Recognition of Solids

A net of a cube is made up of six equal squares, which on folding form the six faces of the cube in such a way that no two squares overlap and no face of the cube is left open.

On folding the nets (i), (ii) and (v), each of the six squares forms a different face of the cube and a cube is obtained.

On folding the nets (iii) and (iv), two of the squares overlap each other and one face of the cube is left open. So, a cube cannot be formed from them.

Hence, the nets (i), (ii) and (v) can be used to form cubes.

A pattern that can be cut and folded to make a model of a solid shape, is called a net.

Three different nets for making a cube are shown below:

On folding each net into a cube, the following pairs of faces are opposite:

(i) face a comes opposite to face a′.

(ii) face b comes opposite to face b′.

(iii) face c comes opposite to face c′.

(i) The shadow is a square. A cube has six square faces and can cast a square shadow when viewed perpendicular to one of its faces.

Hence, the object is a cube.

(ii) The shadow is a rectangle. A cuboid has rectangular faces and can cast a rectangular shadow when viewed perpendicular to one of its faces.

Hence, the object is a cuboid.

(iii) The shadow is a triangle. A triangular prism can cast a triangular shadow when viewed from one of its triangular ends. Although a triangular shadow alone does not uniquely identify the solid, the intended object here is a prism.

Hence, the object is a prism.

By Euler's formula,

V + F - E = 2

In the first column,

F = a, V = 6 and E = 12

⇒ 6 + a - 12 = 2

⇒ a - 6 = 2

⇒ a = 2 + 6

⇒ a = 8

In the second column,

F = 5, V = b and E = 9

⇒ b + 5 - 9 = 2

⇒ b - 4 = 2

⇒ b = 2 + 4

⇒ b = 6

In the third column,

F = 20, V = 12 and E = c

⇒ 12 + 20 - c = 2

⇒ 32 - c = 2

⇒ c = 32 - 2

⇒ c = 30

In the fourth column,

F = 6, V = d and E = 12

⇒ d + 6 - 12 = 2

⇒ d - 6 = 2

⇒ d = 2 + 6

⇒ d = 8

Hence, a = 8, b = 6, c = 30 and d = 8.

Opposite faces of a die have a total of seven dots, so the pairs of numbers on opposite faces are 1 and 6, 2 and 5, and 3 and 4.

(i) On folding the net, in the row of four squares, the first square comes opposite to the third square and the second square comes opposite to the fourth square. Also, the square above the row comes opposite to the square below the row.

The second and the fourth squares have 2 and 5 dots, and 2 + 5 = 7.

The third square has 4 dots, so the first square must have 7 - 4 = 3 dots.

The square below the row has 6 dots, so the square above the row must have 7 - 6 = 1 dot.

The completed net is shown below:

Hence, the blanks are filled with 1 and 3.

(ii) On folding the net, the face with 1 dot comes opposite to the face at the right end of the bottom row, the face with 2 dots comes opposite to the face at the left of the face with 3 dots, and the face with 3 dots comes opposite to the face at the left end of the bottom row.

Number on the face opposite to the face with 1 dot = 7 - 1 = 6

Number on the face opposite to the face with 2 dots = 7 - 2 = 5

Number on the face opposite to the face with 3 dots = 7 - 3 = 4

The completed net is shown below:

Hence, the blanks are filled with 5, 4 and 6.

(i) The given net is made up of six equal squares, which on folding form six square faces of equal size.

A cube has six equal square faces.

Hence, the solid is a cube.

(ii) The given net is made up of six rectangles, which on folding form six rectangular faces at right angles to each other, with the opposite faces being equal.

A cuboid has six rectangular faces at right angles to each other.

Hence, the solid is a cuboid.

Steps:

1. Measure the length and the breadth of the classroom and the sizes of the different objects in it. Let the classroom be 9 m long and 7 m broad, with 12 dual desks for 24 students, one teacher's table, one blackboard, one door and two windows.

2. Choose a suitable scale. As the classroom is 9 m long, take scale : 1 cm = 1 m.

3. Convert the actual measurements into map measurements using the scale:

Object	Actual size	Size on the map
Classroom	9 m × 7 m	9 cm × 7 cm
Blackboard	3 m long	3 cm long
Teacher's table	1.5 m × 0.75 m	1.5 cm × 0.75 cm
Student's desk (dual)	1.2 m × 0.6 m	1.2 cm × 0.6 cm
Door	1 m wide	1 cm wide
Window	1.5 m wide	1.5 cm wide

4. Draw a rectangle of length 9 cm and breadth 7 cm to represent the classroom.

5. Choose a symbol for each object. Draw the blackboard in the middle of the front wall and the teacher's table in front of it. Draw the 12 desks in 4 rows of 3 desks each, leaving aisles between them. Draw the door on the right wall near the front and the two windows on the left wall.

6. Write the scale and the meaning of each symbol (legend) on the map.

The map of the classroom is shown below:

Steps:

1. Find the length and the breadth of the school compound and the sizes of its various features. Let the school compound be 150 m long and 100 m broad, with a main building, a playground, a garden, a canteen, a parking area, a main gate and a path from the gate to the main building.

2. Choose a suitable scale. As the compound is 150 m long, take scale : 1 cm = 10 m.

3. Convert the actual measurements into map measurements using the scale. For example,

Length of the compound on the map = $\frac{150}{10}$ cm = 15 cm

The other measurements are converted in the same way:

Feature	Actual size	Size on the map
School compound	150 m × 100 m	15 cm × 10 cm
Main building	60 m × 20 m	6 cm × 2 cm
Playground	70 m × 50 m	7 cm × 5 cm
Garden	30 m × 20 m	3 cm × 2 cm
Canteen	20 m × 15 m	2 cm × 1.5 cm
Parking area	30 m × 15 m	3 cm × 1.5 cm
Main gate	6 m wide	0.6 cm wide

4. Draw a rectangle of length 15 cm and breadth 10 cm to represent the boundary of the school compound.

5. Choose a symbol for each feature. Draw the main gate on the front (south) wall and the path from the gate to the main building. Draw the main building near the north wall, the canteen in the north-east corner, the playground on the east side, and the garden and the parking area on the west side.

