If F = 10, V = 15 and E = 20. Then, we know Using Euler’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 Euler’s formula. Class 8 Maths Chapter 10 Exercise 10.3 Solution in Video for more answers vist to: https://www.tiRead more
If F = 10, V = 15 and E = 20.
Then, we know Using Euler’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 Euler’s formula.
Class 8 Maths Chapter 10 Exercise 10.3 Solution in Video
In first column, F = ?, V = 6 and E = 12 Using Euler’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 Euler’s formula, we see F + V – E = 2 F + V – E = 2 ⇒ 5 + V – 9 = 2 ⇒ V – 4 = 2 ⇒ VRead more
In first column, F = ?, V = 6 and E = 12
Using Euler’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 Euler’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 Euler’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.
Class 8 Maths Chapter 10 Exercise 10.3 Solution in Video
(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 – 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 Euler’s formulRead more
(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 – 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 Euler’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.
Class 8 Maths Chapter 10 Exercise 10.3 Solution in Video
No, it can be a cuboid also. Class 8 Maths Chapter 10 Exercise 10.3 Solution in Video for more answers vist to: https://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-10/
No, it can be a cuboid also.
Class 8 Maths Chapter 10 Exercise 10.3 Solution in Video
(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 Solution in Video for more answers vist to: https://www.tiwariacademy.com/ncRead more
(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 Solution in Video
It is possible, only if the number of faces are greater than or equal to 4. Class 8 Maths Chapter 10 Exercise 10.3 Solution in Video for more answers vist to: https://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-10/
It is possible, only if the number of faces are greater than or equal to 4.
Class 8 Maths Chapter 10 Exercise 10.3 Solution in Video
(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 SoluRead more
(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 Solution in Video
Do yourself. See Hindi Medium solutions on Tiwari Academy. Class 8 Maths Chapter 10 Exercise 10.2 Solution in Video for more answers vist to: https://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-10/
Do yourself. See Hindi Medium solutions on Tiwari Academy.
Class 8 Maths Chapter 10 Exercise 10.2 Solution in Video
Can a polyhedron have 10 faces, 20 edges and 15 vertices?
If F = 10, V = 15 and E = 20. Then, we know Using Euler’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 Euler’s formula. Class 8 Maths Chapter 10 Exercise 10.3 Solution in Video for more answers vist to: https://www.tiRead more
If F = 10, V = 15 and E = 20.
Then, we know Using Euler’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 Euler’s formula.
Class 8 Maths Chapter 10 Exercise 10.3 Solution in Video
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-10/
Using Euler’s formula, find the unknown:
In first column, F = ?, V = 6 and E = 12 Using Euler’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 Euler’s formula, we see F + V – E = 2 F + V – E = 2 ⇒ 5 + V – 9 = 2 ⇒ V – 4 = 2 ⇒ VRead more
In first column, F = ?, V = 6 and E = 12
Using Euler’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 Euler’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 Euler’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.
Class 8 Maths Chapter 10 Exercise 10.3 Solution in Video
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-10/
Verify Euler’s formula for these solids.
(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 – 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 Euler’s formulRead more
(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 – 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 Euler’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.
Class 8 Maths Chapter 10 Exercise 10.3 Solution in Video
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-10/
Is a square prism same as a cube? Explain.
No, it can be a cuboid also. Class 8 Maths Chapter 10 Exercise 10.3 Solution in Video for more answers vist to: https://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-10/
No, it can be a cuboid also.
Class 8 Maths Chapter 10 Exercise 10.3 Solution in Video
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-10/
(i) How are prisms and cylinders alike? (ii) How are pyramids and cones alike?
(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 Solution in Video for more answers vist to: https://www.tiwariacademy.com/ncRead more
(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 Solution in Video
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-10/
Is it possible to have a polyhedron with any given number of faces? (Hint: Think of a pyramid)
It is possible, only if the number of faces are greater than or equal to 4. Class 8 Maths Chapter 10 Exercise 10.3 Solution in Video for more answers vist to: https://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-10/
It is possible, only if the number of faces are greater than or equal to 4.
Class 8 Maths Chapter 10 Exercise 10.3 Solution in Video
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-10/
Can a polygon have for its faces: (i) 3 triangles (ii) 4 triangles (iii) a square and four triangles
(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 SoluRead more
(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 Solution in Video
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-10/
Draw a map of your class room using proper scale and symbols for different objects.
Do yourself. See Hindi Medium solutions on Tiwari Academy. Class 8 Maths Chapter 10 Exercise 10.2 Solution in Video for more answers vist to: https://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-10/
Do yourself. See Hindi Medium solutions on Tiwari Academy.
Class 8 Maths Chapter 10 Exercise 10.2 Solution in Video
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-10/