CBSE Class 8

These NCERT Solutions for Class 8 Maths Chapter 14 Factorization Ex 14.3 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 8 Maths Chapter 14 Factorization Exercise 14.3

Question 1.
Carry out the following divisions.
(i) 28x⁴ ÷ 56x
(ii) – 36y³ ÷ 9y²
(iii) 66pq²r³ ÷ 11qr²
(iv) 34x³y³z³ ÷ 51xy²z³
(v) 12a⁸b⁸ ÷ (- 6a⁶b⁴)
Answer:
(i) 28x⁴ ÷ 56x
= \(\frac{2 \times 2 \times 7 \times x^{4}}{2 \times 2 \times 2 \times 7 \times x}\)
= \(\frac{x^{4-1}}{2}=\frac{x^{3}}{2}=\frac{1}{2} x^{2}\)

(ii) – 36y³ ÷ 9y²
= \(\frac{-2 \times 2 \times 3 \times 3 \times y^{3}}{3 \times 3 \times y^{2}}\)
= – 4y^3-2
= -4y

(iii) 66pq²r³ ÷ 11qr²
= \(\frac{2 \times 3 \times 11 \times p \times q^{2} \times r^{3}}{11 \times q \times r^{2}}\)
= 2 x 3 x p x q^2-1 x r^3-2
= 6 x p x q x r = 6pqr

(iv) 34x²y²z² ÷ 51xy²z³
= \(\frac{2 \times 17 \times x^{3} \times y^{3} \times z^{3}}{3 \times 17 \times x \times y^{2} \times z^{3}}\)
= \(\frac{2}{3}\) x x^3-1 x y^3-2 x z^3-3
= \(\frac{2}{3}\) x x² x y x z⁰
= \(\frac{2 x^{2} y}{3}\)

(v) 12a⁸b⁸ ÷ (-6 a⁶ b⁴)
= \(\frac{2 \times 2 \times 3 \times a^{8} \times b^{8}}{-2 \times 3 \times a^{6} \times b^{4}}\)
= -2 x a^8-6 x b^8-4
= – 2 x a² x b⁴
= – 2a²b⁴

Question 2.
Divide the given polynomial by the given monomial.
(i) (5x² – 6x) ÷ 3x
(ii) (3y⁸ – 4y⁶ + 5y⁴) ÷ y⁴
(iii) 8 (x³y²z² + x²y³z² + x²y²z³) ÷ 4x²y²z²
(iv) (x³ + 2x² + 3x) ÷ 2x
(v) (p³q⁶– p⁶q³) ÷ p³q³
Answer:
(i) (5x² – 6x) ÷ 3x

(ii) (3y⁸ – 4y⁶ + 5y⁴) ÷ y⁴

=3y⁴ – 4y^6-4 + 5y^4-4
=3y⁴ – 4y² – 5y°
= 3y⁴ – 4y² + 5

(iii) 8(x³y²z² + x²y³z² + x²y²z³) 4x²y²z²

= [x^3-2 x y^2-2 x z^2-2 + x^2-2 x y^3-2 x z^2-2 + x^2-2 x y^2-2 x z^3-2]
= 2 (x x y° x z° + x° x y x z° + x° x y° x z)
= 2 (x + y + z)
[using a° = 1]

(iv) x³ + 2x² + 3x ÷ 2x

(v) (p³q⁶– p⁶q³) ÷ p³q³

= p^3-3 x q^6-3 – p^6-3 x q^3-3
= p⁰ x q³ – p³ x q⁰
= p³ x q³

Question 3.
Work out the following divisions:
(i) (10x – 25) ÷ 5
(ii) (10x-25) ÷ (2x-5)
(iii) 10y(6y + 21) ÷ 5 (2y + 7)
(iv) 9x2y2(3z – 24) ÷ 27xy (z – 8)
(v) 96abc (3a – 12) (5b – 30) ÷ 144 (a – 4) (b – 6)
Answer:
(i) (10x – 25) ÷ 5

= 2x – 5

(ii) (10x – 25) ÷ (2x – 5)

(iii) 10y (6y + 21) ÷ 5 (2y + 7)

[Taking 3 as common from 6y + 21]
= 2 x y x 3
= 6y

(iv) 9x2y2(3z – 24) ÷ 27xy (z – 8)

= x^2-1 x y^2-1 = xy.