6. Write the scale, the direction (North) and the meaning of each symbol (legend) on the map.

The map of the school compound is shown below:

Given, scale : 1 cm = 12 km

∴ Actual distance represented by 13.8 cm = 13.8 × 12 km = 165.6 km

Hence, the actual distance between the two cities = 165.6 km.

For a polyhedron, the relation between V, E and F is given by Euler's formula as:

V + F - E = 2

Hence, option 1 is the correct option.

Given:

E = 12, F = 6

For a convex polyhedron, by Euler's formula,

V + F - E = 2

⇒ V + 6 - 12 = 2

⇒ V - 6 = 2

⇒ V = 2 + 6

⇒ V = 8

Hence, option 4 is the correct option.

The given net is made up of six equal squares. On folding it, the six squares form the six equal faces of a solid.

A cube has six equal square faces.

So, the given figure shows a net pattern of a cube.

Hence, option 2 is the correct option.

The given net is made up of four equal equilateral triangles. On folding it, the middle triangle forms the base and the other three triangles meet at a common vertex above it.

So, the solid formed has 4 triangular faces, i.e., it is a triangular pyramid.

Hence, option 4 is the correct option.

Number of unit cubes along the length = 4

Number of unit cubes along the breadth = 4

Number of unit cubes along the height = 2

∴ Number of unit cubes = 4 × 4 × 2 = 32

Hence, option 4 is the correct option.

The given solid has two identical hexagonal faces, one at the top and one at the bottom, joined by rectangular faces.

A solid having two identical hexagonal ends joined by rectangular faces is a hexagonal prism.

Hence, option 2 is the correct option.

A cube has six faces, twelve edges and eight vertices.

A solid having 6 faces is called a hexahedron, and a cube is a hexahedron.

A hexagonal prism has 8 faces, 18 edges and 12 vertices.

So, the given 3D figure is a cube as well as a hexahedron.

Hence, option 4 is the correct option.

An isometric sheet is made up of dots arranged in such a way that on joining the nearest dots, equilateral triangles are formed.

Hence, option 3 is the correct option.

A square pyramid has 1 square base and 4 triangular faces.

Number of faces = 1 + 4 = 5

It has 4 vertices at the corners of the base and 1 vertex at the top.

Number of vertices = 4 + 1 = 5

Hence, option 4 is the correct option.

A cuboid has six faces, twelve edges and eight vertices (corners).

Hence, option 1 is the correct option.

By observing the given net, we find that on folding it, a rectangular solid having dimensions 3 cm × 2 cm × 6 cm is formed.

So, Statement 1 is true.

A cuboid has six rectangular faces. Each pair of opposite faces has the same dimensions and shape.

So, Statement 2 is true.

∴ Both the Statements are true.

Hence, option 1 is the correct option.

When three cubes of side 2 cm are placed side by side, the resulting solid has:

Length = 2 cm + 2 cm + 2 cm = 6 cm

Breadth = 2 cm

Height = 2 cm

As its length, breadth and height are not all equal, the resulting solid is a cuboid of dimensions 6 cm × 2 cm × 2 cm and not a cube of side 6 cm.

So, Statement 1 is false.

The length, breadth and height of a cuboid need not all be equal.

So, Statement 2 is true.

∴ Statement 1 is false, and Statement 2 is true.

Hence, option 4 is the correct option.

Number of cubes that can be placed along the length = $\frac{100}{10}$ = 10

Number of cubes that can be placed along the breadth = $\frac{50}{10}$ = 5

Number of cubes that can be placed along the height = $\frac{55}{10}$ = 5.5, i.e., only 5 complete cubes can be placed, as a part of a cube cannot be fitted.

∴ Number of cubes that can be fitted in the carton = 10 × 5 × 5 = 250

So, Assertion (A) is false.

Volume of carton = 100 × 50 × 55 cm³ = 275000 cm³

Volume of each cube = 10 × 10 × 10 cm³ = 1000 cm³

$\frac{\text{Volume of cuboidical carton}}{\text{Volume of each cube}} = \frac{275000}{1000}$ = 275, which is not the number of cubes that can actually be fitted in the carton.

So, Reason (R) is false.

∴ Both A and R are false.

Hence, option 4 is the correct option.

In an ordinary die, opposite faces always have a total of seven dots on them.

So, Reason (R) is true.

From the figure, the numbers on the two faces shown are 5 and 6.

Number on the face opposite to the face having 5 = 7 - 5 = 2

Number on the face opposite to the face having 6 = 7 - 6 = 1

Sum of the two numbers on the opposite faces = 2 + 1 = 3

So, Assertion (A) is true.

∴ Both A and R are true.

Hence, option 3 is the correct option.

The given diagram is made up of a row of four rectangles of widths 3 cm, 5 cm, 3 cm and 5 cm, each of height 4 cm, and two rectangles of dimensions 5 cm × 3 cm, one above and one below the row.

On folding it, the four rectangles in the row form the four side faces and the other two rectangles form the top and the bottom faces of a cuboid of dimensions 5 cm × 3 cm × 4 cm.

So, the given diagram is the net of a solid (cuboid) and Assertion (A) is true.

The line segment joining two opposite vertices of a rectangular solid, which passes through its interior, is called its space diagonal.

So, Reason (R) is true.

∴ Both A and R are true.

Hence, option 3 is the correct option.