Author name: Prasanna

Haryana State Board HBSE 7th Class Maths Solutions Chapter 12 Algebraic Expressions Ex 12.3 Textbook Exercise Questions and Answers.

Haryana Board 7th Class Maths Solutions Chapter 12 Algebraic Expressions Exercise 12.3

Question 1.
If m = 2, find the value of:
(i) m – 2
(ii) 3m – 5
(iii) 9 – 5m
(iv) 3m² – 2m – 7
(v) \(\frac{5 m}{2}\) – 4
Solution:
(i) m – 2 – 2 – 2 = 0
(ii) 3m – 5 = 3(2) – 5 = 6 – 5 = 1
(iii) 9 – 5 m = 9 – 5(2) = 9 – 10 = – 1.
(iv) 3m² – 2m – 7
= 3(2)² – 2(2) – 7
= 3 × 4 – 4 – 7
= 12 – 4 – 7 = 8 – 7 = 1
(v) \(\frac{5 m}{2}\) – 4 = \(\frac{5 \times 2}{2}\) – 4
= \(\frac{10}{2}\) – 4 = 5 – 4 = 1

Question 2.
If P = – 2, find the value of;
(i) 4p + 7
(ii) – 3p² + 4p + 7
(iii) – 2p² – 3p² + 4p + 7, when p = – 2.
Solution:
(i) 4p + 7 = 4(- 2) + 7 = – 8 + 7
= -1

(ii) – 3(-p)² + 4(-p) + 7
= – 3(- 2)² + 4(- 2) + 7
= – 3 × 4 + 4 × -2 + 7
= – 12 – 8 + 7
= – 13.

(iii)- 2p² – 3p² + 4p + 7
= -2(-2)² – 3(- 2)² + 4 × (- 2) + 7
= -2 × 4 – 3 × 4 + 4 × – 2 + 7
= -8 – 12 – 8 + 7
= -21.

Question 3.
Find the value of the following expressions, when x = – 1.
(i) 2x- 7
(ii) -x + 2
(iii) x² + 2x + 1
(iv) 2x² – x – 2.
Solution:
(i) 2x – 7 = 2(-1) – 7 = -2 – 7
= -9

(ii) -x + 2 = – (-1) + 2
= 1 + 2 = 3

(iii) x² + 2x + 1
= (- 1)² + 2(- 1) + 1
= 1 – 2 + 1 = 0.

(iv) 2x² – x – 2 = 2(- 1)² – (- 1) – 2
= 2 × 1 + 1 – 2
= 2 + 1 – 2
= 3 – 2 = 1.

Question 4.
If a = 2, b = – 2, find the value of:
(i) a² + b²
(ii) a² + ab + b²
(iii) a² – b2².
Solution:
(i) a² + b² = (2)² + (- 2)²
= 4 + 4
= 8

(ii) a² + ab + b² = (2)² + 2 x -2 + (-2)²
= 4 – 4 + 4
= 8 – 4 = 4.

(iii) a² – b² = (2)² – (- 2)²
= 4 – 4
= 0.

Question 5.
When a = 0, b = -1, find the value of the given expressions:
(i) 2a+2b
(ii) 2a² + b² + 1
(Hi) 2a²b + 2ab² + ab
(iv) a² + ab + 2.
Solution:
(i) 2a + 2b = 2 × 0 + 2(- 1)
= 0 + (- 2) = – 2

(ii) 2a² + b² + 1 = 2(0)² + (- 1)² + 1
= 2 × 0 + 1 + 1
= 1 + 1 = 2.

(iii) 2a²b + 2ab² + ab
= 2(0)² (- 1) + 2 × 0(- 1)² + 0 × – 1
= 2 × 0 × (-1) + 2 × 0 × 1 + 0
= 0 + 0 + 0 = 0

(iv) a² + ab + 2
a = 0
b = -1
= (0)² + (0) (- 1) + 2
= 0 + 0 + 2 = 2.

Question 6.
Simplify the expressions and find the value of x is 2.
(i) x + 7 + 4(x – 5)
(ii) 3(x + 2) + 5x – 7
(iii) 6x + 5(x-2)
(iv) 4(2x – 1) + 3x + 11
Solution:
(i) x + 7 + 4(x – 5)
= x + 7 + 4x – 20
= 2 + 7 + 4 × 2 – 20
= 9 + 8 – 20
= 17 – 20 = – 3.

(ii) 3(x + 2) + 5x – 7
= 3x + 6 + 5x – 7
= 3 × 2 + 6 + 5 × 2 – 7
= 6 + 6 + 10 – 7
= 12 + 10 – 7
= 22 – 7 = 15.

(iii) 6x + 5(x – 2) = 6x + 5x – 10
= 6 × 2 + 5 × 2 – 10
= 12 + 10 – 10
= 22 – 10 = 12.

(iv) 4(2x – 1) + 3x + 11
= 8x – 4 + 3x + 11
= 8 x 2 – 4 + 3 x 2 + 11
= 16 – 4 + 6 + 11
= 12 + 6 + 11
= 18 + 11
= 29.

Question 7.
Simplify these expressions and find their yalue of x = 3, a = – 1, b = – 2,
(i) 3x – 5 -x + 9
(ii) 2 – 8x + 4x + 4
(iii) 3a + 5 – 8a + 1
(iv) 10 – 3b – 4 – 5b
(v) 2a – 2b – 4 – 5 + a.
Solution:
(i)3x – 5 – x + 9
= 3 × 3 – 5 – 3 + 9
= 9 – 5 – 3 + 9
= 4 – 3 + 9
= 1 + 9 = 10.

(ii) 2 – 8x + 4x + 4
= 2 – 8 × 3 + 4 × 3 + 4
= 2 – 24 + 12 + 4
= 18 – 24 = – 6.

(iii) 3a + 5 – 8a + 1
= 3(- 1) + 5 – 8(- 1) + 1
= -3 + 5 + 8 + 1
= 2 + 8 + 1 = 11.

(iv) 10 – 3b – 4 – 5b
= 10 – 3(-2) – 4 – 5(-2)
= 10 + 6 – 4 + 10
= 16 – 4 + 10 = 12 + 10
= 22.

(v) 2a -26-4-5 + a
= 2(-1) – 2(-2) – 4 – 5 + (-1)
= -2 + 4 – 4 – 5 – 1
= 2 – 9 – 1
= 2 – 10 = -8.

Question 8.
(a) If z = 10, find the value of z³ – 3(z -10).
(6) If p = -10 find the value of p² – 2p – 100.
Solution:
(a) z³ – 3(z -10)
= (10)³ – 3(10-10)
= 1000 – 3 x 0
= 1000 – 0 = 1000

(b) p² – 2p – 100 = (- 10)² – 2 x (- 10) – 100
= 100 + 20 – 100
= 120 – 100 = 20.

Question 9.
What should be the value of a if the value of 2x² + x – a equals to 5, when x = 0 ?
Solution:
2x² + x – a = 5
2(0)² + 0 – a = 5
or 2 x 0 + 0 – a = 5
or -a = 5 or a = -5

Question 10.
Simplify the expression and find its value when a = 5 and b = – 3, 2(a² + ab) + 3 – ab.
Solution:
2(a² + ab) + 3 – ab
= 2a² + 2ab + 3 – ab
= 2a² + ab + 3
= 2 × (5)² + (5) × (-3) + 3
= 2 × 25 + (- 15) + 3
= 50 – 15 + 3
= 53 – 15 = 38.

Haryana Board HBSE 6th Class Sanskrit Solutions रुचिरा भाग 1

• Chapter 1 शब्द परिचयः 1
• Chapter 2 शब्द परिचयः 2
• Chapter 3 शब्द परिचयः 3
• Chapter 4 विद्यालयः
• Chapter 5 वृक्षाः
• Chapter 6 समुद्रतटः
• Chapter 7 बकस्य प्रतिकारः
• Chapter 8 सूक्तिस्तबकः
• Chapter 9 क्रीडास्पर्धा
• Chapter 10 कृषिकाः कर्मवीराः
• Chapter 11 पुष्पोत्सवः
• Chapter 12 दशमः त्वम असि
• Chapter 13 विमानयानं रचयाम
• Chapter 14 अहह आः च
• Chapter 15 मातुलचन्द्र

Haryana State Board HBSE 7th Class Maths Solutions Chapter 12 Algebraic Expressions Ex 12.2 Textbook Exercise Questions and Answers.

Haryana Board 7th Class Maths Solutions Chapter 12 Algebraic Expressions Exercise 12.2

Question 1.
Simplify combining like terms :
(i) 21b – 32 + 7b – 20b
(ii) – z², + 13z² -5z + 7z³ – 15z
(iii) p-(p-q)-q- (q – p)
(iv) 3a – 2b – ab – (a -b + ab) + 3ab + b-a
(v) 5x²y – 5x² + 3yx² – 3y² + x² – y² + 8xy² – 3y²
(vi) (3y² + 5y – 4) – (8y – y² – 4)
Solution:
(i) 21b – 32 + 7b – 20b
Rearranging terms:
= 21b + 7b – 20b – 32
= 28b – 20b – 32 = 8b – 32.

