NCERT Solution for Visualising Solid Shapes

Sol: (a) → (iii) → (iv)

(b) → (i) → (v)

(c) → (iv) → (ii)

(d) → (v) → (iii)

(e) → (ii) → (i)

Solution:

Answer the following.

(a) Colour the map as follows: Blue-water, red-fire station, orange-library, yellow-schools, Green park, Pink-College, Purple-Hospital, Brown-Cemetery.

(b) Mark a green X' at the intersection of Road 'C' and Nehru Road, Green `Y' at the intersection of Gandhi Road and Road A.

(c) In red, draw a short street route from library to the bus depot.

(d) Which is further east, the city park or the market?

(e) Which is further south, the primary school or the Sr. Secondary School?

Sol: The shaded (coloured) map according to the required directions is given in part (a). Here various colours are as under

(b) Do as directed.

(c) Do as directed.

(d) City park

(e) Senior Secondary school

Sol: An activity, do it yourself.

Sol: It is an activity. Please do it yourself.

Sol: It is an activity. Please do it yourself.

(i) 3 triangles?

(ii) 4 triangles?

(iii) a square and four triangles?

Sol: A polyhedron is bounded by four or more than four polygonal faces.

(i) No, it is not possible that a polyhedron has 3 triangles for its faces.

(ii) Yes, 4 triangles can be the faces of a polyhedron.

(iii) Yes, a square and 4 triangles can be the faces of a polyhedron.

Hint: Think of a pyramid.

Sol: Yes, it can be possible only if the number of faces is four or more than four.

Sol: Since, a prism is a polyhedron having two of its faces congruent and parallel, where as other faces are parallelogram.

(i) No, a nail is not a prism.

(ii) Yes, unsharpened pencil is a prism.

(iii) No, table weight is not a prism.

(iv) Yes, box is a prism.

(ii) How are pyramids and cones alike?

Sol: (i) Both of the prisms and cylinders have their base and top as congruent faces and parallel to each other. Also, a prism becomes a cylinder as the number of sides of its base becomes larger and larger.

(ii) The pyramid and cones are alike becomes their lateral faces meet at a vertex. Also a pyramid becomes a cone as the number of sides of its base becomes larger and larger.

Sol: No, not always, because it can be a cuboid also.

Sol: (i) In figure (i), we have

F = 7, V = 10 and E = 15

∴F + V = 7 + 10 = 17

F + V – E = 17 – 15 = 2

i.e. F + V – E = 2

Thus, Euler’s formula is verified.

(ii) In figure (ii), we have

F = 9, V = 9 and E = 16

F + V = 9 + 9 = 18

and F + V – E = 18 – 16 = 2

i.e. F + V – E = 2

Thus, Euler's formula is verified.

Sol: (i) Here, V = 6 and E = 12

Since F + V – E = 2

∴F + 6 – 12 = 2

or F – 6 = 2

or F = 2 + 6 = 8

(ii) Here, F = 5 and E = 9

Since F + V – E = 2

or 5 + V – 9 = 2

or V – 4 = 2

or v = 2 + 4 = 6

(iii) Here F = 20 V = 12

Since, F + V – E = 2

∴20 + 12 – E = 2

or 20 + 12 – E = 2

32 – E = 2

or E = 32 – 2 = 30

Sol: Here, F = 10, E = 20 and V = 15

We have:

F + V – E = 2

10 + 15 – 20 = 2

or 25 – 20 = 2

or 5 = 2 which is not true

i.e. F + V – E 2

Thus, such a polyhedron is not possible.