Rd Sharma 2019 2020 for Class 8 Maths Chapter 19 - Visualising Shapes

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Rd Sharma 2019 2020 Solutions for Class 8 Maths Chapter 19 Visualising Shapes are provided here with simple step-by-step explanations. These solutions for Visualising Shapes are extremely popular among Class 8 students for Maths Visualising Shapes Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rd Sharma 2019 2020 Book of Class 8 Maths Chapter 19 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rd Sharma 2019 2020 Solutions. All Rd Sharma 2019 2020 Solutions for class Class 8 Maths are prepared by experts and are 100% accurate.

Page No 19.10:

Question 7:

Using Euler's formula find the unknown:

Faces	?	5	20
Vertices	6	?	12
Edges	12	9	?

Answer:

$We know that the Euler's formula is: F+V = E+2 (i) The number of vertices V is 6 and the number of edges E is 12. Using Euler's formula: F+6 = 12+2 F+6 = 14 F = 14-6 F = 8 So, the number of faces in this polyhedron is 8. (ii) Faces, F = 5 E dges, E = 9. W e have to find the number of vertices. Putting these values in Euler's formula: 5+V = 9+2 5+V = 11 V = 11-5 V = 6 So, the number of vertices in this polyhedron is 6. (iii) N umber of faces F = 20 N umber of vertices V = 12 Using Euler's formula: 20+12 = E+2 32 = E+2 E+2 = 32 E = 32-2 E = 30. So, the number of edges in this polyhedron is 30.$

Page No 19.12:

Question 1:

Which among the following are nets for a cube?

Answer:

$To create a cube, we need six equal faces that enclose a closed box. In the given figure, only (iv), (v) and (vi) are such nets that enclose a box when we fold each face from the edge.$

Page No 19.13:

Question 2:

Name the polyhedron that can be made by folding each net:

Answer:

$(i) If we fold the given figure from the edges, we'll get a pyramid with a square base.$

$(ii) If we fold the given polyhedron from the edges, we'll get a triangular prism. \frac{}{}$

$(iii) If we fold the given polyhedron from the edges, we'll get a triangular prism.$

$(iv) If we fold the given polyhedron from the edges, we'll get a hexagonal prism.$

$(v) If we fold the given net from the edges, we'll get a hexagonal pyramid.$

$(vi) If we fold the given net from the edges, we'll get a cuboid.$

Page No 19.13:

Question 3:

Dice are cubes where the numbers on the opposite faces must total 7. Which of the following are dice?

Answer:

$Among the given figures, only figure (i) is a dice. This is because if we fold the given net from the edges, we'll get a cube in which the sum of the opposite faces is 7.$

Page No 19.13:

Question 4:

Draw nets for each of the following polyhedrons:

Answer:

(i)

(ii)

(iii)

(iv)

Page No 19.14:

Question 5:

Match the following figures:

Answer:

$(a) The given figure is a cuboid with sides 4, 4 and 6 units. Area of a rectangle =length×width ∴ Area of the rectangular face sith sides 4 and 4 =4×4=16 And, area of the other face with sides 4 and 6 =4×6=24 Thus, the net for the given figure will have four faces with area 24 and two faces with area 16. Observe net (iv) satisfies this. Thus, the net of figure (a) is net (iv).$

$(b) The given figure is a cuboid with sides 3, 3 and 8. Area of a rectangle =length×width ∴ Area of the rectangular face sith sides 3 and 3 =3×3=9 And the area of the other face with sides 3 and 8 =3×8=24 Thus, the net for given figure will have four faces with area 24 and two faces with area 9. Observe that net (i) satisfies this. Thus, the net of figure (b) is net (i).$