(v) 96abc (3a – 12) (5b – 30) ÷ 144 (a – 4) (b – 6)

Question 4.
Divide as directed.
(i) 5(2x+1) (3x +5) ÷ (2x+1)
(ii) 26xy(x + 5) (y-4) ÷ 13x(y-4)
(iii) 52pqr (p + q) (q + r) (r + p) ÷ 104pq (q + r) (r + p)
(iv) 20 (y + 4) (y² + 5y + 3) ÷ 5(y + 4)
(v) x(x + 1)(x + 2)(x + 3) ÷ x(x + 1)
Answer:
(i) 5(2x + 1) (3x + 5) ÷ (2x + 1)
= \(\frac{5(2 x+1)(3 x+5)}{(2 x+1)}\)
= 5 (3x + 5)

(ii) 26xy (x + 5) (y – 4) ÷ 13x (y – 4)
= \(\frac{2 \times 13 \times x \times y \times(x+5) \times(y-4)}{13 \times x \times(y-4)}\)
= 2 x y x (x + 5) = 2y (x + 5)

(iii) 52pqr (p +q) (q + r) (r + p) ÷ 104pq (q + r) (r + p)
= \(\frac{2 \times 2 \times 13 \times p \times q \times r \times(p+q) \times(q+r) \times(r+9)}{2 \times 2 \times 2 \times 13 \times p \times q \times(q+r) \times(r+p)}\)
_ 2x2xl3xpxqxrx(p + q)x(q + r)x(r + 9) 2x2x2xl3xpxqx(q + r)x(r + p)
= \(\frac{r(p+q)}{2}=\frac{1}{2}\)r(p+q)

(iv) 20(y + 4) (y² + 5y + 3) ÷ 5 (y + 4)
= \(\frac{4 \times 5(y+4)\left(y^{2}+5 y+3\right)}{5 \times(y+4)}\)
= 4 (y² + 5y + 3)

(v) x(x+ 1) (x + 2) (x + 3) ÷ x(x + 1)
= \(\frac{x(x+1)(x+2)(x+3)}{x(x}\)
= (x + 2)(x + 3)

Question 5.
Factorise the expressions and divide them as directed.
(i) (y² + 7y + 10) – (y + 5)
(ii) (m² – 14m – 32) + (m + 2)
(iii) (5p² – 25p + 20) 4- (p – 1)
(iv) 4yz(z² + 6z – 16) -f- 2y (z + 8)
(v) 5pq (p² – q²) -f- 2p (p + q)
(vi) 12xy (9x² – 16y²) -J- 4xy (3x + 4y)
(vii) 39y3 (50y² – 98) + 26y² (5y + 7)
Answer:
(i) y² + 7y + 10
[10 = 2 x 5
[7 = 2 + 5]
= y² + 5y + 2y + 10
= y( y + 5) + 2 (y + 5) = (y + 5) (y + 2)
∴ (y² + 7y + 10) ÷ (y + 5)
\(\frac{(y+5)(y+2)}{(y+5)}\) = y + 2

(ii) m² – 14m – 32
– 32 = -16 x 2
-14 = -16 + 2
= m (m – 16) + 2(m – 16)
= (m – 16) (m + 2)
(m² – 14m – 32) ÷ (m + 2)
= \(\frac{(m-16)(m+2)}{m+2}\)
= m – 16

(iii) 5p² – 25p + 20
= 5[p² – 5p + 4]
= 5 [p² – 4p – p + 4]
= 5[p(p – 4)- 1 (p – 4)
= 5(p – 4) (p – 1)
∴ 5p² – 25 p + 20 ÷ (p – 1)
= \(\frac{5(p-4)(p-1)}{(p-1)}\)
= 5 (p -4)

