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# NCERT Solutions for Class 8 Maths Exercise 10.2 ### myCBSEguide App

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NCERT solutions for Class 8 Maths Visualising Solid Shapes ## NCERT Solutions for Class 8 Maths Visualising Solid Shapes

Class –VIII Mathematics (Ex. 10.2)

NCERT SOLUTION

1. Can a polygon have for its faces:

(i) 3 triangles

(ii) 4 triangles

(iii) a square and four triangles

Ans. (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.

###### 2. Is it possible to have a polyhedron with any given number of faces? (Hint: Think of a pyramid)

Ans. It is possible, only if the number of faces are greater than or equal to 4.

NCERT Solutions for Class 8 Maths Exercise 10.2

###### 3. Which are prisms among the following: Ans. Figure (ii) unsharpened pencil and figure (iv) a box are prisms.

NCERT Solutions for Class 8 Maths Exercise 10.2

###### 4. (i) How are prisms and cylinders alike?

(ii) How are pyramids and cones alike?

Ans. (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.

###### 5. Is a square prism same as a cube? Explain.

Ans. No, it can be a cuboid also.

NCERT Solutions for Class 8 Maths Exercise 10.2

###### 6. Verify Euler’s formula for these solids. Ans. (i) Here, figure (i) contains 7 faces, 10 vertices and 15 edges.

Using Eucler’s formula, we see

F + V – E = 2

Putting F = 7, V = 10 and E = 15,

F + V – E = 2 7 + 10 – 5 = 2 17 – 15 = 2 2 = 2 L.H.S. = R.H.S.

(ii)  Here, figure (ii) contains 9 faces, 9 vertices and 16 edges.

Using Eucler’s formula, we see

F + V – E = 2

F + V – E = 2 9 + 9 – 16 = 2 18 – 16 = 2 2 = 2 L.H.S. = R.H.S.

Hence verified Eucler’s formula.

NCERT Solutions for Class 8 Maths Exercise 10.2

###### 7. Using Euler’s formula, find the unknown:
 Faces ? 5 20 Vertices 6 ? 12 Edges 12 9 ?

Ans. In first column, F = ?, V = 6 and

E = 12

Using Eucler’s formula, we see

F + V – E = 2

F + V – E = 2 F + 6 – 12 = 2 F – 6 = 2 F = 2 + 6 = 8

Hence there are 8 faces.

In second column, F = 5, V = ? and E = 9

Using Eucler’s formula, we see

F + V – E = 2

F + V – E = 2 5 + V – 9 = 2 V – 4 = 2 V = 2 + 4 = 6

Hence there are 6 vertices.

In third column, F = 20, V = 12 and E = ?

Using Eucler’s formula, we see

F + V – E = 2

F + V – E = 2 20 + 12 – E = 2 32 – E = 2 E = 32 – 2 = 30

Hence there are 30 edges.

NCERT Solutions for Class 8 Maths Exercise 10.2

###### 8. Can a polyhedron have 10 faces, 20 edges and 15 vertices?

Ans. If F = 10, V = 15 and E = 20.

Then, we know Using Eucler’s formula,

F + V – E = 2

L.H.S. = F + V – E

= 10 + 15 – 20

= 25 – 20

= 5

R.H.S.  = 2 L.H.S. R.H.S.

Therefore, it does not follow Eucler’s formula.

So polyhedron cannot have 10 faces, 20 edges and 15 vertices.

## NCERT Solutions for Class 8 Maths Exercise 10.2

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