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These NCERT Solutions for Class 8 Maths Chapter 10 Visualizing Solid Shapes InText Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 8 Maths Chapter 10 Visualizing Solid Shapes InText Questions

NCERT Intext Question Page No. 153
Question 1.
Match the following: (First one is done for you)
Answer:

Question 2.
Match the following pictures (objects) with their shapes:

Answer:
(i) A triangular field adjoining a square field.
(ii) A cone taken out of a cylinder.
(iii) A hemisphere surmounted on a cone.
(iv) A circular path around a circular ground.
(v) Two rectangular cross paths inside a rectangular park.

NCERT Intext Question Page No. 165
Question 1.
Tabulate the number of faces, edges and vertices for the following polyhedrons: (Here ‘V’ stands for number of vertices, ‘F’ stands for number of faces and ‘E’ stands for number of edges).
Answer:

These NCERT Solutions for Class 8 Maths Chapter 10 Visualizing Solid Shapes Ex 10.3 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 8 Maths Chapter 10 Visualizing Solid Shapes Exercise 10.3

Question 1.
Can a polyhedron have for its faces.
(i) 3 triangles
(ii) 4 triangles
(iii) a square and four triangles?
Answer:
(i) No, it is not possible
(ii) Yes, four triangles can be the faces of a polyhedron.
(iii) Yes, a square and four triangles can be the faces.

Question 2.
Is it possible to have a polyhedron with any given number of faces?
(Hint: Think of a pyramid)
Answer:
Possible only if the number of faces is greater than or equal to 4.

Question 3.
Which are prisms among the following?


(i) A nail
(ii) Unsharpened pencil
Answer:
(i) No, a nail is not a prism.
(ii) Yes, unsharpened pencil is a prism.
(iii) No, table weight is not a prism.
(iv) Yes, box is a prism.

Question 4.
(i) How are prisms and cylinders alike?
(ii) How are pyramids and cones alike?
Answer:
(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.

Question 5.
Is a square prism same as a cube?
Explain:
Answer:
No, it can be a cuboid also.

Question 6.
Verify Euler’s formula for these solids.

Answer:
In the given figure,
F = 7, V = 10 and E = 15
F + V – E = 7 + 10 – 15 = 17 – 15
F + V – E = 2
Thus, Euler’s formula is verified.

(ii) In the given figure,
F = 9, V = 9 and E = 16
F + V- E = 9 + 9 – 16 = 18 – 16
Thus, Euler’s formula is verified.

Question 7.
Using Euler’s formula find the unknown.

Answer:
(i) Here V = 6 and E = 12
F + V – E = 2
F + 6 – 12 = 2
F – 6 = 2
F = 6 + 2 = 8

(ii) F = 5, E = 9
F + V – E = 2
5 + V – 9 = 2
V – 4 = 2
V = 2 + 4 = 6

(iii) Here F = 20 V = 12
F + V – E = 2
20 + 12 – E = 2
32 – E = 2
32 – 2 = E
30 = E

Question 8.
Can a polyhedron have 10 faces, 20 edges and 15 vertices?
Answer:
Here, F = 10, E = 20, V = 15
We know that F + V – E = 2
Here, F + V – E =10 + 15 – 20
= 25 – 20
= 15 ≠ 2
∴ Such a polyhedron is not possible.

These NCERT Solutions for Class 8 Maths Chapter 14 Factorization Ex 14.2 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 8 Maths Chapter 14 Factorization Exercise 14.2

Question 1.
Factorise the following expressions.
(i) a² + 8a + 16
(ii) p² – lOp + 25
(iii) 25m² + 30m+ 9
(iv) 49y² + 84yz + 36z²
(v) 4x² – 8x + 4
(vi) 121b² – 88bc + 16c²
(vii) (1 + m)² – 4lm
(viii) a⁴ + 2a²b² + b4
(Hint: Expand (1 + m)² first)
Answer:
(i) a² + 8a + 16
= a² + 2 x a x 4 + 4²
= (a + 4)²
= (a + 4) (a + 4)
(ii) p² – 10p + 25
= p² – 2 x 5p + 5²– (p – 5)²
= (p – 5) (p – 5)