(iv) z² + 6z – 16 = z² + 8z – 2z – 16
-16 = 8 x – 2
6 = 8 + (-2)
= z(z + 8) – 2 (z – 2)
= (z + 8) (z – 2)
∴ 4yz (z² + 6z – 16) 2y (z + 8)
= \(\frac{4 \times y \times z \times(z+8) \times(z-2)}{2 \times y \times(z+8)}\)
4xyxzx(z + 8)x(z-2)

(v) 5pq(p² – q²) ÷ 2p(p + q)
= \(\frac{5 \times p \times q \times(p+q)(p-q)}{2 \times p \times(p+q)}\)
= \(\frac{5 q(p-q)}{2}\)
= \(\frac { 5 }{ 2 }\)q(p-q)

(vi) 12xy (9x² – 16y²)
= 2 x 2 x 3 x x x y x [(3x)² – (4y)²]
= 2 x 2 x 3 x x x y x (3x + 4y) (3x – 4y)
12xy (9x² – 16y²) ÷ 4xy (3x + 4y)
=\(\frac{2 \times 2 \times 3 \times x \times y \times(3 x+4 y) \times(3 x-4 y)}{2 \times 2 \times x \times y \times(3 x+4 y)}\)
= 3 (3x – 4y)

(vii) 39y³ (50y² – 98)
= 3 x 13 x y³ x 2 (25y² – 49)
= 3 x 13 x 2 x y³ [(5y)² – 7²]
= 3 x 13 x 2 x y³ x (5y + 7) (5y – 7)
39y³ (50y² – 98) + 26y² (5y + 7)
= \(\frac{3 \times 13 \times 2 \times y^{3} \times(5 y+7) \times(5 y-7)}{2 \times 13 \times y^{2}(5 y+7)}\)
= 3 x y^3-2 x (5y – 7)
= 3y(5y – 7)

These NCERT Solutions for Class 8 Maths Chapter 11 Mensuration Ex 11.2 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 8 Maths Chapter 11 Mensuration Exercise 11.2

Question 1.
The shape of the top surface of a table is a trapezium. Find its area if its parallel sides are 1 m and 1.2 m and perpendicular distance between them is 0.8 m.

Answer:
Here, = a = 1.2m b = 1m and h = 0.8m
Area of a trapezium = \(\frac{1}{2}\) × h × (a + b)
= \(\frac{1}{2}\) × 0.8 × (1.2 + 1)m² = 0.4 × 2.2 = 0.88 m²

Question 2.
The area of trapezium is 34 cm² and the length of one of the parallel sides is 10 cm and its height is 4 cm. Find the length of the other parallel side.
Answer:
Let the length of the other parallel side be ‘X’, height of a trapezium = 4 m. Length of one parallel side = 10 cm

Area of a trapezium = 34 cm²
\(\frac{1}{2}\) × h × (a + b) = 34
\(\frac{1}{2}\) × 4 × (10 + x) = 34
2 (10 + x) = 34
20 + 2x = 34
2x = 34 – 20
2x = 14
x = \(\frac{14}{2}\) = 7 cm
Hence, length of the other side = 7 cm

Question 3.
Length of the fence of a trapezium shaped field ABCD is 120 m. If BC = 48 m, CD = 17m and AD = 40 m, find the area of this field. Side AB is perpendicular to the parallel sides AD and BC.
Answer:
Given BC = 48m,CD = 17mandAD = 40m
Perimeter of a trapezium = 120 m
AB + BC + CD + DA = 120
AB + 48 + 17 + 40 = 120
AB + 105 = 120
AB = 120 – 105 = 15 m
AB is the height of the trapezium.
Area of a trapezium = \(\frac{1}{2}\) × h × (a + b)
= \(\frac{1}{2}\) × 15 × (48 + 40) m²
= \(\frac{1}{2}\) × 15 × 88 m² = 15 × 44 m² = 660 m²
Area of the given trapezium = 660 m²

Question 4.
The diagonal of quadrilateral shaped field is 24 m and the perpendiculars dropped on it from the remaining opposite vertices are 8 m and 13 m. Find the area of the field.