$(c) The given figure is a cuboid with sides 3, 4 and 6. Area of a rectangle =length×width ∴ A rea of the rectangular face sith sides 3 and 4 =3×4=12, A rea of the rectangular face with sides 4 and 6 =4×6=24 A nd, area of the other face with sides 3 and 6 =3×6=18 Thus, the net for given figure will have two faces with area 24, two faces with area 18 and two faces with area 12. Observe that net (ii) satisfies this. Thus, the net of figure (c) is net (ii).$

$(d) The given figure is a cuboid with sides 3, 3 and 9. Area of a rectangle =length×width Area of the rectangular face with sides 3 and 3 =3×3=9, A nd, area of the other face with sides 3 and 9 =3×9=27 Thus, the net for given figure will have four faces with area 27 and two faces with area 9. Observe that net (iii) satisfies this. Thus, the net of figure (d) is net (iii).$

Page No 19.9:

Question 1:

What is the least number of planes that can enclose a solid? What is the name of the solid?

Answer:

$The least number of planes that can enclose a solid is 4. Tetrahedron is a solid with four planes (faces).$

Page No 19.9:

Question 2:

Can a polyhedron have for its faces:
(i) 3 triangles?
(ii) 4 triangles?
(iii) a square and four triangles?

Answer:

$(i) No, because in order to complete a polyhedron, we need at least four triangular faces. (ii) Yes, a polyhedron with 4 triangular faces is a tetrahedron. (iii) Yes, with the help of a square bottom and four triangle faces, we can form a pyramid.$

Page No 19.9:

Question 3:

Is it possible to have a polyhedron with any given number of faces?

Answer:

$Yes, it is possible to have a polyhedron with any number of faces. The only condition is that there should be at least four faces. This is because there is no possible polyhedron with 3 or less faces.$

Page No 19.9:

Question 4:

Is a square prism same as a cube?

Answer:

$Yes, a square prism and a cube are the same. Both of them have 6 faces, 8 vertices and 12 edges. The only difference is that a cube has 6 equal faces, while a square prism has a shape like a cuboid with two square faces, one at the top and the other at the bottom and with, possibly, 4 rectangular faces in between.$

Page No 19.9:

Question 5:

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

Answer:

$No, because every polyhedron satisfies Euler's formula, given below: F+V=E+2 Here, number of faces F = 10 Number of edges E = 20 N umber of vertices V = 15 So, by Euler's formula: LHS : 10+15 = 25 RHS : 20 + 2 = 22, which is not true because 25≠22 Hence, Eulers formula is not satisfied and no polyhedron may be formed .$

Page No 19.9:

Question 6:

Verify Euler's formula for each of the following polyhedrons:

Answer:

$(i) In the given polyhedron: Edges E=15 Faces F=7 Vertices V=10$

$Now, putting these values in Euler's formula: LHS : F+V = 7+10 = 17 LHS : E + 2 = 15 + 2 = 17 LHS = RHS Hence, the Euler's formula is satisfied.$

$(ii) In the given polyhedron: Edges E=16 Faces F=9 Vertices V=9$

$Now, putting these values in Euler's formula: RHS : F +V = 9 +9 = 18 LHS : E + 2 = 16 + 2 = 18 LHS = RHS Hence, Euler's formula is satisfied.$

$(iii) In the following polyhedron: Edges E=21 Faces F=9 Vertices V=14$

$Now, putting these values in Euler's formula: L H S : F+V = 9+14 = 23 R H S : E+2 =21+2 =23 This is true. Hence, Euler's formula is satisfied.$

$(iv) In the following polyhedron: Edges E=8 Faces F=5 Vertices V=5$

$Now, putting these values in Euler's formula: LHS : F+V = 5 + 5 = 10 RHS : E + 2 =8+2 =10 LHS = RHS Hence, Euler's formula is satisfied.$

$(v) In the following polyhedron: Edges E=16 Faces F=9 Vertices V=9$

$Now, putting these values in Euler's formula: LHS : F+V = 9 + 9 = 18 RHS : E + 2 =16+2 =18 LHS = RHS Hence, Euler's formula is satisfied.$

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