(iii) 25m² + 30m + 9
= (5m)² + 2 x 5m x 3 + 3²
= (5m + 3)²
= (5m + 3) (5m +3)

(iv) 49y² + 84yz + 36z²
= (7y)²+ 2 x 7y x 6z + (6z)²
= (7y + 6z)² = (7y + 6z) (7y + 6z)

(v) 4x² – 8x + 4
= (2x)² – 2 x 2x x 2 + 2²
= (2x – 2)² = 2² (x – 1)²
= 4 (x – 1) (x – 1)

(vi) 12lb² – 88bc + 16c²
= (11b)² – 2 x 11b x 4c + (4c)²
= (11b – 4c)²
= (11b – 4c) (11b – 4c)

(vii) (l + m)² – 4lm
= l² + 2lm + m² – 4lm
= l² + (21m – 41m) + m²
= 1 – 21m + m²
= (1 – m)² = (1 – m) (1 – m)

(viii) a⁴ + 2a² b² + b⁴
= (a²)² + 2 x a² x b² + (b²)²
= (a² + b²)²
= (a² + b²) (a² + b²)

Question 2.
Factorise
(i) 4p² – 9q²
(ii) 63a² – 112b²
(iii) 49x² – 36
(iv) 16x⁵ – 144x³
(v) (l + m)² – (l – m)²
(vi) 9x²y² – 16
(vii) (x² – 2xy + y²) – z²
(viii) 25a² – 4b² + 28bc – 49c²
Answer:
(i) 4p² – 9q²
= (2p)² – (3q)²
= (2p + 3q) (2p – 3q)
[a² – b² = (a + b) (a – b)]

(ii) 63a²-11²b²
= 7 x (9a² – 16b²)
= 7[(3a)² – (4b)² ]
= 7 (3a + 4b) (3a – 4b)
[a² – b² = (a + b) (a – b)]

(iii) 49x² – 36
= (7x)² – 6² = (7x + 6) (7x – 6)
[Using a² – b² = (a + b) (a – b)]

(iv) 16x³ – 144x³
= x³ [16×2 – 144]
= x³ x 16 [x² – 9]
= 16x³ [x² – 3²]
= 16x³ (x + 3) (x – 3)

(v) (l + m)² – (l – m)²
= (l + m + l – m) [l + m – (l – m)]
= 2l [l + m – l + m]
= 2l x 2m = 4lm

(vi) 9x²y² – 16
= (3xy)² – 4²
= (3xy + 4) (3xy – 4)

(vii) (x² – 2xy + y²) – z²
= [x² – 2xy + y²] – z²
= (x – y)² – z²
= (x – y + z) (x – y – z)

(viii) 25a² – 4b² + 28bc – 49c²
= 25a² – [4b² – 28bc + 49c² ]
= (5a)² – [(2b)² – 2 x 2b x 7c + (7c)²]
= (5a)² – (2b – 7c)²
= (5a + 2b – 7c) [5a – (2b – 7c)]
= (5a + 2b – 7c)(5a – 2b + 7c)

Question 3.
Factorise the expressions.
(i) ax² + bx
(ii) 7p² + 21q²
(iii) 2x³ + 2xy² + 2xz²
(iv) am² + bm² + bn² + an²
(v) (lm + 1) + m + 1
(vi) y (y + z) + 9 (y + z)
(vii) 5y² – 20y – 8z + 2yz
(viii) 10ab + 4a + 5b + 2
(ix) 6xy – 4y + 6 – 9x
Answer:
(i) ax² + bx
= a x x x x + b x x
= x (ax + b)

(ii) 7p² + 21q²
= 7 x p x p + 3 x 7 x q x q – 7 (p² + 3q²)

(iii) 2x³ + 2xy² + 2xz²
= 2 x x x x x x + 2 x x x y x y + 2 x x x z x z
= 2x(x xx + yxy + zxz)
= 2x (x² + y² + z²)

(iv) am² + bm² + bn² + an²
= m² (a + b) + n² (b + a)
= m² (a + b) + n² (a + b)
= (a + b) (m² + n²)

(v) (lm + l) + m + l
= l(m + l) + l(m + l)
= (m + l) (l + l)