Answer:
In the given quadrilateral, d = 24 m, h₁ = 13 m and h₂ = 8 m
Area of the quadrilateral = \(\frac{1}{2}\) × d × (h₁ + h₂)
= \(\frac{1}{2}\) × 24 × (13 + 8) m²
= 12 × 21 m² = 252 m²

Question 5.
The diagonals of a rhombus are 7.5 cm and 12 cm. Find its area.
Answer:
Here d₁ = 7.5 cm, d₂ = 12 cm
Area of the rhombus = \(\frac{1}{2}\) × d₁ × d₂
= \(\frac{1}{2}\) × 7.5 × 12 cm² = 7.5 × 6 cm² = 45 cm²

Question 6.
Find the area of a rhombus whose side is 5 cm and whose altitude is 4.8 cm. If one of its diagonals is 8 cm long, find the length of the other diagonal.
Answer
We know that rhombus is also a parallelogram.
Area of the rhombus = Area of a parallelogram = bh = 5 × 4.8 cm² × = 24 cm²
Let the other diagonal be ‘x’
Area of a rhombus = 24 cm²
\(\frac{1}{2}\) × d₁ × d₂ = 24
\(\frac{1}{2}\) × 8 × x = 24
4 × x = 24
x = \(\frac{24}{4}\) = 6 cm
∴ The required diagonal = 6 cm

Question 7.
The floor of a building consists of 3000 tiles which are rhombus shaped and each of its diagonal are 45 cm and 30 cm length.
Find the total cost of polishing the floor if the cost per m² is ₹ 4.
Answer:
Tiles are in the shape of a rhombus.
Here d₁ = 45 cm and d₂ = 30 cm
Area of the rhombus (one tile) = \(\frac{1}{2}\) × d₁ × d₂
= \(\frac{1}{2}\) × 45 × 30 cm² = 45 × 15 = 675 cm²
Total number of tiles = 3000
Area of the floor = 3000 × 675 cm²
= \(\frac{3000 \times 675}{100 \times 100}\)m² = \(\frac{2025}{10}\)m²
Cost of polishing the floor = \(\frac{2025}{10}\) × 4
= 405 × 2 = ₹ 810

Question 8.
wants to buy a trapezium shaped field. Its side along the river is parallel to and twice the side along the road. If the area of this field is 10500 m² and the perpendicular distance between the two parallel sides is 100 m, find the length of the side along the river.

Answer:
Let the length of the side be ‘x’ (along the river)
Length of the other side = \(\frac{1}{2}\) × x m (along the road)
Perpendicular distance between two parallel sides = 100 m
Area of the trapezium field = 10500 m²
\(\frac{1}{2}\) × h (a + b) = 10500
\(\frac{1}{2}\) × 100 (x + \(\frac{1}{2}\)x) = 10500
\(\frac{1}{2}\) × 100 × \(\frac{3 x}{2}\) = 10500
\(\frac{300}{4}\) x = 10500
x = \(\frac{10500}{4}\) × 4 = 35 × 4m = 140 m
∴ Length of the side along the river = 140 m

Question 9.
Top surface of a raised platform is in the shape of a regular octagon as shown in the figure. Find the area of the octagonal surface.

Answer:
The regular octagon can be divided into two trapeziums and one rectangle.
Area of the octagonal surface
= Area of rectangular portion + 2 (Area of trapezoidal portion)
= 11 × 5 + 2[\(\frac { 1 }{ 2 }\) × 4(5 + 11)]
= 55 + 2 × \(\frac { 1 }{ 2 }\) × 4 × 16 m
= 55 + 64 m² = 119 m²
∴ Area of the octagonal surface =119 cm²

Question 10.
There is a pentagonal shaped park as shown in the figure. For finding its area, Joyti and Kavita divided it in two different ways.

