NCERT Solutions of Chapter: Visualising Solid Shapes. NCERT Solutions along with worksheets and notes for Class 8.
Exercise 10.1
(1) 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.
Ans- The corresponding top and front views are matched with object and presented below-
(2) For each of the given solid, the three views are given. Idetity for each solid the corresponding top, front and side views.
Ans- The corresponding top, front and side views for each of the given solid are presented below-
(3) For each given soild, identify th etop view, front view and side view.
Ans- The corresponding top, front and side views for each of the given solid are presented below-
(4) Draw the front view, side view and top view of the given objects.
Ans- The front view, side view and top view of the given objects are presented in the table below-
Exercise 10.2
(1) Look at the given map of the city.
Answer the following.
(a) Colour the map as follows: blue – water, red – fire stattion, 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 St. Secondary School?
Ans-
(a) The required map is showing below-
(b) The required mark of ‘X’ and ‘Y’ are showing below-
(c) The required path is showing below-
(d) The city park is at the further east.
(e) The St. Secondary School is at the further south.
(2) Draw a map of your class room using proper scale and symbols for different objects?
Ans- A map of class room is showing below-
(3) Draw a map of your school compound using proper scale and symbols for various features like play ground mail building, garden etc.
Ans- A map of school compound is showing below-
(4) Draw a map giving instructions to your friend so that she reaches your house without any difficulty.
Ans-A map giving instructions to my friend so that she reaches my house without any difficulty is showing below-
Exercise 10.3
(1) Can a polyhedron have for its faces
(i) 3 triangles? (ii) 4 triangles? (iii) a square and four triangles?
Ans-
(i) No, for polyhedron four faces are necessary.
(ii) Yes, for polyhedron four faces are necessary.
(iii) Yes, for polyhedron four faces are necessary.
(2) Is it possible to have a polyhedron with any given number of faces? ( Hint: Think of a pyramid)
Ans-
Yes, but number of faces should be 4 or more than 4 faces.
(3) Which are prisms among the following?
(i) 
(ii)
A nail Unsharpened pencil
(iii)
(iv) 
A table weight A box
Ans-
(i) A nail is not a prism because its base (point) and top (circle)are not congruent and other faces are not parallelogram.
(ii) An unsharpened pencil is a prism because top (hexagon) and base (hexagon) are congruent and other lateral sides are parallelogram.
(iii) A table weight is not a prism because its base (four sides) and top (point) are not congruent and other faces are not parallelogram.
(iv) A box is a prism because top (rectangle) and base (rectangle) are congruent and other lateral sides are parallelogram.
(4) (i) How are prisms and cylinders alike?
(ii) How are pyramids and cones alike?
Ans-
(i) Prisms and cylinders are alike because top and base of both are congruent.
(ii) Pyramids and cones are alike because base of pyramid is poygon and baso cone is circle.
(5) Is a square prism same as a cube? Explain.
Ans-
A square prism can be a cube or cuboid.
If a square prism has lateral sides square in shape with same measurements then it will be cube.
If a square prism has lateral sides not square in shape then it will be cuboid.
(6) Verify Euler’s formula for these soilds.
(i) 
(ii) 
Ans-
Euler’s Formula faces (F), vertices (V) and edges (E) of polyhedron.
Euler’s Formula
F + V – E = 2
Where,
Faces = F
Vertices = V
Edges = E
(i)
Faces = 7
Vertices = 10
Edges = 15
F + V – E = 2
7 + 10 – 15 =2
2 = 2
LHS = RHS
Hence, Euler’s theorem proved.
(ii)
Faces = 10
Vertices = 9
Edges = 16
F + V – E = 2
10 + 9 – 16 =2
3≠ 2
LHS ≠ RHS
Hence, Euler’s theorem is not proved.
(7) Using Euler’s formula find the unknown.
| Faces | ? | 5 | 20 |
| Vertices | 6 | ? | 12 |
| Edges | 12 | 9 | ? |
Ans-
| Faces | 8 | 5 | 20 |
| Vertices | 6 | 6 | 12 |
| Edges | 12 | 9 | 30 |
When,
Faces = x
Vertices = 6
Edges = 12
F + V – E = 2
x + 6 -12 =2
x -6 = 2
x = 8
when,
Faces = 5
Vertices = x
Edges = 9
F + V – E = 2
5 + x – 9 =2
x – 4 = 2
x = 6
when,
Faces = 20
Vertices = 12
Edges = x
F + V – E = 2
20 + 12 – x =2
32 – x = 2
x = 30
(8) Can a polyhedron have 10 faces, 20 edges and 15 vertices?
Ans-
Faces = 10
Vertices = 15
Edges = 20
Euler’s Theorem
F + V – E = 2
LHS
=10 + 15 – 20
= 25 – 20
= 5
≠
RHS
No, a polyhedron can not have 10 faces, 20 edges and 15 vertices.
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