(vi) y (y + z) + 9 (y + z)
= (y + z) (y + 9)
[(y + z) is common for both terms)]

(vii) 5y² – 20y – 8z + 2yz
= 5y x y – 4 x 5y – 2 x 2 x 2 x z + 2 x y x z
= 5y (y – 4) + 2z (- 4 + y)
= 5y (y – 4) + 2z (y – 4)
= (y – 4) (5y + 2z)

(viii) 10ab + 4a + 5b + 2
= 2 x 5 x a x b + 2 x 2 x a + 5 x b + 2
= 2a (5b + 2) + 1 (5b + 2)
= (5b + 2) (2a + 1)

(ix) 6xy – 4y + 6 – 9x
= 2 x 3 x x x y – 2 x 2 x y + 2 x 3 – 3 x 3 x x
= 2y (3x – 2) + 3 (2 – 3x)
= 2y (3x – 2) – 3 (3x – 2)
= (3x – 2) (2y – 3)

Question 4.
Factorise
(i) a⁴ – b⁴
(ii) p⁴ – 81
(iii) x⁴ – (y + z)⁴
(iv) x²⁴ – (x – z)
(v) a⁴ – 2a²b² + b⁴
Answer:
(i) a⁴ – b⁴
= (a²)² – (b²)²
[(using a² – b² = (a + b) (a – b)]
= (a² + b²) (a² – b²)
= (a² + b²) (a + b) (a – b)

(ii) p⁴ – 81
= (p²)² – 9²
[Using a² – b² = (a + b) (a – b)]
= (p² + 9) (p² – 9)
= (P² + 9) [p² – 32 ]
= (p² + 9) (p + 3) (p – 3)

(iii) x² – (y + z)4
= (x²)² – [(y + z)²]²
[using a² – b² = (a + b) (a – b)
= [x² + (y + z)²] [x² – (y + z)²]
= [x² + (y + z²)] (x + y + z) [x – (y + z)
= [x² + (y + z)²] [x + y + z) (x – y – z)

(iv) x⁴ – (x – z)⁴
= (x²)² – [(x-z)²]²
= [x² + (x – z)²] [x² – (x – z)²]
[Using a² – b² = (a + b) (a – b)]
= [x² + (x – z)² ] [(x + x – z) (x – x + z)]
= [x² + (x – z)²] (2x – z) (z)
= (x² + x² + z² – 2xz) (2x – z) (z)
= (2x² + z² – 2xz) (2x – z) (z)
= z (2x – z) (2x² + z² – 2xz)

(v) a⁴ – 2a² b² + b⁴
= (a²)² – 2 x a² x b² + (b²)² = (a² – b² )²
= (a² – b²) x (a² – b²)
= (a + b) (a – b) (a + b) (a – b)
= (a + b)² (a – b)²

Question 5.
Factorise the following expressions.
(i) p² + 6p + 8
(ii) q² – lOq + 21
(iii) p² + 6p – 16
Answer:
(i) p² + 6p + 8
= p² + 4p + 2p + 8
[8 = 2 x 4 and 2 + 4 = 6]
= p (p + 4) + 2 (p + 4)
= (p + 4) (p + 2)

(ii) q² – 10q + 21
= q² – 7q – 3q + 21
[21 = – 7 x – 3 and – 10 = – 7 – 3]
= q(q – 7) – 3(q – 7)
= (q – 7) (q – 3)

(iii) p² + 6p – 16
= p² + 8p – 2p – 16
[-16 = 8 x – 2 and 8 – 2 = 6]
= p(p + 8) – 2 (p + 8)
= (p + 8) (p – 2)2

These NCERT Solutions for Class 8 Maths Chapter 10 Visualizing Solid Shapes Ex 10.2 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 8 Maths Chapter 10 Visualizing Solid Shapes Exercise 10.2

Question 1.
Look at the given map on a city.
Answer the following.
(a) Colour the map as follows: Blue-water, red-fire station, orange-library, yellow- school. 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 Sr. Secondary School?
Answer:
The shaded (coloured) map according to the required direction is given in part (a). Here various colours are as under


(a) Do as directed
(b) Do as directed
(c) Do as directed
(d) City park
(e) Senior Secondary School

Question 2.
Draw a map of your class using proper scale and symbols for different objects.
Answer:
It is an activity, so do it yourself.