Find the area of this park using both ways. Can you suggest some other way of finding its area?
Answer:
For Joyti’s diagram
The given shape is divided into two
congruent trapeziums
Here a = 30m, b = 15m (parallel sides),
h = \(\frac { 15 }{ 2 }\) m.
Area of a trapezium = \(\frac { 1 }{ 2 }\) × h(a + b)squnits.
Area of one trapezium
= \(\frac { 1 }{ 2 }\) × \(\frac { 15 }{ 2 }\) × (30 + 15) m²
= \(\frac { 1 }{ 2 }\) × \(\frac { 15}{ 2 }\) × 45 m² = \(\frac { 675 }{ 4 }\)²
Area of the pentagonal shape = 2 \(\frac { 675}{ 4 }\)
= \(\frac { 675}{ 2 }\)m² = 337.5 m²

For Kavita’s diagram
The given shape is divided into a square and a triangle.
Here, side of a square is 15 m; height of the triangle is30-15 = 15m
Area of the pentagonal shape = Area of the square + Area of the triangle
= 15 × 15 + \(\frac { 1 }{ 2 }\) × 15 × 15 m²
= 225 + \(\frac { 225 }{ 2 }\)m²
= 225 + 112.5 m²
= 337.5 m²

Another method
Divide the pentagonal shape into three triangles of side 15 m and height 15 m. Area of the pentagonal shape
= 3 × Area of the triangle
= 3 × \(\frac { 1 }{ 2 }\) × 15 × 15 = 337.5 m²

Question 11.
Diagram of the adjacent picture frame has outer dimensions = 24 cm × 28 cm and inner dimensions 16 cm × 20 cm. Find the area of each section of the frame, if the width of each section is same.

Answer:
In trapezium (1) parallel sides are 24 cm and 16 cm.
Height = \(\frac{28-20}{2}\) = 4 cm
Areaofitstrapezium(1) = \(\frac{1}{2}\) × 4 × (24 + 16)
= 2 × 40 = 80 cm²
In trapezium (2) parallel sides are 28 cm and 20 cm.
Height = \(\frac{24-16}{2}=\frac{8}{2}\) = 4 cm
Area of the trapezium (2)
= \(\frac{1}{2}\) × 4 × (28 + 20)cm² = 2 × 48cm² = 96cm²
Area of each section of the frame is 80 cm², 96 cm², 80 cm² and 96 cm²
Note: [Area of 1 section = Area of 3rd section and Area of 2nd section = Area of 4th section]

These NCERT Solutions for Class 8 Maths Chapter 11 Mensuration Ex 11.1 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 8 Maths Chapter 11 Mensuration Exercise 11.1

Question 1.
A square and a rectangular field with measurements as given in the figure have the same perimeter. Which field has a larger area?

Answer:
(a) Side of the square = 60 m.
∴ Perimeter of the square = 4 × side
= 4 × 60 = 240 m
Area of the square = side × side
= 60 × 60 m² = 3600 m²

(b) Length of the rectangle = 80 m
Perimeter of the rectangle = Perimeter of a square
2 (l + b) = 240
2 (80 + b) = 240
160 + 2b = 240
2b = 240 – 160 = 80 ,
b = \(\frac { 80 }{ 2 }\) = 40 m
Area of the rectangle = l × b = 80 × 40m²
= 3200 m²
3600 m² > 3200 m²
∴ Area of the square field is greater.

Question 2.
Mrs Kaushik has a square plot with the measurement as shown in the figure. She wants to construct a house in the middle of the plot. A garden is developed around the house. Find the total cost of developing a garden around the house at the rate of ₹ 55 per m².

Answer:
Side of a square = 25 m
Area of the plot = side × side = 25 × 25 m² = 625 m²

In the given diagram, the construction portion is a rectangle with length = 20 m and breadth = 15 m

Area of the construction portion = Area of a rectangle = 20 × 15 m² = 300 m²

Area of the garden = (625 – 300) m² = 325 m² Cost of developing the garden = ₹ 55 × 325 = ₹ 17,875

Question 3.
The shape of a garden is rectangular in the middle and semi circular at the ends as shown in the diagram. Find the area and the perimeter of this garden (Length of rectangle is 20 – (3.5 + 3.5) metres].

