of a parallelogram = base x altitude

b. h = A

b (5) = 60

b = 60/5

b = 12 cm

The correct answer is option (c) 24 cm²

Explanation:

From the given figure, the diagonal AC divides the rhombus into two triangles of equal area.

Therefore, the area of a triangle ABC = (1/2) bh

= (1/2) (4) (6)

=12 cm²

Therefore, the area of a rhombus ABCD = 2(12) = 24 cm²

The correct answer is option (B) 36 cm²

Explanation:

From the given figure, it is clear that, a quadrilateral ABCD is a parallelogram. Here, the diagonal AC divides the parallelogram into two equal triangles.

Hence, the area of a triangle ABC = (1/2) bh

Here, b = 12 and h = 3

= (1/2) (12)(3)

= 18

Therefore, the area of a parallelogram ABCD = 2 (18) = 36 cm²

The correct answer is option (a) 27x³

Explanation:

The volume of a cube is (side)³

V = (3x)³

V = 27x³

The correct answer is option (c) πr³

Explanation:

The volume of cylinder = πr² h

Given that r = h

Then, the volume of cylinder = πr² (r)

V = πr³

The correct answer is option (c) 1000

Explanation:

We know that, the volume of cube is (side)³

Therefore, the volume of each small cube is (20)³

= 8000 cm³

When it is converted into m³, we get

V = 0.008 m³

It is given that, the volume of the cuboidal box is 2³ = 8 m³

Now, the number of small cubes that can be accommodated in the cuboidal box is

= 8/ 0.008 = 1000

The correct answer is option (d) one third

Explanation:

We know that the curved surface area of a cylinder is 2πrh, when the radius is “r” and height is “h”.

Let “H” be the new height.

When the radius of a cylinder is tripled, then the CSA of a cylinder becomes,

CSA = 2π (3r) H

CSA = 6πr. H

Now, compare the CSA of the cylinder to find the height

6πrH = 2πrh

H = 2πrh / 6πr

H = (1/3)h

Hence, the new height of the cylinder is one-third of the original height.

1cm × 19cm, 2cm × 18cm, 3cm × 17cm, 4cm × 16cm, 5cm × 15cm, 6cm × 14cm, 7cm × 13cm, 8cm × 12cm, 9cm × 11cm, 10cm × 10cm

Length of each side of the base is 10 cm.

Perimeter of rectangular frame = Length of the aluminium strip

⇒ 2(length + breadth) = 40 cm

⇒ length + breadth = 40/2 cm = 20 cm

∴ The possible measurement of rectangular frames are 1 cm × 19 cm, 2 cm × 18 cm, 3 cm × 17 cm, 4 cm × 16 cm, 5 cm × 15 cm, 6 cm × 14 cm, 7 cm × 13 cm, 8 cm × 12 cm, 9 cm × 11 cm, 10 cm × 10 cm

Correct option is: B) 1 : 3

\(\frac {volume \, of \, cone}{volumes \, of\, cylinder} = \frac {\frac 13 \pi r^2h}{\pi r^2h} = \frac 13 = 1 : 3\)

\(\therefore\) The ratio between the volume of a cone and a cylinder is 1 : 3.

Correct option is: B) 1 : 3

Correct option is: C) 4 : 3

Surface area of sphere = \(4 \pi r^2\) 

Surface area of hemisphere = \(3\pi r^2\)

\(\therefore\) \(\frac {Surface \, area \, of\, sphere}{surface \, area \, of \, hemisphere}\) = \(\frac {4 \pi r^2}{3 \pi r^2} = \frac 43 = 4: 3\)

Hence, the ratio between the surface area of sphere and hemisphere is 4 : 3.

Correct option is: C) 4 : 3

Correct option is: D) All the above

Perimeter of square = 4 × side 20

= 4 × side

side = 20/4 = 5m

∴ Side of the square = 5 m

Perimeter of triangle = sum of lengths of all sides of the triangle

∴ Perimeter of ∆le = 10 + 14 + 15 = 39 cm.

Perimeter of regular hexagon = 6 × side of regular hexagon

= 6 × 8 m

∴ Perimeter of regular hexagon = 48 m

Distance traveled by the wheel is one revolution = 2πr

=2 x 22/7 x 28/2=88 cm

and the total distance covered by the wheel = 13.2 × 1000 × 100 cm

= 1320000 cm

So, Number of revolution made by the wheel = 1320000/88

= 15000.

(C) 100 cm

Volume of cube = (side)³ 

∴ 1 = (side)³ 

∴ Side = 1 m 

= 100 cm

Given : Radius (r) = 10 cm, 

Measure of the arc (θ) = 54° 

To find : Area of the sector. 

Area of sector = θ/360 x πr²

= 54/360 × 3.14 × (10)² 

= 3/20 × 3.14 × 100 

= 3 × 3.14 × 5 

= 15 × 3.14 

= 47.1 cm²

∴ The area of the sector is 47.1 cm².