# Visualizing Solid Shapes-NCERT Solutions

Class VIII Math
NCERT Solution for Visualising Solid Shapes
EXERCISE: 10.1
Q1.   For each of the given solid, the two views are given. Match for each solid the corresponding top and front views. The first one is done for you.
Sol:   (a)            →           (iii)           →            (iv)
(b)           →           (i)              →            (v)
(c)           →           (iv)            →            (ii)
(d)           →           (v)             →            (iii)
(e)           →           (ii)             →            (i)
Q2.   For each of the given solid, the three views are given. Identify for each solid the corresponding top, front and side views.
Q3.   For each given solid, identify the top view, front view and side view.
Q4.   Draw the front view, side view and top view of the given objects.
Solution:
EXERCISE: 10.2
Q1.   Look at the given map of a city.
(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
Q2.   Draiv a map of your class using proper scale and symbols for derent objects.
Sol:   An activity, do it yourself.
Q3.   Draw a map of your school compound using proper scale and symbols for various features like play ground main building, garden, etc.
Sol:   It is an activity. Please do it yourself.
Q4.   Draw a map giving instructions to your friend so that she reaches your house without any difficulty.
Sol:   It is an activity. Please do it yourself.
EXERCISE: 10.3 (Page 166)
Q1.   Can a polyhedron have for its faces
(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.
Q2.   Is it possible to have a polyhedron with any given number of faces?
Hint: Think of a pyramid.
Sol:   Yes, it can be possible only if the number of faces is four or more than four.
Q3.   Which are prisms among the following?
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.
Q4.   (i) How are prisms and cylinders alike?
(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.
Q5.   Is a square prism same as a cube? Explain.
Sol:   No, not always, because it can be a cuboid also.
Q6.   Verify Euler's formula for these solids.
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.
Q7.   Using Euler's formula find the unknown.

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
Q8.   Can polyhedron have 10 faces, 20 edges and 15 vertices?
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.