Answer:
Radius of the semicircles = \(\frac { 7 }{ 2 }\)

Length of the rectangle = 20 – (3.5 + 3.5) m = 20 – 7 = 13 m
Breadth of the rectangle = 7m
Area of the rectangle = l × b = 13 × 7 = 91m²
Perimeter of the rectangle = 2 (l + 0) (Breadth will not be consider)
= 2 (13 + 0) m = 2 × 13 = 26 m
Area of the garden = 38.5 + 91m² = 129.5 m²
Perimeter of the garden = 22 + 26 m = 48 m

Question 4.
A flooring tile has the shape of a parallelogram whose base is 24 cm and the corresponding height is 10 cm. How many such tiles are required to cover a floor of area 1080 m²? (If required you can split the tiles in whatever way you want to fill up the corners).
Answer:
Base of the parallelogram = 24 cm = \(\frac{24}{100}\) m
Corresponding height = 10 cm = \(\frac{100}{10}\)
Area of a tile = base × height
= \(\frac{24}{100} \times \frac{10}{100}\) m² = \(\frac{24}{1000}\) m²
Area of the floor = 1080 m²
Number of tiles = \(\frac{\text { Area of the floor }}{\text { Area of one tile }}\)
= \(\frac{1080 \times 1000}{24}\) = 45 × 1000 = 45000 tiles

Question 5.
An ant is moving around a few food pieces of different shapes scattered on the floor. For which food-piece would the ant have to take a longer round? Remember, circumference of a circle can be obtained by using the expression c = 2πr, where r is the radius of the circle.

Answer:
(a) Radius of the semicircle = \(\frac{2.8}{2}\) = 1.4 cm
Perimeter of the semicircle = \(\frac{2 \pi \mathrm{r}}{2}\) + 2r
= \(\frac{22}{7}\) × 1.4 + 2 × 1.4 = 4.4 + 2.8cm
= 7.2 cm

(b) Perimeter of the semicircular part
= \(\frac{1}{2}\) × 2πr = πr = \(\frac{22}{7}\) × 1.4 = 4.4 cm
Perimeter of the rectangular part
= 1.5 + 2.8 + 15 cm = 5.8 cm
Perimeter of the given diagram
= 4.4 cm + 5.8 cm = 10.2 cm

(c) Perimeter of the semicircular part
= \(\frac{2 \pi r}{2}\) πr = πr = \(\frac{22}{7}\) × 1.4 cm = 4.4 cm
Perimeter of the given diagram
= (4.4 + 2 + 2) cm = 8.4 cm
∴ 7.2 cm < 8.4cm < 10.2 cm
Ant has to take a longer round for (b).

These NCERT Solutions for Class 8 Maths Chapter 14 Factorization Ex 14.4 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 8 Maths Chapter 14 Factorization Exercise 14.4

Question 1.
Find and correct the errors in the following mathematical statements.
Answer:
(1) 4(x – 5) = 4x – 5
The given statement is incorrect.
The correct statement is
4(x – 5) = 4x – 20

(2) x (3x + 2) = 3x² + 2
The given statement is incorrect.
The correct statement is x (3x + 2) = 3x² + 2x

(3) 2x + 3y = 5xy
This is an incorrect statement.
The correct statement is 2x + 3y = 2x + 3y

(4) x + 2x + 3x = 5x
This is an incorrect statement.
The correct statement is x + 2x + 3x = 6x

(5) 5y + 2y + y – 7y = 0
This is an incorrect statement.
The correct statement is
5y + 2y + y – 7y = y

(6) 3x + 2x = 5x²
This is an incorrect statement.
The correct statement is 3x + 2x = 5x

(7) (2x)² + 4(2x) + 7 = 2x² + 8x + 7
This given statement is incorrect.
The correct statement is
(2x)² + 4 (2x) + 7 = 4x² + 8x + 7

(8) (2x)² + 5x = 4x + 5x = 9x²
The given statement is incorrect.
The correct statement is (2x)² + 5x = 4x² + 5x

(9) (3x + 2)² = 3x² + 6x + 4
The given statement is incorrect.
The correct statement is
(3x + 2)² = (3x)² + 2 × 3x × 2 + (2)²
(3x + 2)² = 9x² + 12x + 4