(ii) -z² + 13z² – 5z + 7z³ – 15z
Rearranging terms:
= 7z³ – z² + 13z² – 5z – 15z
= 7z³ + 12z² – 20z

(iii) p – (p – q) – q – (q – p)
= P – P + q – q – q + p
Rearranging terms:
= p – p + p + q – q – q
= p – q

(iv) 3a-2b — ab — (a-b+ab) + 3ab + b—a
= 3a – 2b – ab – a + b – ab + 3ab + b – a
Rearranging terms:
= 3a – a – a – 2b + b + b – ab – ab + 3ab
= 3a – 2a – 2b – 2ab + 3ab
= a + ab.

(v) 5x²y — 5x² + 3yx² — 3y² + x² — y² + 8xy² – 3y².
Rearranging terms:
= 5x²y + 3yx² – 5x² + x² – 3y² – y² – 3y² + 8xy².
= 8x²y – 4x² – 7y² + 8xy²

(vi) (3y² + 5y – 4) – (8y -y2 – 4).
= 3y² + 5y – 4 – 8y + y² + 4
Rearranging terms:
= 3y² + y² + 5y – 8y – 4 + 4
= 4y² + 3y.

Question 2.
Add:
(i) 3mn, – 5mn, 8 mn, – 4 mn.
(ii) t – 8tz, 3tz -z, z- t.
(iii) – 7mn + 5, 12mn +2, 9 mn -8, -2mn- 3.
(iv) a + b-3, b-a + 3, a-b + 3.
(v) 14x + 10y – 12xy – 13, 18-7x- 10y + 8xy, 4xy.
(vi) 5m – 7n, 3n – 4m + 2, 2m – 3mn – 5.
(vii) 4x²y, – 3xy², – 5xy², 5x²y.
(viii) 3p²q² – 4pq + 5, – 10p²q²,15+9pq + 7p²q².
(ix) ab – 4a, 4b – ab, 4a – 4b.
(x) x² – y² – 1, y² – 1 – x², 1 – x² – y².
Solution:
(i) 3mn + (- 5mn) + 8 mn + (- 4mn)
= 3mn + 8mn – 5mn – 4mn
[Rearranging Terms]
= 11mn – 9mn – 2mn.

(ii) t – 8tz + 3tz -z + z -t
= t – t – z + z – 8tz + 3tz
[Rearranging Terms]
= – 5 tz.

(iii) – 7mn + 5 + 12mn + 2 + 9mn – 8 + (- 2 mn) – 3
= 12mn + 9mn – 7mn – 2mn + 5 + 2 – 8 – 3
[Rearranging Terms]
= 12mn – 4.

(iv) a + b – 3 + b – a + 3 + a – b + 3
= a- a + a + b + b – b + 3 + 3 – 3
[Rearranging Terms]
= a + b + 3.

(v) 14x + 10y – 12rxy – 13 + 18 – 7x – 10y + 8xy + 4xy
= 14x – 7x – 12xy + 8xy + 4xy + 10y – 10y + 18 – 13
[Rearranging Terms]
= 7x + 5.

(vi) 5m – 7n + 3n – 4m + 2 + 2m – 3mn – 5
= 5m- 4m + 2m -7n + 3n – 3mn -5 + 2
[Rearranging Terms]
= 3 m – 4n – 3 mn – 3.

(vii) 4x²y + (- 3xy²) + (- 5xy²) + 5x²y
= 4x²y + 5xy – 5xy² + 3xy²
[Rearranging Terms]
= 9x²y – 8xy².

(viii) 3p²q² – 4pq + 5 + (- 10 p²q²) + 15 + 9pq + 7p²q²
= 3p²q² -10p²q² + 7p²q² + 9pq – 4pq + 15 + 5 [Rearranging Terms]
= 5pq + 20.

(ix) ab – 4a + 46 – ab + 4a – 46
= ab – ab – 4a + 4a + 46 – 46
[Rearranging Terms]
= 0.

(x) x² – y² – 1 + y² – 1 – x² + 1 – x² – y²
= x² – x² – x² – y² + y² – y² – 1 – 1 + 1.
= x² – y² – 1.

Question 3.
Subtract:
(i) – 5y² from y².
(ii) 6xy from – 12xy
(iii) (a – b) from (a + b)
(iii) a(b – 5) from b(5 – a)
(v) – m² + 5mn from 4m² – 3mn + 8
(vi) -x² + 10x – 5 from 5x – 10
(vii) 5a² – 7ab + 5b² from 3ab – 2a² – 2b²
(viii) 4pq – 5q² – 3p² from 5p² + 3q² – pq
Solution:
(i) – 5y² from y².
= y² – 5y² = – 4y².

(ii) 6xy from – 12xy
= – 12xy – 6xy = – 18xy.

(iii) (a – b) from (a + b)
= (a + b) – (a – b)
= a + b – a + b
= a – a + b + b = 2b.

(iv) a(b- 5) from b (5 – a)
= b(5 – a) – a(b – 5)
= 5b – ab – ab + 5a
= 5a + 5b – 2ab

(v) – m² + 5mn from 4m² – 3mn + 8
= (4m² – 3mn + 8) — (— m²+ 5 mn)
= 4m² – 3mn + 8 + m² – 5 mn
= 5m² – 8mn + 8.

(vi) -x² + 10x – 5 from 5x – 10
= 5x -10 – (-x² + 10x – 5)
= 5x – 10 + x² – 10x + 5
= x² – 5x – 5

(vii) 5a² – 7ab + 5b² from 3ab – 2a² – 2b²
= 3ab – 2a² – 2b² – (5a² – 7ab + 5b²)
= 3ab – 2a² – 2b² – 5a² + 7ab – 5b²
= (- 2a² – 5a²) + (-2b² – 5b²) + (3b + 7ab)
= -7a² – 7b² + 10ab

(viii) 4pq – 5q² – 3p² from 5p² + 3q² – pq
= 5p² + 3q² – pq – (4pq – 5q² – 3p²)
= 5p² + 3q² – pq – 4pq + 5q² + 3p²
= (5p²+ 3p²) + (3q²+5q²) + (-pq – 4pq)
= 8p² + 8q² -5pq.

Question 4.
(a) What should be added to x² + xy + y² to obtain 2x² + 3xy ?
(b) What should be subtracted from 2a + 8b + 10 to get – 3a + 7b + 16 ?
Solution:
(a) (2x² + 3xy) – (x² + xy + y²)
= 2x² + 3xy – x² – xy – y²
= 2x² – x² + 3xy – xy – y²
= x² + 2xy – y².

(b) (2a + 8b + 10)-(-3a + 7b + 16)
= 2a + 8b + 10 + 3a – 7b – 16
= 2a + 3a + 8b – 7b + 10 – 16
= 5a + b – 6

Question 5.
(a) What should be taken away from 3x² – 4y² + 5xy + 20 to obtain – x² – y² + 6xy + 20?
Solution:
(a) (3x² – 4y² + 5xy + 20) – (- x² – y² + 6xy + 20)
= 3x² – 4y² + 5xy + 20 + x² + y² – 6xy – 20
= 3x² + x² – 4y² + y² + 5xy – 6xy + 20 – 20
= 4x² – 3y² – xy.

Question 6.
(a) From the sum of 3x – y + 11 and-y – 11, subtract 3x – y – 11.
(6) From the sum of 4 + 3x and 5 – 4x + 2x², subtract the sum of 3x²– 5x and -x² + 2x + 5.
Solution:
(a) (3x – y + 11) + (-y – 11)
= 3x – y + 11 – y – 11
= 3x – y – y + 11 – 11
= 3x – 2y
Then, subtract 3x -y – 11 from obtained expression:
(3ac – 2y) – (3x-y- 11)
= 3x — 2y — 3x+y + 11
= 3x – 3x – 2y + y + 11
= -y + 11

(b) (4 + 3x) + (5 – 4x + 2x² )
= 4 + 3x + 5 – 4x + 2x²
= 3x – 4x + 4 + 5 + 2x²
= – x + 9 + 2x²
Then, (3x² – 5x) + (- x² + 2x + 5)
= 3x² – 5x – x² + 2x + 5
= 3x² – x² – 5x + 2x + 5
= 2x² – 3x + 5
Then according to question :
= (-x + 9 + 2x²) – (2x² – 3x + 5)
= – x + 9 + 2x² – 2x² + 3x – 5
= – x + 3x + 2x² – 2x² + 9 – 5
= + 2x + 4 = 2x + 4.

Haryana Board HBSE 6th Class Maths Solutions

HBSE 6th Class Maths Solutions in English Medium

HBSE 6th Class Maths Chapter 1 Knowing Our Numbers

• Chapter 1 Knowing Our Numbers Intext Questions
• Chapter 1 Knowing Our Numbers Ex 1.1
• Chapter 1 Knowing Our Numbers Ex 1.2
• Chapter 1 Knowing Our Numbers Ex 1.3

HBSE 6th Class Maths Chapter 2 Whole Numbers

• Chapter 2 Whole Numbers Intext Questions
• Chapter 2 Whole Numbers Ex 2.1
• Chapter 2 Whole Numbers Ex 2.2
• Chapter 2 Whole Numbers Ex 2.3

HBSE 6th Class Maths Chapter 3 Playing With Numbers

• Chapter 3 Playing With Numbers Intext Questions
• Chapter 3 Playing With Numbers Ex 3.1
• Chapter 3 Playing With Numbers Ex 3.2
• Chapter 3 Playing With Numbers Ex 3.3
• Chapter 3 Playing With Numbers Ex 3.4
• Chapter 3 Playing With Numbers Ex 3.5
• Chapter 3 Playing With Numbers Ex 3.6
• Chapter 3 Playing With Numbers Ex 3.7

HBSE 6th Class Maths Chapter 4 Basic Geometrical Ideas

• Chapter 4 Basic Geometrical Ideas Intext Questions
• Chapter 4 Basic Geometrical Ideas Ex 4.1
• Chapter 4 Basic Geometrical Ideas Ex 4.2
• Chapter 4 Basic Geometrical Ideas Ex 4.3
• Chapter 4 Basic Geometrical Ideas Ex 4.4
• Chapter 4 Basic Geometrical Ideas Ex 4.5
• Chapter 4 Basic Geometrical Ideas Ex 4.6

HBSE 6th Class Maths Chapter 5 Understanding Elementary Shapes

• Chapter 5 Understanding Elementary Shapes Intext Questions
• Chapter 5 Understanding Elementary Shapes Ex 5.1
• Chapter 5 Understanding Elementary Shapes Ex 5.2
• Chapter 5 Understanding Elementary Shapes Ex 5.3
• Chapter 5 Understanding Elementary Shapes Ex 5.4
• Chapter 5 Understanding Elementary Shapes Ex 5.5
• Chapter 5 Understanding Elementary Shapes Ex 5.6
• Chapter 5 Understanding Elementary Shapes Ex 5.7
• Chapter 5 Understanding Elementary Shapes Ex 5.8
• Chapter 5 Understanding Elementary Shapes Ex 5.9

HBSE 6th Class Maths Chapter 6 Integers

• Chapter 6 Integers InText Questions
• Chapter 6 Integers Ex 6.1
• Chapter 6 Integers Ex 6.2
• Chapter 6 Integers Ex 6.3

HBSE 6th Class Maths Chapter 7 Fractions

• Chapter 7 Fractions Intext Questions
• Chapter 7 Fractions Ex 7.1
• Chapter 7 Fractions Ex 7.2
• Chapter 7 Fractions Ex 7.3
• Chapter 7 Fractions Ex 7.4
• Chapter 7 Fractions Ex 7.5
• Chapter 7 Fractions Ex 7.6

HBSE 6th Class Maths Chapter 8 Decimals

• Chapter 8 Decimals Intext Questions
• Chapter 8 Decimals Ex 8.1
• Chapter 8 Decimals Ex 8.2
• Chapter 8 Decimals Ex 8.3
• Chapter 8 Decimals Ex 8.4
• Chapter 8 Decimals Ex 8.5
• Chapter 8 Decimals Ex 8.6

HBSE 6th Class Maths Chapter 9 Data Handling

• Chapter 9 Data Handling Intext Questions
• Chapter 9 Data Handling Ex 9.1
• Chapter 9 Data Handling Ex 9.2
• Chapter 9 Data Handling Ex 9.3
• Chapter 9 Data Handling Ex 9.4

HBSE 6th Class Maths Chapter 10 Mensuration

• Chapter 10 Mensuration Intext Questions
• Chapter 10 Mensuration Ex 10.1
• Chapter 10 Mensuration Ex 10.2
• Chapter 10 Mensuration Ex 10.3

HBSE 6th Class Maths Chapter 11 Algebra

• Chapter 11 Algebra Intext Questions
• Chapter 11 Algebra Ex 11.1
• Chapter 11 Algebra Ex 11.2
• Chapter 11 Algebra Ex 11.3
• Chapter 11 Algebra Ex 11.4
• Chapter 11 Algebra Ex 11.5

HBSE 6th Class Maths Chapter 12 Ratio and Proportion

• Chapter 12 Ratio and Proportion Intext Questions
• Chapter 12 Ratio and Proportion Ex 12.1
• Chapter 12 Ratio and Proportion Ex 12.2
• Chapter 12 Ratio and Proportion Ex 12.3

HBSE 6th Class Maths Chapter 13 Symmetry

• Chapter 13 Symmetry Intext Questions
• Chapter 13 Symmetry Ex 13.1
• Chapter 13 Symmetry Ex 13.2
• Chapter 13 Symmetry Ex 13.3

HBSE 6th Class Maths Chapter 14 Practical Geometry

• Chapter 14 Practical Geometry Intext Questions
• Chapter 14 Practical Geometry Ex 14.1
• Chapter 14 Practical Geometry Ex 14.2
• Chapter 14 Practical Geometry Ex 14.3
• Chapter 14 Practical Geometry Ex 14.4
• Chapter 14 Practical Geometry Ex 14.5
• Chapter 14 Practical Geometry Ex 14.6

HBSE 6th Class Maths Solutions in Hindi Medium

HBSE 6th Class Maths Chapter 1 अपनी संख्याओं की जानकारी

• Chapter 1 अपनी संख्याओं की जानकारी Intext Questions
• Chapter 1 अपनी संख्याओं की जानकारी Ex 1.1
• Chapter 1 अपनी संख्याओं की जानकारी Ex 1.2
• Chapter 1 अपनी संख्याओं की जानकारी Ex 1.3

HBSE 6th Class Maths Chapter 2 पूर्ण संख्याएँ

• Chapter 2 पूर्ण संख्याएँ Intext Questions
• Chapter 2 पूर्ण संख्याएँ Ex 2.1
• Chapter 2 पूर्ण संख्याएँ Ex 2.2
• Chapter 2 पूर्ण संख्याएँ Ex 2.3

HBSE 6th Class Maths Chapter 3 संख्याओं के साथ खेलना

• Chapter 3 संख्याओं के साथ खेलना Intext Questions
• Chapter 3 संख्याओं के साथ खेलना Ex 3.1
• Chapter 3 संख्याओं के साथ खेलना Ex 3.2
• Chapter 3 संख्याओं के साथ खेलना Ex 3.3
• Chapter 3 संख्याओं के साथ खेलना Ex 3.4
• Chapter 3 संख्याओं के साथ खेलना Ex 3.5
• Chapter 3 संख्याओं के साथ खेलना Ex 3.6
• Chapter 3 संख्याओं के साथ खेलना Ex 3.7

HBSE 6th Class Maths Chapter 4 आधारभूत ज्यामितीय अवधारणाएँ

• Chapter 4 आधारभूत ज्यामितीय अवधारणाएँ Intext Questions
• Chapter 4 आधारभूत ज्यामितीय अवधारणाएँ Ex 4.1
• Chapter 4 आधारभूत ज्यामितीय अवधारणाएँ Ex 4.2
• Chapter 4 आधारभूत ज्यामितीय अवधारणाएँ Ex 4.3
• Chapter 4 आधारभूत ज्यामितीय अवधारणाएँ Ex 4.4
• Chapter 4 आधारभूत ज्यामितीय अवधारणाएँ Ex 4.5
• Chapter 4 आधारभूत ज्यामितीय अवधारणाएँ Ex 4.6

HBSE 6th Class Maths Chapter 5 प्रारंभिक आकारों को समझना

• Chapter 5 प्रारंभिक आकारों को समझना Intext Questions
• Chapter 5 प्रारंभिक आकारों को समझना Ex 5.1
• Chapter 5 प्रारंभिक आकारों को समझना Ex 5.2
• Chapter 5 प्रारंभिक आकारों को समझना Ex 5.3
• Chapter 5 प्रारंभिक आकारों को समझना Ex 5.4
• Chapter 5 प्रारंभिक आकारों को समझना Ex 5.5
• Chapter 5 प्रारंभिक आकारों को समझना Ex 5.6
• Chapter 5 प्रारंभिक आकारों को समझना Ex 5.7
• Chapter 5 प्रारंभिक आकारों को समझना Ex 5.8
• Chapter 5 प्रारंभिक आकारों को समझना Ex 5.9

HBSE 6th Class Maths Chapter 6 पूर्णांक

• Chapter 6 पूर्णांक Intext Questions
• Chapter 6 पूर्णांक Ex 6.1
• Chapter 6 पूर्णांक Ex 6.2
• Chapter 6 पूर्णांक Ex 6.3

HBSE 6th Class Maths Chapter 7 भिन्न

• Chapter 7 भिन्न Ex 7.1
• Chapter 7 भिन्न Ex 7.2
• Chapter 7 भिन्न Ex 7.3
• Chapter 7 भिन्न Ex 7.4
• Chapter 7 भिन्न Ex 7.5
• Chapter 7 भिन्न Ex 7.6

HBSE 6th Class Maths Chapter 8 दशमलव

• Chapter 8 दशमलव Intext Questions
• Chapter 8 दशमलव Ex 8.1
• Chapter 8 दशमलव Ex 8.2
• Chapter 8 दशमलव Ex 8.3
• Chapter 8 दशमलव Ex 8.4
• Chapter 8 दशमलव Ex 8.5
• Chapter 8 दशमलव Ex 8.6

HBSE 6th Class Maths Chapter 9 आँकड़ों का प्रबंधन

• Chapter 9 आँकड़ों का प्रबंधन Intext Questions
• Chapter 9 आँकड़ों का प्रबंधन Ex 9.1
• Chapter 9 आँकड़ों का प्रबंधन Ex 9.2
• Chapter 9 आँकड़ों का प्रबंधन Ex 9.3
• Chapter 9 आँकड़ों का प्रबंधन Ex 9.4

HBSE 6th Class Maths Chapter 10 क्षेत्रमिति

• Chapter 10 क्षेत्रमिति Intext Questions
• Chapter 10 क्षेत्रमिति Ex 10.1
• Chapter 10 क्षेत्रमिति Ex 10.2
• Chapter 10 क्षेत्रमिति Ex 10.3

HBSE 6th Class Maths Chapter 11 बीजगणित

• Chapter 11 बीजगणित Intext Questions
• Chapter 11 बीजगणित Ex 11.1
• Chapter 11 बीजगणित Ex 11.2
• Chapter 11 बीजगणित Ex 11.3
• Chapter 11 बीजगणित Ex 11.4
• Chapter 11 बीजगणित Ex 11.5

HBSE 6th Class Maths Chapter 12 अनुपात और समानुपात

• Chapter 12 अनुपात और समानुपात Intext Questions
• Chapter 12 अनुपात और समानुपात Ex 12.1
• Chapter 12 अनुपात और समानुपात Ex 12.2
• Chapter 12 अनुपात और समानुपात Ex 12.3

HBSE 6th Class Maths Chapter 13 सममिति

• Chapter 13 सममिति Intext Questions
• Chapter 13 सममिति Ex 13.1
• Chapter 13 सममिति Ex 13.2
• Chapter 13 सममिति Ex 13.3

HBSE 6th Class Maths Chapter 14 प्रायोगिक ज्यामिती

• Chapter 14 प्रायोगिक ज्यामिती Intext Questions
• Chapter 14 प्रायोगिक ज्यामिती Ex 14.1
• Chapter 14 प्रायोगिक ज्यामिती Ex 14.2
• Chapter 14 प्रायोगिक ज्यामिती Ex 14.3
• Chapter 14 प्रायोगिक ज्यामिती Ex 14.4
• Chapter 14 प्रायोगिक ज्यामिती Ex 14.5
• Chapter 14 प्रायोगिक ज्यामिती Ex 14.6

Haryana State Board HBSE 7th Class Maths Solutions Chapter 12 Algebraic Expressions Ex 12.1 Textbook Exercise Questions and Answers.

Haryana Board 7th Class Maths Solutions Chapter 12 Algebraic Expressions Exercise 12.1

Question 1.
Get the algebraic expressions in the following cases using variables, constants and arithmetic operations.
(i) Subtraction of z from y.
(ii) One half of the sum of number x and y.
(iii) The number z multiplied by itself.
(iv) One fourth of the product of number p and q.
(v) Number x and y both squared and added.
(vi) Number 5 added to three times and product of number m and n.
(vii) Product of numbers y and z subtracted from 10.
(viii) Sum of numbers a and b subtracted from their product.
Solution:
(i) y -z
(ii) \(\frac{1}{2}\) (x + y)
(iii) z x z =z²
(iv) \(\frac{1}{4}\) (p x q) = \(\frac{1}{4}\) (pq) = \(\frac{pq}{4}\)
(v) x² + y²
(vi) 3mn + 5
(vii) 10 – yz
(viii) ab – (a + b)

Question 2.
(i) Identify the terms and their factors in the foUowing expressions. Show the terms and factors by tree diagrams.
(a) x – 3
(b) 1 + x + x²
(c) y – y³
(d) 5xy² + 7x²y
(e) -ab + 2b² – 3a²
(ii) Identify terms and factors in the expressions given below:
(a) -4x + 5
(b) -4x + 5y
(e) 5y + 3y²
(d) xy + 2x²y²
(e) pq+q
(f) 1.2ab – 2.4b + 3.6a
(g.) \(\frac{3}{4} x+\frac{1}{4} h\)
(h) \(\frac{l+m}{5}\)
[Hint: Separate l and m terms]
(j) 0.1p² + 0.2q²
(j) \(\frac{3}{4}\) (a – b) + \(\frac{7}{4}\)
[Hint: Open the brackets]
Solution:
(i) (a) x – 3: In this expression x – 3 consists two terms x and -3.

(ii) (a) – 4x + 5.
The expression (- 4x + 5) consists of two terms – 4x and 5. The term – 4x is product of – 4 and x. And the term 5 has only one factor that is 5.
Terms are – 4x and 5.
Factors are-4 andx, of-4x and factor 5 of 5.

(b) -4x + 5y
In the expression -4x + 5y The terms are – 4x and 5y and factors are- 4 and x of-4x and 5 andy of 5y.

(c) 5y + 3y²
In the expression 5y + 3y²
The terms are 5y and 3y² and factors are 5 and y of 5y, 3, y and y of 3y².

(d) xy + 2x²y²
In the expression xy + 2x²y²
The terms are xy and 2x²y²
And factors are x and y of xy and 2, x, x, y and y of 2x²y².

(e) pq + q
In the expression pq + q.
The terms are pq and q.
The factors are p and q of pq and q of q because q has only one factor.

(f) In the expression 1.2ab – 2.4b + 3.6a
The terms are 1.2ab, 2.4b and 3.6a and
factors are 1.2, a and b of 1.2ab, 2.4, and 6 of 2.4b; 3.6 and a of 3.6a.

(g) \(\frac{3}{4} x+\frac{1}{4}\)

(i) 0.1p² + 0.2q²
In this expression 0.1 p² + 0.2 q²
The terms are 0.1 p² and 0.2 q² and factors are 0.1, p, p of 0.1 p² and 0.2, q, q of 0.2 q².

Question 3.
Identify the numerical co¬efficients of terms other than constants in the following expressions:
(i) 5-3t²
(ii) 1 + t + t² + t³
(iii) x + 2xy + 3y
(iv) 100 m + 1000 n
(v) – p²q² + 7pq
(vi) 1.2a + 0.8 b
(vii) 3.14 r²
(viii) 2(l + b)
(ix) 0.1 y + 0.01 y².
Solution:

Question 4.
(a) Identify terms which contain x and give the co-efficient of x.
(i) y²x + y
(ii) 13y² – 8yx
(iii) x + y + 2
(iv) 5 + z + zx
(v) 1 + x +xy
(vi) 12xy² + 25
(vii) 7x + xy²
(b) Identify terms which contains y² and give the co-efficient of y².
(i) 8 – xy²
(ii) 5y² + 7x
(iii) 2x²y – 15xy² + 7y²
Solution:
(a)

(b)

Question 5.
Classify into monomials, binomials and trinomials. Give reasons for your answer:
(i) 4y – 7z
(ii) y²
(iii) x + y – xy
(iv) 100
(v) ab-a-b
(vi) 5 – 3t
(vii) 4p²q – 4pq²
(viii) 7mn
(ix) z² – 3z + 8
(x) a² + b²
(xi) z² + z
(xii) 1 + x + x²
Solution:
(i) 4y – 7z is binomial. It contain 2 terms.
(ii) y² is monomial. It contains 1 term.
(iii) x + y – xy is trinomial.lt contains 3 terms.
(iv) 100 is monomial. It contains 1 term.
(v) ab-a-b is trinomial.lt contains3 terms
(vi) 5 – 3t is binomial. It contains 2 terms.
(vii) 4p²q – 4pq² is binomial.lt contains 2 terms.
(viii) 7mn is monomial. It contains 1 term.
(ix) z² – 3z + 8 is trinomial. It contains 3 terms.
(x) a² + b² is binomial. It contain 2 terms.
(xi) z² + z is binomial. It contains 2 terms.
(xii) 1 + x + x² is trinomial. It contains 3 terms.

Question 6.
State whether the given pair of terms is of like or unlike terms.
(i) 1,100
(ii) -7x, \(\frac{5}{2}\)x
(iii) – 29x, – 29y
(iv) 14xy, 42xy
(v) 4m²p,4mp²
(vi) 12xz, 12x²z²
Solution:

Question 7.
Identify like terms in the following:
(a) – xy², – 4yx², 8x², 2xy², 7y, – 11x², – 100x, -11yx, 20x²y, -6x², y, 2xy, 3x.
(b) 10pq, 7p, 8q, – p²q², – 7qp, – 100q, – 23, 12q²p², -5P², 41, 2405p, 78qp, 13p²q, – 9pq², qp², 701P².
Solution:

Haryana State Board HBSE 7th Class Maths Solutions Chapter 11 Perimeter and Area InText Questions and Answers.

Haryana Board 7th Class Maths Solutions Chapter 11 Perimeter and Area InText Questions

Try These (Page 205)

Question 1.
What would you need to find area or perimeter, to answer the following?
1. how much space does a black-board occupy?
2. What is the length of wire require to fence a rectangular flower bed?
3. What distance would you cover by taking two rounds of a triangular park?
4. How much plastic sheet do you need to cover a rectangular swimming pool?
Solution:
1. Perimeter of rectangular black board = 2 x (l + b)
2. Length of the wire = Perimeter of rectangular flower bed = 2 x (l + b)
3. Perimeter of rectangular park = 2(l + b).
4. Perimeter ofrectangular swimming pool = 2(l+b).

Try These (Page 206)

Question 1.
Experiment with several such shapes and cut-outs. You might find it useful to draw these shapes on squared sheets and compute their area and perimeter.
You have seen that increase in perimeter does not mean that area will also increase.
Question 2. Give two examples where the area increases as the perimeter increases.
Question 3. Give two examples where the area does not increase when perimeter increases.
Solution:
1. Area of rectangle = l × b
= 3 cm × 2 cm
= 6 cm².

Perimeter of rectangle = 2 (l + b)
= 2 (3 + 2) = 2 × 5 = 10 cm. 2.

2. (a) Area of rectangle = l × b
= 4 cm × 3 cm = 12 cm².

Perimeter of rectangle = 2(l + b)
= 2 (4 cm + 3 cm)
= 2 × 7 cm = 14 cm.

(b) Area of square (Fig.) = Side × side = 2 cm × 2 cm = 4 cm².
Perimeter of square = 4 × side = 4 × 2 cm = 8 cm.

3. Area of square = 3 cm × 3 cm = 9 cm²
Perimeter of square = 4 × 3 cm = 12 cm.
Hence area decreases as the perimeter increases.

Try These (Page 210)

Question 1.
Each of the following rectangles of the length 6 cm and the breadth 4 cm is composed of congurent polygons. Find the area of each polygon.

Solution:
Since, the two parts are congruent to each other.
So, the area of any part is equal to the area of the other part.
Therefore the area of congurent polygons 1
= \(\frac{1}{2}\) (The area of rectangle)
= \(\frac{1}{2}\) (6 cm × 4 cm) = \(\frac{1}{2}\) × 24 cm² = 12 cm²

Try These (Page 212)

Question 1.
Find the area of following parallelograms :
(i)

(ii) In a parallelogram ABCD, AB = 7.2 cm and the prependicular from con AB is 4.5 cm.
Solution:
Area of parallelogram = base × height
(i) Here, base = 8 cm,
height = 3.5 cm
∴ Area = 8 × 3.5 = 28.0 cm²

(ii) base = 8 cm,
height = 2.5 cm
Area = 8 × 2.5 = 20.0 cm²

(iii) Here base = 7.2 cm
height = 4.5 cm
(ii) Area of parallelogram = base × height
= 7.5 × 2.5 = 32.40cm²

Try These (Page 213)

Question 1.
Try the above activity with dirrerent types of triangles.
Solution:
Area of parallelogram = base × height
or, Area of parallelogram
= [ \(\frac{1}{2}\) × b × h + \(\frac{1}{2}\) × b × h ]
i.e. In a parallelogram,
Area of each triangle = \(\frac{1}{2}\) (b × h)

Question 2.
Take different paralelograms. Divide each of the parallelograms into two triangles by cutting along any of its diagonals. Are the triangles congruent?
Solution:
In a parallelogram have two triangle, ΔABD and ΔBDC.

In a parallellogram,
AB = DC andAD = BC
and AB || DC and AD || BC.
Hence, AB = DC, B = BD (common)
and AD = BC
∴ ∆ABD ≅ ∆BDC (SSS congruency)
So, by CPCT all angles are equal.

Try These (Page 219)

Question 1.
(a) Which square has the larger perimeter ?
(b) Which is larger, perimeter of smaller square or the circumference of the circle?

Solution:
(a) Perimeter of square = 4 × side According to fig. outer square is larger perimeter than inner square.
(b) Perimeter or circumference of circle
= 2πr [π = 22/7]
and perimeter of square = 4 × side
If side of square = diameter of circle = 14 cm
hence radius of circle = \(\frac{14}{2}\) = 7 cm
Now, Perimeter of square = 4 × 7 = 28 cm
and circumference of circle = 2 × \(\frac{22}{7}\) × 7
= 44 cm.
Hence, perimeter of square is smaller than circumference of circle.

Try These (Page 222)

Question 1.
Draw circles of different radii on a graph paper. Find the area by counting the number of squares. Also find the area by using the formula. Compare the two amswer.


Solution:
(i) radius of circle = 1 cm
d = 2cm
hence, side of square = 2 cm
Area of square = (side)²
= (2)² = 4 cm²

(ii) r = 2 cm, ∴ d = 4 cm
Area of Square = (4)² = 16cm²

(iii) r = 3 cm, ∴ d = 6 cm
Area of Square = (6)² = 36 cm²
If the area by counting the number of square and compare the answer, we get.
Area circle = Area of Square

Try These (Page 225)

Question 1.
Convert the following:
(i) 50 cm² in mm²
(ii) 50 ha in m²
(Hi) 110 m² in cm²
(iv) 1,000 cm² in m²
Solution:
(i) ∵ 50 cm² = 50 × 100 mm²
= 5000 mm²

(ii) ∵ 1 hectare = 10,000 m²
∴ 2 hectare = 2 × 10,000 m²
= 20,000 m²

(iii) ∵ 1 m² = 10,000 cm²
∴ 10 m² = 10 × 10,000 cm²
= 1,00,000 cm²

(iv) ∵ 10000 cm² = 1 m²
∴ 1² = \(\frac{1}{10000}\)m²
∴ 1000 cm² = \(\frac{1}{10000}\) m² × 1,000
= \(\frac{1}{10}\)cm² = 0.1 m²

Haryana State Board HBSE 7th Class Maths Solutions Chapter 11 Perimeter and Area Ex 11.3 Textbook Exercise Questions and Answers.

Haryana Board 7th Class Maths Solutions Chapter 11 Perimeter and Area  Exercise 11.3

Question 1.
Find the circumference of the circles with the following radii :
(a) 10 cm
(b) 28 mm
(c) 20 cm[Take π=22/7]
Solution:
Circumference of the circle = 2πr
or C = 2πr
(a) r = 10 cm
C = 2πr
= 2 × \(\frac{22}{7}\) × 10 = \(\frac{440}{7}\) = 62.86

(b) r = 28 mm
C = 2 × \(\frac{22}{7}\) x 28 = 176cm

(c) r = 20 cm
C = 2 × \(\frac{22}{7}\) x 20
= \(\frac{880}{7}\) = 125.7 cm

Question 2.
Find the area of the following circles:
(a) radius = 14mm
(b) diameter =49 m
(c) radius =5 cm [Take π= 22/7]
Solution:
Area of circle = πr², where π= \(\frac{22}{7}\)
(a) r = 14mm
A = \(\frac{22}{7}\) × (14)²
=\(\frac{22}{7}\) × 14 × 14 = 316 mm².

(b) d = 49 cm, r = \(\frac{49m}{2}\)
∴ A = \(\frac{22}{7} \times \frac{49}{2} \times \frac{49}{2}\)
= 1886.5m².

(c) r = 5cm
∴ A= \(\frac{22}{7}\) × (5)²
= \(\frac{22}{7}\) × 5 × 5 = 78.57 cm².

Question 3.
If the circumference of a circular sheet is 154 m, find its radius. Also find the area of the sheet.
[Tube π = 22/7]
Solution:
Circumference of circular sheet = 2πr
or, 154m = 2 × \(\frac{22}{7}\) × r
or, \(\frac{44}{7}\) r = 154 m.
or, 44r = 154 × 7
∴ r = \(\frac{154 \times 7}{44}=\frac{49}{2}\)
r = 24.50cm
Area of circular sheet = 2πr²
= \(\frac{22}{7}\) × 24.50 × 24.50 = 1886.5 cm².

Question 4.
A gardener wants to fence a circular gardezi of thc diameter 21 ni. Find the length of the rope he needs to purchase, if lic makes 2 rounds of feiice. Also find the cost of the rope, ¡lit cost Rs. 4 per meter.
[Take π = 22/7]
Solution:
d = 21 m
∴ r = \(\frac{21}{2}\)
Circumference of circular
garden = 2πr
or,length of the rope
= Circumference of garden = 2πr = 2 × \(\frac{22}{7} \times \frac{21}{2}\) = 66m
∴ length of the 2 rounds of the fence
= 2 × 66 rn = 132 m
The cost of 132 m rope = 132 × 4 = Rs.528.

Question 5.
From a circular sheet of radius 4 cm, a circle of radius 3 cm is removed (Fig. 11.30). Find the area of the remaining sheet.
[Take π = 22/7]
im 1
Solution:
Radius of the circular sheet (R) = 4 cm
Radius of the removed circular sheet (inner circle)
(r) = 3 cm
Since
Area of a circle = π × (Radius)²
Area of the remaining sheet = Area of outer circle – Area of inner circle = πR² – πr² = π[R² – r²]
= π(R + r) (R – r)
= 3.14 (4 + 3) (4 – 3) cm²
= 3.14 (7) (1) cm² = 21.98 cm²

Question 6.
Saima wants to put a lace on the edge of a circular table cover of diameter 1.5 m. Find the length of the lace required and also find its cost if one meter of the lace cost Rs. 15. [Take π = 3.14]

Solution:
d = 1.5 m,
r = \(\frac{1.5}{2}\) m
=> Circumference of the table cover = 2πr
= 2 × 3.14 × \(\frac{1.5}{2}\)m
= 4.71m
∴ Length of the lace required = 4.71 m
∴ Cost of the lace = Rs. 4.71 × 15 = Rs. 70.65.

Question 7.
Find the perimeter of the adjoining Fig. , which is a semicircle including its diameter.

Solution:
Perimeter or circumference of semicircle = \(\frac{2 \pi r}{2}\)
= πr
= \(\frac{22}{7}\) × 10 cm
= \(\frac{220}{7}\) cm = 31.43 cm.

Question 8.
Find the cost of polishing a circular table-top of diameter 1.6 m, if the rate of polishing is Rs. 15 per sq. m.
[Take π = 3.14]
Solution:
Diameter of the table-top = 1.6 m
∴ Radius = \(\frac{1.6}{2}\)
= 0.8 m
∴ Area of the table-top = πr²
= 3.14 × (0.8)² m²
= 3.14 × 0.64 m²
= 2.0096 m²
∴ Cost of polishing the table-top
= Rs. 2.0096 × 15
= Rs. 30.144.

Question 9.
Shazli took a wire of length 44 cm and bent into the shape of a circle. Find the radius of that circle. Also find the area. If the same wire is bent into the shape of a square, what will be the length of each of its side ? Which figure encloses more area circle or square ? [Take π = 22 / 7]

Solution:
Circumference of circle = 2πr
or, 44 cm = 2 × \(\frac{22}{7}\) x r,
or, 44 r = 44 × 7
∴ r = \(\frac{44 \times 7}{44}\) = 7 cm
∴ radius of circle = 7 cm
∴ area of circle = πr²
= \(\frac{22}{7}\) × 7 × 7 = 154 cm²
Now, circumference of circle = perimeter of square
or 44 cm = 4 × side,
4 × side = 44 cm
∴ side = \(\frac{44}{4}\) = 11 cm.
Area of square = side × side = 11 cm × 11 cm = 121 cm².
Hence, area of circle is larger than area of square.
More area of circle = 154 – 121 = 33 cm².

Question 10.
From a circular card sheet of radius 14 cm, two circles of radius 3.5 cm and a rectangle of length cm and breadth 1 m are removed, (as shown in Fig.) Find the area of the remaining sheet.

[Take π = 22/7]
Solution:
From the figure,
Area of circular card sheet = πr²
= \(\frac{22}{7}\) × 14 × 14 [∵ r = 14 cm]
= 616 cm².
Area of two circles

Area of rectangle
= l × b = 3 × 1 = 3 cm².
[∵ l = 3 cm, b = 1 cm]
hence area of the remaining sheet
= Area of circular card sheet – (Area of two circles + area of rectangle)
= 616 cm² – (77 cm² + 3 cm²)
= 616 cm² – 80 cm² = 536 cm².

Question 11.
A circle of radius 2 cm is cut out from a square piece of an aluminium sheet of side 6 cm. What is the area of the left over aluminium sheet ?

[Take π = 3.14}
Solution:
Area of square piece = side × side
= 6 × 6 = 36 cm².
Area of circle = πr² = 3.14 × 2 × 2
= 12.56 cm².
Area of the left over aluminium sheet = Area of square piece – Area of circle
= 36 cm² – 12.56 cm² = 23.44 cm².

Question 12.
The circumference of a circle is 31.4 cm. Find the radius and the area of the circle ? [Take π = 3.14]
Solution:
Circumference of a circle = 31.4 cm
2πr = 31.4 cm
2 × \(\frac{22}{7}\) × r = 31.4cm
∴ r = 31.4 × \(\frac{7}{22} \times \frac{1}{2} \mathrm{cm}\)
r = 5 cm
Area of the circle = πr²
= 3.14 × 5 × 5 cm² = 3.14 × 25 cm² = 78.5 cm².

Question 13.
A circular flower bed is surrounded by a path 4 m wide. The diameter of the flower bed is 66 m. What is the area of this path ?
[Take π = 3.14]

Solution:
Wide of path = 4 m
Outer circular diameter = 66 m + 4 + 4 = 74 m
∴ r = \(\frac{74}{2}\)= 37 m.
Inner circular diameter = 66 m
∴ r = \(\frac{66}{2}\)= 33 m.
The area of this path = Area of outer circular flower bed – Area of inner circular flower, bed.
= π[(37 m)² – (33 m)²]
= \(\frac{22}{7}\) [1369 – 1089] m²
= \(\frac{22}{7}\) × 280 m² = 880 m²
or, 3.14 × 280 m² = 879.2 m².

Question 14.
A circular flower garden has an area of about 314 m². A sprinkler at the center of the garden can cover an area that has a radius of 12 m. Will the sprinkler water the entire garden ? [Take π =3.14]
Solution:
Area of circular flower garden = πr²
314 m² = πr²
\(r^{2}=\frac{314 \mathrm{~m}^{2}}{\pi}=\frac{314}{3.14}=\frac{31400}{314}=100 \mathrm{~m}^{2}\)
∴ r = 10 m
But radius of sprinkler water = 12 m
∴ Area of sprinkler water = πr²
= 3.14 × 12 × 12 = 452.16 m².
Yes, because the area of sprinkler water is greater than area of circular flower garden.

Question 15.
Find the circumference of the inner and the outer circle as shown in
[Take π = 3.14]

Solution:
Radius of the outer circle = 19 m
∴ Circumference of the outer circle = 2π r = 2 × 3.14 × 19 m = 119.32 m
Radius of the inner circle = 19 m – 10 m = 9 m
∴ Circumference of the inner circle = 2π r = 2 × 3.14 × 9m = 56.52 m.

Question 16.
How many times the wheel of radius 28 cm must rotate to go 352 m.
[Take π = 22/7]
Sol. Radius of the wheel (r) = 28 cm
Circumference of the wheel = 2πr cm = 2x \(\frac{22}{7}\) × 28cm = 176 cm
∴ Number of times the wheel must rotate to go 352 m
= \(\frac{352 \times 100}{176}\) = 200

Question 17.
The minute hand of a circular clock is 15 cm long. How far does the tip of the minute hand move in 1 hour.
[Take π = 3.14]

Solution:
Circumference of circular clock = 2πr
= 2 ×3.14 × 15 [π = 15 cm]
= 94.20 cm
1 hour = 60 minute
In one hour the tip of the minute hand complete the one revolution
= The circumference of the circular clock.
hence the distance of the moving minute hand = 94.20 cm.

Haryana State Board HBSE 7th Class Maths Solutions Chapter 11 Perimeter and Area Ex 11.4 Textbook Exercise Questions and Answers.

Haryana Board 7th Class Maths Solutions Chapter 11 Perimeter and Area  Exercise 11.4

Question 1.
A garden is 90 m long and 75 m broad. A path 5 m wide is to be built outside and around it. Find the area of the path. Also find the area of the garden in hectare.

Solution:
Area of the garden = l x b
= 90 x 75 m² = 6750 m²
= \(\frac{6750}{10000}\)hectare = 0.675 hectare.
Area of the rect angle PQRS = l x b
= (90 + 5 + 5) x (75 + 5 + 5)
= 100 m x 85 m = 8500 m²
= \(\frac{8500}{10000}\) hectare = 0.850 hectare
∴ Area of the path = Area of the recangle PQRS – Area of the rectangle ΔBCD
= 0.850 hect – 0.675 hect.
∴ 8500-6750 = 1750 m²
= 0.175 hectare.

Question 2.
A 3m wide path runs outside around a rectangular park of the length 125 m and the breadth 65 m. Find the area of the path.

Solution:
For outside rectangle
Length = 125 + 3 + 3 = 131 m
Breadth = 65+ 3 + 3 = 71 m
Area of external rectangle = 131 x 71 = 9301 m².
For internal rectangle
Length = 125 m
Breadth = 65 m
Area of inner rectangle
= 125 x 65 = 8125 m²
Area of path = 9301 – 8125 = 1176 m².

Question 3.
A picture is
painted on a cardboard 8 cm long and 5 cm wide such that there is a margin of 1.5 cm along each of its sides. Find the total area of the margin.

Solution:
Let us represent the card board of length 8 cm and width 5 cm by the rectangle ABCD, and the painting by the rectangle EFGH as shown in Fig.
∵ AB = 8 cm, BC = 5 cm
∴ EF = (8-2 x 1.5) cm
= (8 – 3) cm = 5 cm
and FG = (5 — 2 x 1.5) cm
= (5 – 3) cm = 2 cm
∴ area of margin
= Area of rectangle ABCD – Area of rectangle EFGH.
= 8 cm x 5 cm – 5 cm x 2 cm
= 40 cm² – 10 cm² = 30 cm².

Question 4.
A verandah of width 2.25 m is constructed all along outside room which 5.5 m long and 4 m wide. Find
(i) the area of the verandah.
(ii) the cost of cementing the floor of the verandah at the rate ofRs. 200 per m².

Solution:
Let us represent the room 5.5 m long and 4 m wide by the rectangle ABCD and the verandah 1.25 m wide all along the outside of the room by the rectangle EFGH as shown in Fig. 11.44.
Now, EF = 2.25 m + 5.5 m + 2.25 m = 10 m
and FG = 2.25 + 4 m + 2.25 m = S.5 m
Therefore the area of the verandah,
= Area of rectangle EFGH – Area of rectangle ABCD
= 10 m x 8.5 m – 5.5 m x 4 m
= 85.0 m² – 22.0 m² = 63 m².
Rate of cementing of the floor = Rs. 200 per sq. m.
∴ total cost of cementing = Rs. 200 x 63
= Rs. 6000. =Rs. 12,600.

Question 5.
A path 1 m wide is built along the border inside a square garden of side 30 m. Find
(i) the area of the path
(ii) the cost of planting grass in the remaining portion of the garden at the rate of Rs. 40 per m³.

Solution:
Let us represent the square garden of side 30 m by the square ABCD and the path along the border by shaded region as shown as Fig. 11.48.
Area of the remaining portion of the garden = Area of square EFGH
Now, side of square EFGH
= (30 – 2 x 1) m = 28 m
So, the area of the remaining portion of garden = 28 m. x 28 m = 784 m².
Because, rate of covering the remaining
portion of the garden by grass
= Rs, 40.00 per m².
∴ Cost of covering it by grass = Rs. 40.00 x 784 = Rs. 31360.00.

Question 6.
Two cross
roads, each of width 10 m, cut at right angles through the centre of a rectangular park of ?
the length 700 m and the breadth 300 m and parallel to its sides. Find the area of the roads. Also find the area of park excluding cross roads. Give the answer in hectares.

Solution:
Length = 700 m
Breadth = 300 m
Area of road parallel to length
= 700 x 10 = 7000 m²
Area of road parallel to breadth
= 300 x 10 = 3000 m²
Area-of common square = 10 x 10 = 100 m²
∴ Area of roads = 7000 + 3000-100 = 9900 m².
Now, the area of park excluding the cross
road
= (Area of rectangular park – Area of roads)
= [(700 x 300) m²-9900 m²]
= (210000 m²-9900 m²)
= 200,100 m².
∵ 10000 m² = 1 hectare
∴ 1m² = \(\frac{1}{10000}\) hectare
∴ 200100 m² = \(\frac{1}{10000}\) x 200100
= \(\frac{2001}{100}\) = 20.01 hec
= 20.01 hectare.

Question 7.
Through „ a rectangular Held of the length 90 m and the breadth 60 m, two roads are constructed which are parallel to the sides and cut each other at right angles through the centre of the fields. If the width of each road is 3 m, find
(i) the area covered by the roads
(ii) the cost of constructing the roads at the rate ofRs. 110 per cm².

Solution:
For field Length = 90 m
Breadth = 60 m
width of road = 3m
Area of road parallel to length
= 90 x 3 = 270 m²
Area of road parallel to breadth
= 60 x 3 = 180 m²
Area of common square = 3 x 3 = 9 m²

(i) Area of roads = 270 + 180 – 9 = 441 m²
(ii) Rate of construction = Rs. 110 per m²
Cost of constructing roads = 110 x 441 = Rs. 48510.

Question 8.
Pragya wrapped a cord around a circular pipe of the radius 4 cm (shown below) and cut off the length required of the card. Then she wrapped it around the square box of the side 4 cm (also shown). Did she have any cord left ?

Solution:
Circumference of circulare pipe
= 2πr
= 2 x 3.14 x 4 [∵ r = 4]
= 25.12 cm
Perimeter of circular pipe
= 4 x side
= 4 x 4 cm = 16 cm
Cordleft = 25.12 — 16 = 9.12cm

Question 9.
The Fig.
11.49, represents a rectangular lawn with a circular flower bed in the middle. Find.

(i) the area of the whole land
(ii) the area of the flower bed
(iii) the area of the lawn excluding the area of the flower-bed
(iv) the circumference of the flower bed.
Solution:
(i) The area of whole rectangular lawn = l x b
length = 5m
breadth = 10 m
∴ area of the whole rectangular lawn
= l x b = 5 x 10 = 50 m².

(ii) Area of circular flower bed = πr²
= 22 x (1)²
[r = \(\frac{\mathrm{d}}{2}=\frac{2}{2}\) = 1m]
= 22 m².

(iii) Area of lawn excluding area of flower bed
= [(Area of rectangular whole land) – (Area of circular flower bed)]
= (50 – 20) m² = 30 m².

(iv) Circumference of flower-bed = 2πr
= 2 x \(\frac{22}{7}\) x 1 = \(\frac{44}{7}\) = 6.28 m (approx.)

Question 10.
Find the area of shaded portion.

Solution:
(i) Area of rectangle ABCD = l x b
length = 10 cm
breadth = 18 cm
∴ area of rectangle = 10 x 18 = 180 cm².
Now,area of ΔEAF
= \(\frac{1}{2}\) b x h = \(\frac{1}{2}\) x 6 x 10
= \(\frac{60}{2}\) = 30 cm²,

and area of of ΔEBC
= \(\frac{1}{2}\) b x h = \(\frac{1}{2}\) x 8 x 10
= \(\frac{80}{2}\) = 40 cm²,
Area of shaded portion
= [(Area of rectangle ΔBCD) – (Area of ΔEAF + area of ΔEBC)]
= [(180)-(40 + 30)3 = [180 — 70] = 110 cm²,

(ii) Area of squares PQRS = Side x side side = 20 cm
∴ area of square = 20 x 20 = 400 cm².
Now,area of ΔTSU
= \(\frac{1}{2}\) b x h = \(\frac{1}{2}\) x 10 x 10
= \(\frac{100}{2}\) = 50cm²
Now,area of ΔQPT
= \(\frac{1}{2}\) b x h = \(\frac{1}{2}\) x 10 x 20
= \(\frac{200}{2}\) = 100cm²
andarea of ΔQRU
= \(\frac{1}{2}\) b x h = \(\frac{1}{2}\) x 10 x 20
= \(\frac{200}{2}\) = 100cm²

Area of shaded portion = [(Area of square PQRS) – (Area of ΔTSU) + (Area of ΔQPT) + (Area of ΔQRU)]
= [(400) – (50 + 100 + 100)] cm²
= (400 – 250) cm² = 150 cm².

Question 11.
Find the area of the quadrilateral ABCD. Here, AC = 22 cm, BM = 3 cm, DN = 3 cm, and BM ⊥ AC, DN ⊥ AC.

Solution:
In AABC, height = 3 cm base = 22 cm
∴ area of ΔABC = \(\frac{1}{2}\) x b x h
= \(\frac{1}{2}\) x 22 x 3
= \(\frac{66}{2}\) = 33 m²
And area ΔADC,
height = 3 cm
base = 22 cm
∴ area of ΔADC = \(\frac{1}{2}\) x b x h
\(\frac{1}{2}\) x 22 x 3 = \(\frac{66}{2}\) = 33 m²
∴ Area of quadrilaterial ABCD = Area of ΔABC + Area ΔADC
= (33 + 33) cm² = 66 cm².

Haryana State Board HBSE 7th Class Maths Solutions Chapter 11 Perimeter and Area Ex 11.2 Textbook Exercise Questions and Answers.

Haryana Board 7th Class Maths Solutions Chapter 11 Perimeter and Area  Exercise 11.2

Question 1.
Find the area of each of the following parallelograms :

Solution:
(a) Length of base (6) = 7 cm
height (h) = 4 cm
∴ Area of the parallelogram
=6 x A = (7 x 4) cm² = 28 cm².

(b) The length of base = 5 cm height (h) = 3 cm
∴ Area of the parallelogram
= b x h = (5 x 3) cm² = 15 cm².

(c) The length of base = 2.5 cm
height (h) = 3.5 cm
∴ Area of the parallelogram
=b x h = (2.5 x 3.5) cm² = 8.75 cm².

(d) The length of base (6) = 5 cm
height (h) = 4.8 cm
∴ Area of the parallelogram
=b x h = (5 x 4.8) cm² = 24.0 cm².

(e) The length of base (b) = 2 cm
height (h) = 4.4 cm
∴ Area of the parallelogram
=b x h = (2 x 4.4) cm² = 24.0 cm².

Question 2.
Find the area of each of the following triangles:

Solution:
(a) The length of base (b) = 4 cm
height (h) = 3
∴ Area of triangle = \(\frac{1}{2}\) x b x h = \(\frac{1}{2}\) x 3 x 4 = \(\frac{12}{2}\) = 6 cm².

(b) base (b) = 4 cm
height (h) = 3
∴ Area of triangle = \(\frac{1}{2}\) x b x h = \(\frac{1}{2}\) x 3 x 4 = \(\frac{16}{2}\) = 8 cm².

(c) base (b) = 3 cm
height (h) = 4
∴ Area of triangle = \(\frac{1}{2}\) x b x h = \(\frac{1}{2}\) x 3 x 4 = \(\frac{12}{2}\) = 6 cm².

(d) base (b) = 3 cm
height (h) = 2
∴ Area of triangle = \(\frac{1}{2}\) x b x h = \(\frac{1}{2}\) x 3 x 2 = \(\frac{6}{2}\) = 3 cm².

Question 3.
Find the missing values :

Solution:
(a) Base = 20 cm,
Area of the parallellogram = 246 cm²
Height = \(\frac{\text { Area }}{\text { Base }}=\frac{246}{20}\) cm
= 12.3 cm
∴ The missing value (height) = 12.3 cm²

(b) Height = 15 cm,
∴ Area of the parallelogram = 154.5 cm²
Area of the parallelogram = b x h
⇒ 154.5 = b x 15
⇒ b = \(\frac{154.5}{15}\) = 10.3 cm
∴ The missing value (base)
= 10.3 cm

(c) Height = 8.4 cm,
Area of the parallelogram
= 48.72 cm²
Let the base of the parallelogram = b cm
∴ b x h = 48.72
⇒ b x 8.4 = 48.72
b = \(\frac{48.72}{8.4}\) = 5.8 cm
Thus, the missing value (base) = 5.8 cm

(d) Base (6) = 15.6 cm,
Area of the parallellogram
= 16.38 cm²
Let the height be = h cm
b x h = 16.38 cm²
⇒ 15.6 x h = 16.38 cm²
⇒ h = \(\frac{16.38 \mathrm{~cm}^{2}}{15.6 \mathrm{~cm}}\) = 1.05 cm.
Thus the missing value (height) = 1.05 cm

Question 4.
Find the missing value.

(i) Here,
Base(b) = 15 cm
Area of Triangle = 87cm²
Let Height = h
∴ Area of Triangle = \(\frac{1}{2}\) x b x h
⇒ 87 = \(\frac{1}{2}\) x 15 x h
⇒ 87 x 2 = 15h
⇒ \(\frac{87 \times 2}{15}\) = h
⇒ h = \(\frac{174}{15}\)
= 11.6 cm.
The missing value (height) = 11.6 cm

(ii) Here,
Height (h) = 31.4 mm
Area of Triangle = 1256 mm²
Let, Base = b mm
∴ \(\frac{1}{2}\) x b x h = 1256
⇒ \(\frac{1}{2}\) x 6 x 31.4 = 1256
⇒ b = \(\frac{2 \times 1256}{31.4}\)
∴ b = 80 mm
The missing value Base (b)
= 80 mm.

(iii) Here,
Base (b) = 22 cm
Area of Triangle = 170.5 cm2
Let, Height = h cm
∴ \(\frac{1}{2}\) x b x h = 170.5 cm²
⇒ \(\frac{1}{2}\) x 22 x h = 170.5 cm²
⇒ h = \(\frac{2 \times 170.5}{22}\) cm²
= 15.5 cm.

Question 5.
PQRS is a parallelogram. QM is the height from Q to SR and QN is the height from Q to PS. If SR = 12 cm and QM = 7.6 cm. Find:

(a) the area of the parallelogram PQRS.
(b) QN, if PS = 8 cm.
Solution:
(a) QM (height) = 7.6 cm
SR (base) = 12 cm
∴ Area of parallelogram PQRS
= b x h = (12 x 7.6) cm² = 91.2 cm².

(b) QN (height) = ?
PS (base) = 8 cm.
and area of parallelogram PQRS = 91.2 cm²
∴ Area of parallelogram PQRS = base x height
91.2 cm² = 8 cm x h
or h = \(\frac{91.2 \mathrm{~cm}^{2}}{8 \mathrm{~cm}}\)
(QN)height = 11.4 cm.

Question 6.
DL and BM are the heights on sides AB & AD respectively of parallelogram ABCD. If the area of parallelogram is 1470
cm² and AB = 35 cm and AD = 49 cm. Find the length of BM and DL.

Solution:
In parallelogram ABCD,
AB (base) = 35 cm
∴ Area of parallelogram = 1470 cm²
DL (height) = ?
Area of parallelogram = b x h
h = \(\frac{1470}{35}\)
or, height (DL) = 42 cm
And AD(base) = 49 cm
BM (height) = ?
Area of parallelogram = 1470 cm²
∴ Area of parallelogram = b x h
1470 = 49 x h
or, h = \(\frac{1470}{49}\)
or, height (BM) = 30 cm.

Question 7.
ΔABC is right angle at A. AD is per-pendicular BC. If AB = 5 cm, BC = 13 cm and AC = 12 cm. Find the area of AABC. Also find the length of AD.

Solution:
In a right angled AABC,
base = 12 cm,
height = 5 cm
∴ Area of a right angled triangle = \(\frac{1}{2}\) x b x h
= \(\frac{1}{2}\) x 12 x 5
= 30 cm².
If base = 13 cm,
height (AD) = ?
Area = \(\frac{1}{2}\) xb x h = \(\frac{1}{2}\) x 13 x AD
or 30 cm² = \(\frac{13 \mathrm{AD}}{2}\)
or 13 AD = 30 x 2 = 60 cm²
∴ AD = \(\frac{60 \mathrm{~cm}^{2}}{13 \mathrm{~cm}}\) = 4.62cm
And, Second method : From Hero’s formula,

Question 8.
AABC is isosceles with AB = AC =
7,5 cm, and BC = 9 cm.
The height AD from A to BC is 6 cm. Find the area of ∆ABC. What will be the height from C to AB 8 i.e., CE.

Solution:
In a ∆ABC, base = 9 cm
height = 6 cm
∴ Area of AABC = \(\frac{1}{2}\) x b x h
= \(\frac{1}{2}\) x 9 x 6 = 27 cm².
Now, In a ∆ABC, if base = 7.5 cm and
height = ?
∴ Area of a ABC = \(\frac{1}{2}\) x base x height
27cm² = \(\frac{1}{2}\) x 7.5 cm CE
or 7.5 cm x CE = 27 cm² x 2
or CE = \(\frac{54 \mathrm{~cm}^{2}}{7.5 \mathrm{~cm}}\)
∴ height, CE = 7.2

Haryana State Board HBSE 7th Class Maths Solutions Chapter 11 Perimeter and Area Ex 11.1 Textbook Exercise Questions and Answers.

Haryana Board 7th Class Maths Solutions Chapter 11 Perimeter and Area  Exercise 11.1

Question 1.
The length and breadth of a rectangular piece of land is 500 m and 300 m respectively. Find:
(i) its area
(ii) the cost of the land ofl m2 is Rs. 10000.
Solution:
(i) The area of rectangular piece of land = l x b = 500 m x 300 m = 150000 m².
(ii) The cost of 1 m² of the land = Rs. 10000
∴ the cost of 150000 m²
= Rs. 10000 x 150000 m² = Rs. 1,50,00,00,000.

Question 2.
Find the area of a square park whose perimeter is 320 m.
Solution:
Perimeter of square = 4 x side
320 m = 4 x side
or, 4 x side = 320 m
or side = \(\frac{320}{4}\)m = 80 m.
So, area of square = side x side
= 80 m x 80 m = 6400 m².

Question 3.
Find the breadth of the rectangular plot of land, if its area is 440 m². and the length is 22 m. Also find its perimeter.
Solution:
The area of rectangular plot = 440 sq. m.
Length = 22 m
Breadth = ?
So, the area of rectangular plot = l x b
440 sq. m. = 22 m x breadth (b)
or, 22 m x breadth = 44 sq. m
or b = \(\frac{440 \mathrm{~m}^{2}}{22 \mathrm{~m}}\) = 20 m.
Perimeter of rectangular plot of land = 2 (l+b)
= 2 x (22 m + 20 m)
= 2 x 42 = 84 m

Question 4.
The perimeter of rectangular sheet is 100 cm. If the length is 35 cm, find its breadth. Also find its area.
Solution:
Perimeter of rectangular sheet = 100 cm
Length = 35 cm
Breadth = ?
Area of rectangular sheet = ?
∴ Perimeter of rectangular sheet = 2 x (l + b)
100 m = 2(35 + 6)
or, 100 cm = 70 + 2b
or, 100 – 70 = 2b
or, 30 = 2b
or, b = \(\frac{30}{2}\) = 15
∴ breath = 15 cm
And, Area = l x 6 = 35 x 15
∴ Area = 525 cm2.

Question 5.
The area of square park is the same as of a rectangular park. If the side of the square park is 60 m and the length of the rectangular park is 90 m, find the breadth of the rectangular park.
Solution:
The area of the square park
= side x side = 60 m x 60 m = 3600 m².
So, according to question,
Area of square park
= area of rectangular park
∴ 3600 m²
= area of rectangular park
Now, Area of rectangular park = l x b
3600 m²
= 90 x b
or, b = \(\frac{3600 \mathrm{~m}^{2}}{90 \mathrm{~m}}\) = 40 cm
Hence the breadth of the rectangular park = 40 m.

Question 6.
A wire is in the shape of a rectangle. Its length is 40 cm and breadth is 22 cm. If the same wire is rebent in the shape of a square, what will be the measure of each side. Also find which shape enclose more area.
Solution:
length of rectangle = 40 cm
breadth of rectangle = 22 cm
Since, perimeter of rectangle = perimeter of square
Hence 2(1 + 6) = 4 x side
or, 4 x side = 2(1 + 6)
∴ side = \(\frac{2(1+\mathrm{b})}{4}=\frac{2 \times(40 \mathrm{~cm}+22 \mathrm{~cm})}{4}=\frac{2 \times 62 \mathrm{~cm}}{4}\)
Side of square =31 cm.
Now, Area of rectangle = l x b
= 40 cm x 22 cm = 880 cm².
Area of square = side x side = 31 cm x 31 cm
= 961 cm².
Hence, square encloses more area.

Question 7.
The perimeter of a rectangle is 130 cm. If the breadth of the rectangle is 30cm, find its length. Also find the area of the rectangle.
Solution:
Perimeter of rectangle = 130 cm
Length = ?
Breadth = 30 cm
Area of rectangle = ?
Now,Perimeter of rectangle = 2(l + b)
130 = 2(l + 30)
or, 130 = 2l + 60
or, 130 – 60 = 2l
or, 70 = 2l
or, l = \(\frac{70}{2}\) = 35 cm.
∴ Length = 35 cm
And, Area of rectangle = l x b
= 35 x 30
= 1050 cm².

Question 8.
A door of length 2 m and breadth 1 m is fitted in a wall. The length of the wall is 4.5 m and the breadth is 3.6 m. Find the cost of white washing the wall, if the rate of white washing of the wall is Rs. 20 per m².

Solution:
Area of a door = l x b
= 1 m x 0.5 = 0.5 m².
Area of the wall = l x b
= 4.5m x 3.6m=16.20m².
∴ Area of the = 16.20 – 0.5 = 15.70 m².
rest of the wall
Now, ∵ The rate of white washing of the 1 m² wall = 20 Rs.
∴ The rate of white washing of the 15.70 m2 wall = 15.70 x 20 = Rs. 314.