Exercise-10.3
class 8 maths chapter 10 exercise 10.3 (1)
1. Can a polyhedron have for its faces
(i) 3 triangles? (ii) 4 triangles?
(iii) a square and four triangles?
Solution to class 8 maths chapter 10 exercise 10.3 (1)
(i) No, a polyhedron cannot have 3 triangles for its faces.
(ii) Yes, a polyhedron can have four triangles which is known as pyramid on triangular base
(iii) Yes, a polyhedron has its faces a square and four triangles which makes a pyramid on square base.
class 8 maths chapter 10 exercise 10.3 (2)
2. Is it possible to have a polyhedron with any given number of faces? (Hint: Think of a pyramid).
Solution to class 8 maths chapter 10 exercise 10.3 (2)
It is possible, only if the number of faces are greater than or equal to 4.
class 8 maths chapter 10 exercise 10.3 (3)
3. Which are prisms among the following?
Solution to class 8 maths chapter 10 exercise 10.3 (3)
Figure (ii) unsharpened pencil
and figure (iv) a box are prisms.
class 8 maths chapter 10 exercise 10.3 (4)
4. (i) How are prisms and cylinders alike?
(ii) How are pyramids and cones alike?
Solution to class 8 maths chapter 10 exercise 10.3 (4)
(i) A prism becomes a cylinder as the number of sides of its base becomes larger and larger
(ii) A pyramid becomes a cone as the number of sides of its base becomes larger and larger.
class 8 maths chapter 10 exercise 10.3 (5)
5. Is a square prism same as a cube? Explain.
Solution to class 8 maths chapter 10 exercise 10.3 (5)
No, a square prism may be cube or cuboid.
class 8 maths chapter 10 exercise 10.3 (6)
6. Verify Euler’s formula for these solids.
Solution to class 8 maths chapter 10 exercise 10.3 (6)
(i)Here, figure (i) contains 7 faces, 10 vertices and 15 edges. Using Euler’s formula, we see F+ V=E+2 Putting F = 7, V = 10 and E = 15,
F+ V=E+2
⇒7+ 10 = 15 +2
⇒17 =17
L.H.S. = RHS.
(ii)Here, figure (ii) contains 9 faces, 9 vertices and 16 edges. Using Euler’s formula, we see F+ V=E+2 Putting F = 9, V = 9 and E = 16,
F+ V=E+2
⇒9+ 9 = 16 +2
⇒18 =18
L.H.S. = RHS.
class 8 maths chapter 10 exercise 10.3 (7)
7. Using Euler’s formula find the unknown.
Solution to class 8 maths chapter 10 exercise 10.3 (7)
In first column F=? V=6 and E=12
Using Euler’s formula, F+ V=E+2
F+ 6=12+2
⇒F+6=14
⇒F=14 –6=8
Hence there are 8 faces.
In second column F=5 V=? and E=9
Using Euler’s formula, F+ V=E+2
5+ V=9+2
⇒V+5=11
⇒V=11 –5=6
Hence there are 6 vertices.
In third column F=20 V=12 and E=?
Using Euler’s formula, F+ V=E+2
20+ 12=E+2
⇒32=E+2
⇒E=32–2=30
Hence there are 30 edges.
class 8 maths chapter 10 exercise 10.3 (8)
8. Can a polyhedron have 10 faces, 20 edges and 15 vertices?
If Faces=10, Edges=20 and Vertices=15
Then using Euler’s formula, F+ V=E+2
L.H.S=10+15=25
R.H.S=20+2=22
L.H.S≠R.H.S
a polyhedron can not have 10 faces, 20 edges and 15 vertices.
Since it doesn’t follow Euler’s formula.