(10) Substituting x = – 3 in
(a) x² + 5x + 4
= (- 3)² + 5 (-3) + 4
= 9 + 2 + 4
= 15
The given statement is incorrect.
The correct statement is x² + 5x + 4
= (-3)² + 5 (- 3) + 4
= 9 – 15 + 4
= 13 – 15 = – 2
∴ x² + 5x + 4 = -2 at x = – 3

(b) x² – 5x + 4
= (-3)² – 5 (-3) + 4
= 9 – 15 + 4 = -2
The given statement is incorrect.
The correct statement is
x² – 5x + 4
= (-3)² – 5 (-3) + 4
= 9 + 15 +4 = 28

(c) x² + 5x = (-3)² + 5 (-3)
= – 9 – 15 = – 24
The given statement is incorrect.
The correct statement is
x² + 5x
= (-3)² + 5(—3)
= 9 – 15 = -6

(11) (y – 3)² = y² – 9
The given statement is incorrect. (y-3)²
= y² – 2 (y) (3) + 3²
= y² – 6y + 9
The correct statement is (y – 3)² = y² – 6y + 9

(12) (z + 5)² = z² + 25
The given statement is incorrect.
(z + 5)² = z² + 2 × z × 5 + 5²
= z² + 10z + 25
The correct statement is (z + 5)² = z² + 10z + 25

(13) (2a + 3b) (a – b) = 2a² – 3b²
The given statement is incorrect.
(2a + 3b) (a – b)
= 2a (a – b) + 3b (a – b)
= 2a² – 2ab + 3ab – 3b²
= 2a² + ab – 3b²
The correct statement is
(2a + 3b) (a – b) = 2a² + ab – 3b²
(14) (a + 4) (a + 2) = a² + 8
The given statement is incorrect.
(a + 4) (a + 2)
= a (a + 2) + 4 (a + 2)
= a² + 2a + 4a + 8
= a² + 6a + 8
The correct statement is (a + 4) (a + 2) = a² + 6a + 8

(15) (a – 4) (a – 2) = a² – 8
The given statement is incorrect, (a – 4) (a – 2)
= a(a – 2) – 4(a – 2)
= a² – 2a – 4a + 8
= a² – 6a + 8
The correct statement is (a – 4) (a – 2) = a² – 6a + 8

(16) \(\frac{3 x^{2}}{3 x^{2}}\) = 0
The given statement is incorrect.
The correct statement is
\(\frac{3 x^{2}}{3 x^{2}}\) = 1

(17) \(\frac{3 x^{2}+1}{3 x^{2}}\) = 1 + 1 = 2
\(\frac{3 x^{2}+1}{3 x^{2}}=\frac{3 x^{2}}{3 x^{2}}+\frac{1}{3 x^{2}}=1+\frac{1}{3 x^{2}}\)
The given statement is incorrect.
The correct statement is
\(\frac{3 x^{2}+1}{3 x^{2}}=1+\frac{1}{3 x^{2}}\)

(18) \(\frac{3 x}{3 x+2}=\frac{1}{2}\)
The given statement is incorrect.
The correct statement is
\(\frac{3 x}{3 x+2}=\frac{3 x}{3 x+2}\)

(19) \(\frac{3}{4 x+3}=\frac{1}{4 x}\)
The given statement is incorrect.
The correct statement is \(\frac{3}{4 x+3}=\frac{3}{4 x+3}\)

(20) \(\frac{4 x+5}{4 x}\) = 5
The given statement is incorrect.
\(\frac{4 x+5}{4 x}=\frac{4 x}{4 x}+\frac{5}{4 x}=1+\frac{5}{4 x}\)
The correct statement is
\(\frac{4 x+5}{4 x}=1+\frac{5}{4 x}\)

(21) = 7x
The given statement is incorrect.
\(\frac{7 \mathrm{x}+5}{5}=\frac{7 \mathrm{x}}{5}+\frac{5}{5}=\frac{7 \mathrm{x}}{5}+1\)
The correct statement is
\(\frac{7 x+5}{5}=\frac{7 x}{5}+1\)

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.