Question 3.
Draw a map of your school compound using proper scale and symbols for various features like play ground, main building, garden, etc.
Answer:
It is an activity. Please do it yourself.

Question 4.
Draw a map giving instructions to your friend so that she could reach your house without any difficulty.
Answer:
It is an activity. Please do it yourself.

Faces, Edges and Vertices
Note:
I. A polygon is a 2-D shape made of line segments only.
II. A polyhedron is a 3-D shape made of plane faces.
III. Plural of polyhedron is polyhedra.
IV. A cube, cuboid, pyramid, prism, etc. are polyhedra whereas spheres, cones and cylinders are not polyhydra.

A cuboid has 6 faces, 12 edges and 8 vertices.

Convex polyhedrons: You will recall the concept of convex polygons. The idea of convex polyhedron is similar.

Regular Polyhedron: A convex polyhedron is said to be regular if its faces are made up of regular polygons and the same number of faces meet at each vertex.

Prism: Prism are polyhedra whose lateral base and top are congruent polygons and the other faces are parallelograms.

Pyramids: Pyramids are polyhedra whose base is a polygon and whose lateral faces are in the shape of triangles at a common vertex.

Note: The prisms or a pyramids is named after its bases.

Euler’s Formula: It is a relation between the faces, edges and vertices of a polyhedra. It is given by
F + V = E + 2
is F + V – E = 2
where, F → Number of faces
V → Number of vertices
E → Number of edges.

These NCERT Solutions for Class 8 Maths Chapter 14 Factorization Ex 14.1 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 8 Maths Chapter 14 Factorization Exercise 14.1

Question 1.
Find the common factors of the given terms.
(i) 12x, 36
(ii) 2y, 22xy
(iii) 14pq, 28p²q²
(iv) 2x, 3x², 4
(v) 6 abc, 24ab², 12 a²b
(vi) 16x³, – 4x², 32x
(vii) 10 pq, 20 qr, 30rp
(viii) 3x²y³, 10x³, 6x²y²z
Answer:
(i) 12x = 2 x 2 x 3 x x
36 = 2 x 2 x 3 x 3
∴ The common factor = 2 x 2 x 3 = 12

(ii) 2y = 2 x y
22y = 2 x 11 x y
∴ The common factor = 2 x y = 2y

(iii) 14pq = 2 x 7 x p x q
28 p²q² = 2 x 2 x 7 x p x p x q x q
∴ The common factor
= 2 x 7 x p x q = 14pq

(iv) 2x = 2 x x
3x² = 3 x x x x
4 = 2 x 2
∴ Common factor = 1
(1 is a factor of every term)

(v) 6abc = 2 x 3 x a x b x c
24ab² = 2 x 2 x 2 x 3 x a x b x b
12a²b = 2 x 2 x 3 x a x a x b
∴ The common factor
= 2 x 3 x a x b = 6ab

(vi) 16x³ = 2 x 2 x 2x 2 x x x x x x
-4x² = -1 x 2 x 2 x x x x
32x = 2 x 2 x 2 x 2 x 2 x x
∴ The common factor = 2 x 2 x x = 4x

(vii) 10pq = 2 x 5 x p x q
20qr = 2 x 2 x 5x q x r
30rp = 2 x 3 x 5 x r x p
∴  The common factor = 2 x 5 = 10

(viii) 3x²y³ = 3 x x x x x y x y x y
10x³y² = 2 x 5 x x x x x x x y x y
6x²y²z = 2 x 3 x x x x x y x y x z
∴  The common factor = x x x x y x y
= x²y²

Question 2.
Factorise the following expressions.
(i) 7x – 42
(ii) 6p – 12q
(iii) 7a² + 14a
(iv) – 16 z + 20 z³
(v) 20 l² m + 30 alm
(vi) 5x²y – 15xy²
(vii) 10a² + 15 b² + 20c²
(viii) – 4a² + 4ab – 4ca
(ix) x²yz + xyz² xy²z xyz²
(x) ax²y + bxy² + cxyz
Answer:
(i) 7x – 42
=7 x x – 2 x 3 x 7
= 7 (x – 2 x 3)
= 7 (x – 6)

(ii) 6p – 12q
= 2 x 3p – 2 x 2 x 3 x q
= 2 x 3 [p – 2 x q]
= 6 (p – 2q)

(iii) 7a² + 14a
=7 x a x a + 2 x 7 x a
= 7a (a + 2)

(iv) – 16z + 20 z³
= -2 x 2 x 2 x 2 x z + 2 x 2 x 5 x z x z x z
= 2 x 2 x z [-2x² + 5 x z x z]
= 4z (-4 + 5z²)

(v) 20 l²m + 30 alm
= 2 x 2 x 5 x l x l x m + 2 x 3 x 5 x a x l x m
= 2 x 5 x l x m[2 x l + 3a]
= 10lm (2l + 3a)

(vi) 5x²y-15xy²
=5 x x x x x y – 3 x 5 x x x y x x
= 5xy (x – 3y)

(vii) 10a² – 15b² + 20c²
= 2 x 5 x a x a – 3 x 5 x b x b + 2 x 2 x 5 x c x c
= 5[2 x a x a – 3 x b x b + 2 x 2 x c x c]
= 5 (2a² – 3b² + 4c²)

(viii) -4a² + 4ab – 4ca
= -2 x 2 x a x a + 2 x 2 x a x b – 2 x 2 x c x a
= 2 x 2 x a (-a + b – c)
= 4a (-a + b – c)

(ix) x²yz + xy²z + xyz²
= x x x x y x z + x x y x y x z + x x y x z x z
= x x y x z[x + y + z]
= xyz (x + y + z)

(x) ax²y + bxy² + cxyz
= a x x x x x y + b x x x y x y + c x x x y x z
= x x y[a x x + b x y + c x z]
= xy (ax + by + cz)

Question 3.
Factorise
(i) x² + xy + 8x + 8y
(ii) 15xy – 6x + 5y – 2
(iii) ax + bx – ay – by
(iv) 15pq + 15 + 9q – 25p
(v) z – 7 + 7xy – xyz
Answer:
(i) x² + xy + 8x + 8y
= x(x + y) + 8 (x + y)
= (x + y) (x + 8)

(ii) 15xy – 6x + 5y – 2
= 3x (5y – 2) + 1 (5y – 2)
= (5y – 2) (3x + 1)

(iii) ax + bx – ay – by
= x (a + b) – y (a + b)
= (a + b) (x – y)

(iv) 15pq + 15 + 9q + 25p
= 15pq + 25p + 9q + 15
= 5p (3q + 5) + 3 (3q + 5)
(Re-arranging the terms)
= (3q + 5) (5p + 3)

(v) z – 7 + 7xy – xyz
= z – xyz + 7xy – 7
= z( 1 – xy) + (7 (xy – 1)
(Re-arranging the terms)
= z (1 – xy) – 7 (1 – xy)
= (1 – xy) (z- 7)

These NCERT Solutions for Class 8 Maths Chapter 10 Visualizing Solid Shapes Ex 10.1 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 8 Maths Chapter 10 Visualizing Solid Shapes Exercise 10.1

Question 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.

Answer:
(a) (iii) → (iv)
(b) (i) → (v)
(c) (iv) → (ii)
(d) (v) → (iii)
(e) (ii) → (i)

Question 2.
For each of the given solid, the three views are given. Identify for each solid the corresponding top, front and side views.

(a) (i) Front (ii) Side (iii) Top
(b) (i) Side (ii) Front (iii) Top
(c) (i) Front (ii) Side (iii) Top
(d) (i) Front (ii) Side (iii) Top

Question 3.
For each given solid, identify the top view, front view and side view.


Answer:
(a) (i) Top (ii) Front (iii) Side
(b) (i) Side (ii) Front (iii) Top
(c) (i) Top (ii) Side (iii) Front
(d) (i) Side (ii) Front (iii) Top
(e) (i) Front (ii) Top (iii) Side

Question 4.
Draw the front view, side view and top view of the given objects.

Answer: