Class 7

Well-designed AP 7th Class Maths Textbook Solutions Chapter 11 Exponents and Powers Exercise 11.2 offers step-by-step explanations to help students understand problem-solving strategies.

Exponents and Powers Class 7 Exercise 11.2 Solutions – 7th Class Maths 11.2 Exercise Solutions

Question 1.
Using laws of exponents, simplify and write the answer in exponential form:
i) 3² × 3⁴ × 3⁸
Solution:
3² × 3⁴ × 3⁸ = 3²⁺⁴⁺⁸ = 3¹⁴
(∵ a^m × aⁿ × a^p = a^m+n+p

ii) 6¹⁵ ÷ 6¹⁰
Solution:
6¹⁵ ÷ 6¹⁰ = 6^{15 – 10} = 6⁵
(∵ a^m ÷ aⁿ = a^m-n)

iii) a³ × a²
Solution:
a³ × a² = a^3-2 = a⁵
(∵ a^m × aⁿ = a^m-n)

iv) 7^x × 7²
Solution:
7^x × 7² = 7^x+2
(∵ a^m × aⁿ =a^m+n)
= 7^x+2

v) (5²)³ ÷ 5³
Solution:
(5²)³ ÷ 5³
= \(\frac{\left(5^2\right)^3}{5^3}\) = \(\frac{5^6}{5^3}\) (∵ (a^m) = a^m×n)
= 5^6-3 = 5³ = 125 \(\left(\frac{a^m}{a^n}=a^{m-n}\right)\)

vi) 2⁵ × 5⁵
Solution:
2⁵ × 5⁵ = (2 × 5)³ [∵ a^m × b^m = (a × b)^m]
=10⁵ = 10 × 10 × 10 × 10 × 10 = 100000

vii) a⁴ × b⁴
Solution:
a⁴ × b¹ = (a × b)⁴ = (ab)⁴

viii) (3⁴)³
Solution:
(3⁴)³ = 3^4×3 = 3¹²
[∵ (a^m)ⁿ = a^mn]

ix) (2²⁰ ÷ 2¹⁵) × 2³
Solution:
(2²⁰ ÷ 2¹⁵) × 2³ = \(\left(\frac{2^{20}}{2^{15}}\right)\) × 2³
= (2^20-15) × 2³ \(\left(\frac{a^m}{a^n}=a^{m-n}\right)\)
= 2⁵ × 2³
= 2⁵⁺³ (∵ a^m × aⁿ = a^m+n)
= 2⁸

x) 8^t ÷ 8²
Solution:
8^t ÷ 8² = 8^t-2
(∵ a^m ÷ aⁿ = ^m-n)

ix) (2²⁰ ÷ 2¹⁵) × 2³
Solution:
(2²⁰ ÷ 2¹⁵) × 2³ = \(\left(\frac{2^{20}}{2^{15}}\right) \times 2^3\)
= (2^{20 – 15} × 2³ \(\left( \frac{a^m}{a^n}=a^{m-n}\right)\)
= 2⁵ × 2³
= 2⁵⁺³ (∵ a^m × aⁿ = a^m+n)
= 2⁸

x) 8^t ÷ 8²
Solution:
8^t ÷ 8² = 8^t-2 (∵ a^m ÷ aⁿ = a^m-n)

Question 2.
Simplify and express each of the following in exponential form :


v) \(\frac{3^7}{3^4 \times 3^3}\)
Solution:
\(\frac{3^7}{3^4 \times 3^3}\) = \(\frac{3^7}{3^{4+3}}\) (∵ a^m × aⁿ = a^m+n)
= \(\frac{3^7}{3^7}\) \(\left(\frac{a^m}{a^n}=a^{m-n}\right)\)
= 3^7–7 = 3° (∵ a° = 1)
= 1

vi) 2° + 3° + 4°
Solution:
2° + 3° + 4° = 1 + 1 + 1 = 3 (∵ a° = 1)

viii) (3° + 2°) × 5°
Solution:
(3° + 2°) × 5° = (1 + 1) × 1 = 2 × 1 = 2 (∵ a° = 1)

ix) \(\frac{2^8 \times a^5}{4^3 \times a^3}\)
Solution:
\(\frac{2^8 \times a^5}{4^3 \times a^3}\) = \(\frac{2^8 \times a^5}{\left(2^2\right)^3 \times a^3}\) (∵ (a^m)ⁿ = a^mn)
= \(\frac{2^8 \times \mathrm{a}^5}{2^6 \times \mathrm{a}^3}\)
= 2^8–6 × a^5–3 \(\left(\frac{a^m}{a^n}=a^{m-n}\right)\)
= 2² × a²
= (2 × a)² [∵ a^m × b^m = (a × b)^m]
= (2a)²

x) \(\left(\frac{a^5}{a^3}\right) \times a^8\)
Solution:
\(\left(\frac{a^5}{a^3}\right) \times a^8\) = a⁸ = (a^5–3) × a⁸
\(\left(\frac{x^m}{x^n}=x^{m-n}\right)\)
= a² × a⁸ = a²⁺⁸ = a¹⁰ (∵ X^m × Xⁿ = X^m+n)

xi) \(\frac{4^5 \times a^8 b^3}{4^5 \times a^5 b^2}\)
Solution:
\(\frac{4^5 \times a^8 b^3}{4^5 \times a^5 b^2}\)
= 4^5–5 × a^8–5 × b^3–2 \(\left(\frac{a^m}{a^n}=a^{m-n}\right)\)
= 4⁰ × a³ × b¹ = 1 × a³ × b = a³b (∵ a° = 1)

xii) (2³ × 2)²
Solution:
(2³ × 2)² = (2³ × 2¹)² (∵ a^m × aⁿ = a^m+n)
= (2³⁺¹)²
= (2⁴)² = 2^4×2 (∵ (a^m)ⁿ = a^m×n)
= 2⁸

Question 3.
Say true or false and justify your answer :
i) 10 × 10¹¹ = 100¹¹
Solution:
10 × 10¹¹ = 100¹¹
LHS 10¹ × 10¹¹ = 10¹⁺¹¹ =10¹² (∵ a^m × aⁿ = a^m+n)
RHS 100¹¹
10¹² ≠ 100¹¹
∴ 10 × 10¹¹ ≠ 100¹¹
∴ False

ii) 2³ > 5²
Solution:
2³ > 5²
2³ = 8; 5² = 25
8 > 25 (False)
But 8 < 25
∴ False

iii) 2³ × 3² = 6⁵
Solution:
2³ × 3² = 6⁵
LHS 2³ × 3² = 8 × 9 = 72
RHS 6⁵ = 6 × 6 × 6 × 6 × 6 = 7776
7776 > 72
∴ False

iv) 3° = (1000)°
Solution:
3° = (1000)°
LHS 3° = 1
RHS (1000)° = 1
∴ LHS = RHS
∴ True

Question 4.
Express each of the following as a product of prime factors only in exponential form :
i) 108 × 192
Solution:
108 × 192

108 = 2 × 2 × 3 × 3 × 3

192 = 2 × 2 × 2 × 2 × 2 × 2 × 3
108 × 192 = 2 × 2 × 3 × 3 × 3 × 2 × 2 × 2 × 2 × 2 × 2 × 3 = 2⁸ × 3⁴

ii) 270
Solution:
270

270 = 2 × 5 × 3 × 3 × 3
270 = 2 × 5 × 3³

iii) 729 × 64
Solution:
729 × 64

729 = 3 × 3 × 3 × 3 × 3 × 3 ⇒ 729 = 3⁶

64 = 2 × 2 × 2 × 2 × 2 × 2
64 = 2⁶; 729 × 64 = 3⁶ × 2⁶ = (3 × 2)⁶ = 6⁶

iv) 768
Solution:
768

768 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3
768 = 2⁸ × 3

Question 5.
Simplify:
i) \(\frac{\left(2^5\right)^3 \times 7^3}{8^3 \times 7}\)
Solution:
\(\frac{\left(2^5\right)^3 \times 7^3}{8^3 \times 7}\) = \(\frac{2^{5 \times 2} \times 7^3}{\left(2^3\right)^3 \times 7^1}\) (∵(a^m)ⁿ – a^mn)
= \(\frac{2^{15} \times 7^3}{2^9 \times 7^1}\) = 2^10-4 × 7^3-1
\(\left(\frac{a^m}{a^n}=a^{m-n}\right)\)
= 2¹ × 7² – 2 × 7² = 2 × 49 = 98

ii) \(\frac{25 \times 5^2 \times t^2}{10^3 \times t^4}\)
Solution:

iii) \(\frac{3^5 \times 10^5 \times 25}{5^7 \times 6^5}\)
Solution:

Well-designed AP 7th Class Maths Textbook Solutions Chapter 11 Exponents and Powers Exercise 11.1 offers step-by-step explanations to help students understand problem-solving strategies.

Exponents and Powers Class 7 Exercise 11.1 Solutions – 7th Class Maths 11.1 Exercise Solutions

Question 1.
1. Find the value of :
i) 2⁶
Solution:
2⁶ = 2 × 2 × 2 × 2 × 2 × 2 = 64

ii) 9³
Solution:
9³ = 9 × 9 × 9 = 81 × 9 = 729

iii) 11²
Solution:
11² = 11 × 11 = 121

iv) 5⁴
Solution:
5⁴ = 5 × 5 × 5 × 5 = 25 × 25 = 625

Question 2.
Express the following in exponential form :
i) 6 × 6 × 6 × 6
Solution:
6 × 6 × 6 × 6 = 6⁴

ii) t × t
Solution:
t × t = t²

iii) b × b × b × b
Solution:
b × b × b × b = b⁴

iv) 5 × 5 × 7 × 7 × 7
Solution:
5 × 5 × 7 × 7 × 7 = 5² × 7³

v) 2 × 2 × a × a
Solution:
2 × 2 × a × a = 2² × a²

vi) a × a × a × c × c × c × c × d
Solution:
a × a × a × c × c × c × c × d = a³ × c¹ × d

Question 3.
Express each of the following numbers using exponential notation :
i) 512
Solution:
512

512 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2
512 = 2⁹

ii) 343
Solution:

343
343 = 7 × 7 × 7
343 = 7³

iii) 729
Solution:

729
729 = 3 × 3 × 3 × 3 × 3 × 3
729 = 3⁶

iv) 3125
Solution:
3125

3125 = 5 × 5 × 5 × 5 × 5 = 5⁵

Question 4.
Identify the greater number, wherever possible, in each of the following.
i) 4³ or 3⁴
Solution:
4³ or 3⁴ = (2²)³ or 3⁴
= 2⁶ or 3⁴ = 64 or 81
81 > 64
∴ 3⁴ > 4³

ii) 5³ or 3⁵
Solution:
5³ or 3⁵
5³ = 5 × 5 × 5 = 125
3⁵ = 3 × 3 × 3 × 3 × 3 = 243
∴ 3⁵ > 5³

iii) 2⁸ or 8²
Solution:
2⁸ or 8²
2⁸ = (2⁴)² = 16² = 256
8² = 8 × 8 = 64
256 > 64
∴ 2⁸ > 8²

iv) 100² or 2¹⁰⁰
Solution:
100² or 2¹⁰⁰
(10²)² or 2¹⁰⁰
10⁴ or 2¹⁰⁰
10,000 or (2¹⁰)¹⁰
10,000 or (1024)¹⁰
∴ 2¹⁰⁰ >100²

v) 2¹⁰or 10²
Solution:
2¹⁰ or 10²
2¹⁰ = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 1024
10² = 100
1024 > 100
∴ 2¹⁰ > 10²

Question 5.
Express each of the following as product of powers of their prime factors:
i) 648
Solution:
648

648 = 2 × 2 × 2 × 3 × 3 × 3 × 3
648 = 2³ × 3⁴

ii) 405
Solution:
405

405 = 5 × 3 × 3 × 3 × 3
405 = 5 × 3⁴

iii) 540
Solution:
540

540 = 2 × 2 × 5 × 3 × 3 × 3
540 = 2² × 5 × 3³

iv) 3,600
Solution:
3,600

3600 = 2 × 2 × 2 × 2 × 5 × 5 × 3 × 3
3600 = 2⁴ × 5² × 3³

Question 6.
Simplify :
i) 2 × 10³
Solution:
2 × 10³ = 2 × 1000 = 2000

ii) 7² × 2²
Solution:
7² × 2² = (7 × 2)² [∵ a^m × b^m = (ab)^m]
= 14² = 14 × 14 = 196

iii) 2³ × 5
Solution:
2³ × 5 = 8 × 4 = 40

iv) 3 × 4⁴
Solution:
3 × 4⁴ = 3 × 256 = 768

v) 0 × 10²
Solution:
0 × 10² = 0 × 100 = 0

vi) 5² × 3³
Solution:
5² × 3³ = 25 × 27 = 675

vii) 2⁴ × 3²
Solution:
2⁴ × 3² = 16 × 9 = 144

viii) 3² × 10⁴
Solution:
3² × 10⁴ = 9 × 10000 = 90,000

Question 7.
Simplify :
i) (-4)³
Solution:
(-4)³ = -4× -4 × -4 = -64

ii) (-3) × (-2)³
Solution:
(-3) × (-2)³ = -3 × -8 = 24

iii) (-3)² × (-5)²
Solution:
(-3)² × (-5)² = 9 × 25 = 225
(OR)
(-3)² × (-5)² = (-3 × -5)²
∵ a^m × b^m = (a × b)^m
= (15)² = 15 × 15 = 225

iv) (-2)³ × (-10)³
Solution:
(-2)³ × (-10)³ = -8 × -1000 = 8000
(OR)
(-2)³ × (-10)³ =(-2 × -10)³
∵ a^m × b^m = (a × b)^m
= (20)²
= 20 × 20 × 20 = 8000

Question 8.
Compare the following numbers :
i) 2.7 × 10¹²; 1.5 × 10⁵
Solution:
2.7 × 10¹² = \(\frac{27}{10^1}\) × 10¹²
= 27 × 10^12-1
∵ \(\frac{a^m}{a^n}\) = a^m-n
= 27 × 10¹¹

1.5 × 10⁸ = \(\frac{15}{10^1}\) × 10⁸
= 15 × 10^{8 – 1}
∵ \(\frac{a^m}{a^n}\) = a^m-n
= 15 × 10⁷
∴ 2.7 × 10¹¹ > 1.5 × 10⁷
2.7 × 10¹² > 1.5 × 10⁸

ii) 4 × 10¹⁴ ; 3 × 10¹⁷
Solution:
4 × 10¹⁴ = 4 × 10¹⁴
3 × 10¹⁷ = 3 × 10³ × 10¹⁴
= 3 × 1000 × 10¹⁴ = 3000 × 10⁴
∴ 3000 × 10¹⁴ > 4 × 10⁴
∴ 4 × 10¹⁴ > 3 × 10¹⁷

Well-designed AP 7th Class Maths Textbook Solutions Chapter 10 Algebraic Expressions InText Questions offers step-by-step explanations to help students understand problem-solving strategies.

AP 7th Class Maths 10th Chapter Algebraic Expressions InText Questions

Try These（Page No: 76)

Question 1.
Describe how the following expressions are obtained :
7xy + 5, x²y, 4x² – 5x
Solution:
7xy + 5 : In 7xy + 5, we first obtain xy, multiply it by 7 to get 7xy and add 5 to 7xy to get the expression.
x²y : In x²y, we first obtain x × x = x², multiply it by y to get x²y.
4x² – 5x : In 4 x² – 5x, we first obtain x² and multiply it by 4 to get 4x². From 4x², we subtract the product of 5 and x as 5x.

Try These (Page No. 78)

Question 1.
What are the terms in the following expressions? Show how the terms are formed. Draw a tree diagram for each expression :
8y + 3x², 7mn – 4, 2x²y
Solution:
8y + 3x² : In 8y + 3x², the terms are 8y and 3x²,

Question 2.
Write three expression each having 4 terms.
Solution:
Three algebraic expressions each having 4 terms are :
i) 3x⁵ + 2x³ + 7x² + 6
ii) 2x² + 3x² + 6x + 9
iii) x⁴ + 6x³ + 2x + 3

Try These (Page No. 78)

Question 1.
Identify the coefficients of the terms of following expressions :
4x – 3y, a + b + 5, 2y + 5, 2xy
Solution:
Given expression 4x – 3y

Term	Coefficient
4x	4
-3y	-3

Given expression a + b + 5

Term	Coefficient
A	1
b	1
5	constant

In 2y + 5, the terms are 2y and 5
In 2 xy, the term is 2xy.

Try These (Page No. 82)

Question 1.
Group the like terms together from the following :
12x, 12, -25x, -25, -25y, 1, x, 12y, y
Solution:
Given terms 12x, 12, -25x, -25, -25y, 1, x, 12 y, y.
Like term: 12x – 25x, x
-25y, 12y, y; 12 – 25, 1

Try These (Page No. 82)

Question 1.
Classify the following expressions as a monomial, a binomial or a trinomial:
a, a + b, ab + a + b, ab + a + b – 5, xy, xy + 5, 5x² – x + 2, 4pq – 3q + 5p, 7, 4m – 7n + 10, 4mn + 7
Solution:

Well-designed AP 7th Class Maths Textbook Solutions Chapter 10 Algebraic Expressions Exercise 10.2 offers step-by-step explanations to help students understand problem-solving strategies.

Algebraic Expressions Class 7 Exercise 10.2 Solutions – 7th Class Maths 10.2 Exercise Solutions

Question 1.
If m = 2, find the value of :
i) m – 2
Solution:
Given m = 2
m – 2 = 2 – 2 = 0
∴ The value of m – 2 at m = 2 is 0.

ii) 3m – 5
Solution:
3m – 5
3(2) – 5 = 6 – 5 = 1
∴ The value of 3m – 5 at m = 2 is 1.

iii) 9 – 5m
Solution:
9 – 5m
= 9 – 5(2) = 9 – 10 = – 1
∴ The value of 9 – 5m at m = 2 is -1.

iv) 3m² – 2m – 7
Solution:
3m² – 2m – 7
3(2)² – 2(2) – 7
=3 × 4 – 4 – 7 = 12 – 11 = 1
∴ The value of 3m² – 2m – 7 at m = 2 is 1.

v) \(\frac{5 m}{2}\) – 4
Solution:
\(\frac{5 m}{2}\) – 4 = \(\frac{5 \times 2}{2}\) – 4 = 5 – 4 = 1
∴ The value of \(\frac{5 \mathrm{~m}}{2}\) – 4 at m = 2 is 1 .

Question 2.
If p = – 2, find the value of :
i) 4p + 7
Solution:
Given p = – 2
4p + 7 = 4(-2) + 7 = -8 + 7 = -1

ii) -3p² + 4p + 7
Solution:
-3p² + 4p + 7
= -3(-2)² + 4(-2) + 7
=-3(4) – 8 + 7
= – 12 – 1 = – 13

iii) -2p³ – 3p² + 4p + 7
Solution:
-2p³ – 3p² + 4p + 7
= -2(-2)³ -3(-2)² + 4(-2) + 7
= -2(-8) – 3(4) – 8 + 7
= 16 – 12 – 8 + 7
= 16 – 20 + 7 = 23 – 20 = 3

Question 3.
Find the value of the following expressions, when x = -1:
i) 2x – 7
Solution:
Given x = -1
2x – 7 ⇒ 2(-1) – 7
= – 2 – 7 = – 9

ii)-x + 2
Solution:
-x + 2 ⇒ -(- 1) + 2
= 1 + 2 = 3

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

iv) 2x² – x – 2
Solution:
2x² – x – 2 = 2(-1)² – (-1) – 2
= 2 + 1 – 2
= 3 – 2 = 1

Question 4.
If a = 2, b = -2, find the value of :
i) a² + b²
Solution:
Given a = 2, b = -2
a² + b² = 2² + (-2)²
= 4 + 4 = 8

ii) a² + ab + b²
Solution:
a² + ab + b² =(2)² + 2(-2) + (-2)²
=4 – 4 + 4 = 4

iii) a² – b²
Solution:
a² – b² = 2² – (-2)² = 4 – 4 = 0

Question 5.
When a = 0, b = -1, find the value of the given expressions :
i) 2a + 2b
Solution:
Given a = 0, b = -1
2a + 2b = 2(0) + 2(-1)
= 0 – 2 = -2

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

iii) 2a²b + 2ab² + ab
Solution:
2a²b + 2ab² + ab
=2(0)² (-1) + 2(0)(-1)² + 0(-1)
= 0 + 0 + 0 = 0

iv) a² + ab + 2
Solution:
a² + ab + 2 = 0² + (0)(-1) + 2 = 0 + 2 = 2

Question 6.
Simplify the expressions and find the value if x is equal to 2.
i) x + 7 + 4(x – 5)
Solution:
Given expression
x + 7 + 4(x – 5)
= x + 7 + 4x – 20
=5x + 7 – 20
=5x – 13
At x = 2
5x – 13 = 5(2) – 13
= 10 – 13 = -3

ii) 3(x+2) + 5x – 7
Solution:
Given expression
3(x + 2) + 5x – 7
= 3x + 6 + 5x – 7
= 8x + 6 – 7 = 8x – 1
At x = 2
8x – 1 = 8(2) – 1
= 16 – 1 = 15

iii) 6x +5 (x – 2)
Solution:
Given expression
6x + 5(x – 2)
= 6x + 5x – 10
= 11x – 10
At x = 2
11x – 10 = 11(2) – 10
= 22 – 10 = 12

iv) 4(2x – 1) + 3x + 11
Solution:
Given expression
4(2x – 1) + 3x + 11
= 8x – 4 + 3x + 11
= 11x – 4 + 11
= 11x + 7
At x = 2
11x + 7 = 11(2) + 7
= 22 + 7 = 29

Question 7.
Simplify these expressions and find their values if x = 3, a = -1, b = -2.
i) 3x – 5 – x + 9
Solution:
Given x = 3, a = -1 ; b = -2
3x – 5 – x + 9 = 3x – x – 5 + 9 = 2x + 4
At x = 3
2x + 4 = 2(3) + 4
= 6 + 4 = 10

ii) 2 – 8x + 4x + 4
Solution:
2 – 8x + 4x + 4
= 2 – 4x + 4 = 6 – 4x
At x = 3
6 – 4x = 6 – 4(3)
= 6 – 12 = -6

iii) 3a + 5 – 8a + 1
Solution:
3a + 5 – 8a + 1
= 3a – 8a + 5 + 1 = -5a + 6
At a = -1
-5a + 6 = -5(-1) + 6
=5 + 6 = 11

iv) 10 – 3b – 4 – 5b
Solution:
10 – 3b – 4 – 5b
= 10 – 4 – 3b – 5b = 6 – 8b
At b = -2
6 – 8b = 6 – 8(-2)
= 6 + 16 = 22

v) 2a – 2b – 4 – 5 + a
Solution:
2a – 2b – 4 – 5 + a = 2a + a – 2b – 4 – 5
=3a – 2b – 9
At a = -1, b = -2
3a – 2b – 9 =3(-1) – 2(-2) – 9)
=- 3 + 4 – 9 = 1 – 9 = – 8

Question 8.
i) If z = 10, find the value of z² – 3(z – 10)
Solution:
Given z = 10
Given expression
z³ – 3(z – 10)
= z² – 3z + 30
=10³ – 3(10) + 30
=1000 – 30 + 30
=1000

ii) If p = -10, find the value of p² – 2p – 100
Solution:
Given p = -10
Given expression p² – 2p – 100
=(-10)² – 2(-10) – 100
=100 + 20 – 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:
Given x = 0
Given expression 2x² + x – a = 5
2(0)² + 0 – a = 5
0 – a = 5 ⇒ – a = 5
∴ a = – 5

Question 10.
Simplify the expression and find its value when a = 5 and b = -3.
2(a² + ab) + 3 – ab
Solution:
Given a = 5, b = -3
Given expression
2(a² + ab) + 3 – ab
= 2a² + 2ab + 3 – ab
= 2a² + 2ab – ab + 3
= 2a² + ab + 3
= 2(5)² + 5(-3) + 3
= 2 × 25 – 15 + 3
= 50 – 15 + 3
= 35 + 3 = 38

Well-designed AP 7th Class Maths Textbook Solutions Chapter 10 Algebraic Expressions Exercise 10.1 offers step-by-step explanations to help students understand problem-solving strategies.

Algebraic Expressions Class 7 Exercise 10.1 Solutions – 7th Class Maths 10.1 Exercise Solutions

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 numbers x and y.
iii) The number z multiplied by itself.
iv) One-fourth of the product of numbers p and q.
v) Numbers x and y both squared and added.
vi) Number 5 added to three times the product of numbers 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) Sum of number x and y = x + y
One half of x + y = \(\frac{x+y}{2}\)

iii) z × z = z²

iv) Product of number p and q = p × q = pq
One fourth of pq = \(\frac{pq}{4}\)

v) Square of number x = x²
Square of number y = y²
When x² and y² are added, the sum = x² + y²

vi) Product of number m and n = m × n = mn.
Three times the product of mn = 3 mn
Number 5 is added to 3 mn, then the sum = 3 mn + 5

vii) Product of number y and z = y × z = yz
yz is subtracted from 10, then the difference = 10 – yz

viii) Sum of numbers a and b = a + b
Product of number a and b = ab
According to the problem the algebraic expression is = ab – (a + b) = ab – a – b

Question 2.
i) Identify the terms and their factors in the following expressions. Show the terms and factors by tree diagrams.
a) x – 3
Solution:
Given expression x – 3
Tree Diagram :

b) 1 + x + x²
Solution:
Given expression 1 + x + x²
Tree Diagram :

c) y – y³
Solution:
Given expression y – y³
Tree Diagram :

d) 5xy² + 7x²y
Solution:
Given expression 5xy² + 7x²y
Tree Diagram :

e) -ab + 2b² – 3a²
Solution:
Given expression -ab + 2b² – 3a²
Tree Diagram :

ii) Identify terms and factors in the expressions given below :
a) -4x +5
b) -4x + 5y
c) 5y + 3y²
d) xy + 2x² y²
e) pq + q
f) 1.2 ab – 2.4 b + 3.6 a
g) \(\frac { 3 }{ 4 }\)x + \(\frac { 1 }{ 4 }\)
h) 0.1 p² + 0.2 q²
Solution:

Question 3.
Identify the numerical coefficients 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 + 7 pq
vi) 1.2 a + 0.8 b
vii) 3.14 r²
viii) 2(l +b)
ix) 0.1y + 0.01y²
Solution:

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

b) Identify terms which contain y² and give the coefficient of y².
i) 8 – xy²
ii) 5 y² + 7x
iii) 2 x² y – 15 xy² + 7y²
Solution:

Question 5.
Classify into monomials, binomials and trinomials.
i) 4y – 7z
ii) y²
iii) x + y – xy
iv) 100
v) ab – a – b
vi) 5 – 3t
vii) 4 p²q – 4 pq²
viii) 7 mn
ix) z² – 3z + 8
x) a² + b²
xi) z² + z
xii) 1 + x + x²
Solution:

Question 6.
State whether a given pair of terms is of like or unlike terms.
i) 1,100
ii) -7x, \(\frac { 5 }{ 2 }\)x
iii) -29x, -29y
iv) 14 xy, 42yx
v) 4m² p, 4 mp²
vi) 12 xz, 12 x²z²
Solution:
i) 1, 100 are like terms
ii) -7 x, \(\frac { 5 }{ 2 }\)x are like terms
iii) -29x, -29 y are unlike terms
iv) 14 xy, 42 yx are like terms
v) 4 m² p, 4mp² are unlike terms
vi) 12 xz, 12 x²z² are unlike terms

Question 7.
Identify like terms in the following :
a) -xy², -4 yx², 8x², 2xy², 7y, -11 x², -100 x, -11 yx, 20x² y, -6 x², y, 2 xy, 3x
Solution:
Like terms :
-xy², 2xy²;
-4 yx², 20 x²y;
8 x², -11 x², -6 x²
7y, y
-100 x, 3x
-11yx, 2 xy

b) 10pq, 7p, 8q, – p²q², – 7qp, -100q – 23, 12q²p², -5 p², 41, 2405 p, 78 qp, 13p²q, qp², 701p²
Solution:
Like terms : 10pq, – 7qp, 78qp ;
7p, 2405p;
8q, -100q;
-p²q², 12 q²p²
-23,41;
-5 p², 701p²
13 p²q, q²

Well-designed AP 7th Class Maths Textbook Solutions Chapter 9 Perimeter and Area InText Questions offers step-by-step explanations to help students understand problem-solving strategies.

AP 7th Class Maths 9th Chapter Perimeter and Area InText Questions

Try These (Page No. 48)

Question 1.
Find the area of following parallelograms :

Solution:
In the parallelogram base, b = 8 cm
Height, h = 3.5 cm
Area = b × h = 8 × 3.5 cm² = 28 cm²

Solution:
In the parallelogram base, b = 8 cm
Height, h = 2.5 cm
Area = b × h = 8 × 2.5 cm² = 20 cm²

iii) In a parallelogram ABCD, AB = 7.2 cm and the perpendicular from C on AB is 4.5 cm.

Solution:
In parallelogram ABCD
Base, AB = 7.2 cm.
Height, CE = 4.5 cm
Area = 7.2 × 4.5 × cm² = 32.4 cm²

Try These (page No. 50)

Question 1.
Try the above activity (see Textbook page No. 48) with different types of triangles.
Solution:
Try yourself – Activity Based Question.

Question 2.
Take different parallelograms. Divide each of the parallelograms into two triangles by cutting along any of its diagonals. Are the triangles congruent?
Solution:
Try yourself – Activity Based Question.

Try These (Page No: 62)

Question 1.
In Figure,

a) which square has the larger perimeter?
b) which is larger, perimeter of smaller square or the circumference of the circle?
Solution:
a) Outside square has large perimeter.
b) Circumference of circle is larger by observation.

Try these (Page No : 68)

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 answers.
Solution:
Student’s Activity.

Well-designed AP 7th Class Maths Textbook Solutions Chapter 9 Perimeter and Area Exercise 9.2 offers step-by-step explanations to help students understand problem-solving strategies.

Perimeter and Area Class 7 Exercise 9.1 Solutions – 7th Class Maths 9.2 Exercise Solutions

Question 1.
Find the circumference of the circles with the following radius: (Take π = \(\frac { 22 }{ 7 }\) )
a) 14 cm
Solution:
Given radius, r = 14 cm
Circumference, c = 2πr
= 2 × \(\frac { 22 }{ 7 }\) × 14 cm
= 44 × 2 cm = 88 cm

b) 28 mm
Solution:
Given radius, r = 28 mm
Circumference, c =2πr
= 2 × \(\frac { 22 }{ 7 }\) × 28 mm
= 44 × 4 mm = 176 mm

c) 21 cm
Solution:
Given radius, r = 21 cm
Circumference, c = 2πr
= 2 × \(\frac { 22 }{ 7 }\) × 21 cm
= 44 × 3 cm = 132 cm

Question 2.
Find the area of the following circles, given that :
a) radius = 14 mm (Take π = \(\frac { 22 }{ 7 }\))
Solution:
Given radius, r = 14 mm
Area of circle A =πr²
A = \(\frac { 22 }{ 7 }\) × 14 × 14 mm²
A = 22 × 2 × 14 mm²
A = 44 × 14 mm² = 616 mm²

b) diameter = 49 m
Solution:
Given diameter = 49 m; d = 49 m
r = \(\frac { 49 }{ 2 }\) m
Area of circle A = πr²
A = \(\frac { 22 }{ 7 }\) × \(\frac { 49 }{ 2 }\) × \(\frac { 49 }{ 2 }\) m²
A = \(\frac { 3773 }{ 2 }\)m² = 1886.5m²

c) radius = 5 cm
Solution:
Given radius, r = 5 cm; Area A = πr²
A = \(\frac { 22 }{ 7 }\) × 5 × 5 = \(\frac { 550 }{ 7 }\) cm²
A = 78.5 cm²

Question 3.
If the circumference of a circular sheet is 154 m, find its radius. Also find the area of the sheet. (Take π = \(\frac { 22 }{ 7 }\))
Solution:
Given circumference of circle = 154 m
2πr = 154 m
2 × \(\frac { 22 }{ 7 }\) × r = 154 m
r = 154 × \(\frac { 7 }{ 22 }\) × \(\frac { 1 }{ 2 }\)m
r = \(\frac{7 \times 7}{2}\)m = \(\frac { 49 }{ 2 }\)m
Radius = \(\frac { 49 }{ 2 }\)m = 24.5 m
Area of sheeti = πr²

Question 4.
A gardener wants to fence a circular garden of diameter 21 m. Find the length of the rope he needs to purchase, if he makes 2 rounds of fence. Also find the cost of the rope, if it costs ₹ 4 per meter. (Take π = \(\frac { 22 }{ 7 }\))

Solution:
In a circular garden diameter = 21 m
d = 21 m; r = \(\frac { 21 }{ 2 }\)m
Length of rope to purchase = Circumference of circle
= 2πr = 2 × \(\frac { 22 }{ 7 }\) × \(\frac { 21 }{ 2 }\)m
= 2 × 11 × 3 = 66 m
For two rounds the length of rope required = 2 × 66 m = 132 m
Cost of fencing = ₹ 4 per meter
Total cost = 132 × 4 = ₹ 528

Question 5.
From a circular sheet of radius 4 cm, a circle of radius 3 cm is removed. Find the area of the remaining sheet. (Take π = 3.14)
Solution:
In a circular sheet radius r = 4 cm
Area = πr² = 3.14 × 4 × 4 cm² = 50.24 cm²
One circle of radius 3 in removed area = πr² = 3.14 × 3 × 3 cm² = 28.26 cm²
Area of remaining circular sheet = 50.24 cm² – 28.26 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 costs ₹ 15 . (Take π = 3.14)
Solution:
In a circular table diameter = 1.5 m
radius r = \(\frac{1.5 \mathrm{~m}}{2}\)
Length of lace required = 2πr
= 2 × 3.14 × \(\frac{1.5}{2}\) = 4.71 m
Cost of 1 m lace = ₹ 15
Total cost = 4.71 × 15 = ₹ 70.65

Question 7.
Find the perimeter of the below figure, which is a semicircle including its diameter.

Solution:
In the given semi circle diameter = 10cm
radius, r = \(\frac{10}{2}\) cm = 5 cm
Circumference = πr + 2r = r(π + 2)
= \(5\left(\frac{22}{7}+2\right)\) = 5 × \(\frac{36}{7}\) = 26.7 m.

Question 8.
Find the cost of polishing a circular table-top of diameter 1.6 m, if the rate of polishing is ₹ 15 /m². (Take π = 3.14)
Solution:
In a circular table top diameter = 1.6 m
radius = \(\frac{1.6}{2}\) m = 0.8 m
Area = πr²
= 3.14 × 0.8 × 0.8 m² = 2.0096 m²
Cost of Polishing = ₹ 15/m²
Total cost = ₹ 15 × 2.0096 = ₹30.14 (approximately)

Question 9.
Shazli took a wire of length 44 cm and bent it into the shape of a circle. Find the radius of that circle. Also find its area. If the same iwire is bent into the shape of a square, what will be the length of each of its sides ? Which figure encloses more area, the circle or the square ? (Take π = \(\frac{22}{7}\))
Solution:
Given length of the wire = 44 cm
The circumference of the circle is 44 cm
c = 2πr
44 cm = 2πr
44 = 2 × \(\frac{22}{7}\) × r
r = 44 cm × \(\frac{7}{44}\) = 7 cm
Area of the circle = πr²
= \(\frac{22}{7}\) × 7 cm × 7 cm = 154 cm²
Now, the wire is bent into a square.
Perimeter of square = 44 cm
4 × side = 44 cm
side = \(\frac{44 \mathrm{~cm}}{4}\) = 11 cm
Area of square = side × side
= 11 cm × 11 cm = 121 cm²
Then the circle encloses more area than square.

Question 10.
From a circular card sheet of radius 14 cm , two circles of radius 3.5 cm and a rectangle of length 3 cm and breadth 1 cm are removed. (as shown in the given figure). Find the area of the remaining sheet. (Take π = \(\frac{22}{7}\))

Solution:
Big circle : Radius = 14 cm
Area of circle = πr²
= \(\frac{22}{7}\) × 14 × 14 cm²
= 22 × 2 × 14 cm²
= 44 × 14 cm² = 616 cm²
Small circle : Radius = 3.5 cm
Area of Circle = πr²
= \(\frac{22}{7}\) × 3.5 × 3.5 cm²
= 22 × 0.5 × 3.5 cm²
= 11 × 3.5 cm² = 38.5 cm²
Area of two small circles = 2 × 38.5 cm² = 77 cm²

Rectangle :
l = length = 3 cm; b = breadth = 1 cm
Area of rectangle = l × b
= 3 × 1 cm² = 3 cm²
Required area of the remaining sheet = Area of big circle – Area of small circles – Area of rectangle
= (616 – 77 – 3) cm²
= 616 – 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:
Square :
Side of square shaped aluminium sheet = 6 cm
Area = side × side = 6 × 6 = 36 cm²
Circle : Radius = 2 cm
Area of circle = πr²
= 3.14 × 2 × 2 cm²
= 3.14 × 4 cm² = 12.56 cm²
Area of the left over aluminium sheet = (36 – 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:
Given circumference of circle = 31.4 cm
2πr = 31.4 cm
2 × 3.14 × r = 31.4 cm

∴ Radius of circle = 5 cm
Area of 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 ? (π = 3.14)

Solution:
Small Circle :
Diameter of flower bed, d = 66 m
r = \(\frac { d }{ 2 }\) = \(\frac { 66 }{ 2 }\) = 33 m
Area = πr² = 3.14 × 33 × 33 m²
Given path width = 4 m
∴ Diameter of big circle = 66 + 4 = 74 m
radius, r = \(\frac { 74 }{ 2 }\) = 37 m
Area = πr² = 3.14 × 37 × 37 cm²
∴ Required area of path
= Area of big circle – Area of small circle
= 3.14 × 37 × 37-3.14 × 33 × 33
= 3.14[37 × 37 – 33 × 33] m²
= 3.14[1369 – 1089] m²
= 3.14[280]m² = 879.20 m²

Question 14.
A circular flower garden has an area of 314 m². A sprinkler at the centre 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:
Given circular flower garden has an area = 314 m²
The radius covered by sprinkler of garden = 12 m
Area covered by sprinkler = πr²
= 3.14 × 12 × 12 m² = 452.16 m²
452.16 > 314 m²
∴ The sprinkler covers more area than the guarden.
∴ It covers the entire area of garden.

Question 15.
Find the circumference of the inner and the outer circles, shown in the below figure? (Take π = 3.14 )

Solution:
R = Radius of outer circle = 19 m
width = 10 m
r = radius of small circle = 19 – 10 = 9 m
Circumference of outer circle = 2πr
= 2 × 3.14 × 19 m
= 38 × 3.14 m = 119.32 m
Circumference of smaller circle = 2πr
= 2 × 3.14 × 9 m
= 18 × 3.14 × m = 56.52 m

Question 16.
How many times a wheel of radius 28 cm must rotate to go 352 m? (Take π = \(\frac { 22 }{ 7 }\) )
Solution:
Given total circumference covered by the wheel = 352 m
= 352 × 100 = 35200 m
Given radius of wheel = 28 cm
r = 28 cm
Circumference covered in one rotation = 2πr
= 2 × \(\frac { 22 }{ 7 }\) × 28 cm
= 44 × 4 = 176 cm
Number of times wheel must rotate = \(\frac { 35200 }{176 }\) = 200 times

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:
Given minute hand of a circular clock is 15 cm long.
r = 15 cm
Circumference = 2πr
= 2 × 3.14 × 15 cm
= 30 × 3.14 cm =94.2 cm
∴ Distance covered in 1 hour = 94.2 cm

Well-designed AP 7th Class Maths Textbook Solutions Chapter 9 Perimeter and Area Exercise 9.1 offers step-by-step explanations to help students understand problem-solving strategies.

Perimeter and Area Class 7 Exercise 9.1 Solutions – 7th Class Maths 9.1 Exercise Solutions

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

Solution:
In the given parallelogram
Base, b = 7 cm; Height, h = 4 cm
Area = b × h = 7 × 4 cm² = 28 cm²

Solution:
Base, b = 5 cm; Height, h = 3 cm
Area = b × h = 5 × 3 cm² = 15 cm²

Solution:
Base, b = 2.5 cm; Height, h = 3.5 cm
Area = b × h = 2.5 × 3.5 cm² = 8.75 cm²

Solution:
Base, b = 5 cm; Height, h = 4.8 cm
Area = b × h = 5 × 4.8 cm² = 24 cm²

Base, b = 2 cm; Height, h = 4.4 cm
Area = b × h = 2 × 4.4 cm² = 8.8 cm²

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

Solution:
In the given triangle
Base, b = 4 cm; Height, h = 3 cm
Area = \(\frac { 1 }{ 2 }\) × b × h = \(\frac { 1 }{ 2 }\) × 4 × 3 = 6 cm²

Base, b = 5 cm; Height, h = 3.2 cm
Area = \(\frac { 1 }{ 2 }\) × b × h = \(\frac { 1 }{ 2 }\) × 5 × 3.2 cm² = 5 × 1.6 cm² = 8 cm²

Base, b = 3 cm; Height, h = 4 cm
Area = \(\frac { 1 }{ 2 }\) × b × h = \(\frac { 1 }{ 2 }\) × 3 × 4 cm² = 3 × 2 cm² = 6 cm²

Base, b = 3 cm; Height, h = 2 cm
Area = \(\frac { 1 }{ 2 }\) × b × h = \(\frac { 1 }{ 2 }\) × 3 × 2 cm² = 3 cm²

Question 3.

Solution:
In the parallelogram
a) Base, b = 20 cm
Area = 246 cm²
b × h = 246 cm²
20cm × h = 246 cm²
h = \(\frac{246 \mathrm{~cm}^2}{20 \mathrm{~cm}}\) = \(\frac{24.6}{2}\) cm ⇒ h = 12.3 cm
∴ Height = 12.3 cm

b) Height, h = 15 cm
Area = 154.5 cm²
b × h = 154.5 cm²
b × 15 cm = 154.5 cm²
b = \(\frac{154.5}{15}\) cm ⇒ b = 10.3 cm
∴ Base = 10.3 cm

c) Height = 8.4 cm
Area = 48.72 cm²
b × h = 48.72 cm²
b × 8.4 cm = 48.72 cm²
b = \(\frac{48.72}{8.4}\) cm ⇒ b = 5.8 cm
∴ Base = 5.8 cm

d) Base = 15.6 cm
Area = 16.38 cm²
b × h = 16.38 cm²
15.6 cm × h = 16.38 cm²
h = \(\frac{16.38}{15.6}\) cm ⇒ h = 1.05 cm

Question 4.
Find the missing values:

Solution:
Base, b = 15 cm
Area of triangle = 87 cm²
\(\frac{1}{2}\) × b × h = 87 cm²
\(\frac{1}{2}\) × 15 cm × h = 87 cm²
h = \(\frac{87 \times 2}{15}\) cm = 11.6 cm
Height = 11.6 cm

Height = 31.4 mm
Area of triangle = 1256 mm²
\(\frac{1}{2}\) × b × h = 1256 mm²
\(\frac{1}{2}\) × b × 31.4 mm = 1256 mm²
b = \(\frac{1256 \times 2}{31.4}\) mm = 80 mm
Base = 80 mm

Base = 22 cm
Area of triangle = 170.5 cm²
\(\frac{1}{2}\) × b × h = 170.5 cm²
\(\frac{1}{2}\) × 22 × h = 170.5 cm²
b = \(\frac{170.5}{11}\) cm = 15.5 cm
Height = 15.5 cm.

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

a) the area of the parallegram PQRS.
b) QN, if PS = 8 cm
Solution:
Given PQRS is a parallelogram QM ⊥ SR
QM is the height from Q to SR
QN is the height from Q to PS
SR = 12 cm; QM = 7.6 cm

a) Area of parallelogram PQRS
Area, A = SR × QM
A = 12 × 7.6 cm² ⇒ A = 91.2 cm²

b) PS = 8 cm
Given area = 91.2 cm²
b × h = 91.2 cm²
PS × QN = 91.2 cm²
8 × QN = 91.2 cm²
QN = \(\frac{91.2}{8}\) ⇒ QN = 11.4 cm

Question 6.
DL and BM are the heights on sides AB and AD respectively of parallelogram ABCD (Figure). If the area of the parallelogram is 1470 cm², AB = 35 cm and AD = 49 cm, find the length of BM and DL.

Solution:
Given ABCD is a parallelogram. DL and BM are the heights on sides AB and AD.
Given area of parallelogram = 1470 cm²
AB = 35 cm; AD = 49 cm
When AB is the base, DL is the height of the parallelogram
Area = 1470 cm²
b × h =1470 cm²
AB × DL = 1470 cm²
35 × DL = 1470 cm²
DL = \(\frac{1470}{35}\) cm ⇒ DL = 42 cm

When AD is the Base, BM is the height of the parallelogram
Area = 1470 cm²
b × h = 1470 cm²
AD × BM = 1470 cm²
49 × BM = 1470 cm²
BM = \(\frac{1470}{49}\) ⇒ BM = 30 cm

Question 7.
△ABC is right angled at A (Figure). AD is perpendicular to BC. If AB = 5 cm, BC = 13 cm and AC = 12 cm, Find the area of △ABC. Also find the length of AD.

Solution:
Given ABC is a right angled triangle
AD ⊥ BC
AB = 5 cm; BC = 13 cm; AC = 12 cm
Area of triangle = \(\frac{1}{2}\) × AB × AC
= \(\frac{1}{2}\) × 5 × 12 cm² = 5 × 6 cm² = 30 cm²
Also area of triangle = \(\frac{1}{2}\) × BC × AD
30 = \(\frac{1}{2}\) × 13 × AD
30 × 2 = 13 × AD
AD = \(\frac{60}{13}\)
AD = 4.61 cm ⇒ AD = 4.6 cm

Question 8.
△ABC is isosceles with AB = AC = 7.5 cm and BC = 9 cm (Fig 9.17). The height AD from A to BC, is 6 cm . Find the area of △ABC. What will be the height from C to AB i.e., CE ?

Solution:
△ABC is an Isosceles triangle
AB = AC = 7.5 cm; BC = 9 cm
AD ⊥ BC
AD = 6 cm
Area of triangle = \(\frac{1}{2}\) × BC × AD
= \(\frac{1}{2}\) × 9 × 6 cm²
= 9 × 3 cm² = 27 cm²
Also area of triangle = \(\frac{1}{2}\) × AB × CE
27 = \(\frac{1}{2}\) × 7.5 × CE
2 × 27 = 7.5 × CE
CE = \(\frac{54}{7.5}\)
CE = 7.2 cm

Well-designed AP 7th Class Maths Guide Chapter 8 Rational Numbers InText Questions offers step-by-step explanations to help students understand problem-solving strategies.

AP 7th Class Maths 8th Chapter Rational Numbers InText Questions

Try These (Page No. 4)

Question 1.
Is the number \(\frac{2}{-3}\) rational? Think about it.
Solution:
\(\frac{2}{-3}\) = \(\frac{-2}{3}\) a rational number.

Question 2.
List ten rational numbers.
Solution:
\(\frac{7}{19}\), \(\frac{1}{3}\), \(\frac{7}{4}\), \(\frac{9}{2}\), \(\frac{13}{9}\), \(\frac{1}{4}\), \(\frac{-3}{7}\), \(\frac{-10}{11}\), \(\frac{1}{471}\), \(\frac{7}{4}\)

Try These (Page No. 6)

Try These (Page No. 6)

Question 1.
Is 5 a positive rational number ?
Solution:
Yes, 5 = \(\frac{5}{1}\), a positive rational number.

Question 2.
List five more positive rational numbers.
Solution:
7, \(\frac{16}{7}\), \(\frac{19}{4}\), \(\frac{3}{4}\), \(\frac{12}{81}\)

Try These (Page No. 8)

Question 1.
Is – 8 a negative rational number?
Solution:
Yes,
-8 = \(\frac{-8}{1}\) is a negative rationl number.

Question 2.
List five more negative rational numbers.
Solution:
\(\frac{-7}{2}\), \(\frac{-3}{7}\), \(\frac{-41}{2}\), \(\frac{-30}{17}\), \(\frac{-1}{2}\)

Try Theses (page No. 8)

Question 1.
Which of these are negative rational numbers?
i) \(\frac{-2}{3}\)
ii) \(\frac{5}{7}\)
iii) \(\frac{3}{-5}\)
iv) 0
v) \(\frac{6}{11}\)
vi) \(\frac{-2}{-9}\)
Solution:
i) \(\frac{-2}{3}\) is a negative rational number.
ii) \(\frac{3}{-5}\) = \(\frac{-3}{5}\) is a negative rational number.

TryThese (Page No. 12)

Question 1.
Find the standard form of
i) \(\frac{-18}{45}\)
ii) \(\frac{-12}{18}\)
Solution:
i) \(\frac{-18}{45}\) = \(\frac{-2 \times 3 \times 3}{3 \times 3 \times 5}\) = \(\frac{-2}{5}\)
ii) \(\frac{-12}{18}\) = \(\frac{-2 \times 2 \times 3}{2 \times 3 \times 3}\) = \(\frac{-2}{3}\)

Try These (Page No. 18)

Question 1.
Find the rational numbers between \(\frac{-5}{7}\) and \(\frac{-3}{8}\)
Solution:
Given rational number \(\frac{-5}{7}\) and \(\frac{-3}{8}\)
LCM of 7 and 8 = 7 × 8 = 56
\(\frac{-5}{7}\) × \(\frac{8}{8}\) and \(\frac{-3}{8}\) × \(\frac{7}{7}\)
\(\frac{-40}{56}\) and \(\frac{-21}{56}\)
The five rational numbers are
\(\frac{-22}{56}\), \(\frac{-23}{56}\), \(\frac{-25}{56}\), \(\frac{-31}{56}\), \(\frac{-39}{56}\)

Try These (Page No. 26)

Question 1.
Find \(\frac{-13}{7}\) + \(\frac{6}{7}\), \(\frac{19}{5}\) + \(\left(\frac{-7}{5}\right)\)
Solution:
\(\frac{-13}{7}\) + \(\frac{6}{7}\) = \(\frac{-13+6}{7}\) = \(\frac{-7}{7}\) = -1
\(\frac{19}{5}\) + \(\left(\frac{-7}{5}\right)\)
\(\frac{19}{5}\) + \(\frac{7}{5}\) = \(\frac{19-7}{5}\) = \(\frac{12}{5}\)

Try These (Page No. 26)

Question 1.
Find
i) \(\frac{-3}{7}\) + \(\frac{2}{3}\)
Solution:
\(\frac{-3}{7}\) × \(\frac{3}{3}\) + \(\frac{2}{3}\) × \(\frac{7}{7}\)
\(\frac{-9}{21}\) + \(\frac{14}{21}\) = \(\frac{-9+14}{21}\) = \(\frac{5}{21}\)

ii) \(\frac{-5}{6}\) + \(\frac{-3}{11}\)
Solution:
\(\frac{-5}{6}\) + \(\frac{-3}{11}\) = \(\frac{-5}{6}\) × \(\frac{11}{11}\) – \(\frac{3}{11}\) × \(\frac{6}{6}\)
= \(\frac{-55}{66}\) – \(\frac{18}{66}\) = \(\frac{-55-18}{66}\) = \(\frac{-73}{66}\)

Try These (Page No. 28)

Question 1.
What will be the additive inverse of \(\frac{-3}{9}\)? \(\frac{-9}{11}\)? \(\frac{5}{7}\)?
Solution:
Additive inverse of \(\frac{-3}{9}\) = \(\frac{3}{9}\)
Additive inverse of \(\frac{-9}{11}\) = \(\frac{9}{11}\)
Additive inverse of \(\frac{5}{7}\) = \(\frac{-5}{7}\)
\(\frac{-3}{9}\) + \(\frac{3}{9}\) = 0
\(\frac{-9}{11}\) + \(\frac{9}{11}\) = 0
\(\frac{5}{7}\) – \(\frac{5}{7}\) = 0

Try These (page No. 30)

Question 1.
Find
i) \(\frac{7}{9}\) – \(\frac{2}{5}\)
Solution:
\(\frac{7}{9}\) – \(\frac{2}{5}\)
LCM of 5, 9 = 45
= \(\frac{7}{9}\) × \(\frac{5}{5}\) – \(\frac{2}{5}\) × \(\frac{9}{9}\) = \(\frac{35}{45}\) – \(\frac{18}{45}\) = \(\frac{35-18}{45}\) = \(\frac{17}{45}\)

ii) 2\(\frac{1}{5}\) – \(\frac{(-1)}{3}\)
Solution:
2\(\frac{1}{5}\) – \(\frac{(-1)}{3}\)
\(\frac{11}{5}\) + \(\frac{1}{3}\) = \(\frac{11}{5}\) × \(\frac{3}{3}\) + \(\frac{1}{3}\) × \(\frac{5}{5}\) = \(\frac{33}{15}\) + \(\frac{5}{15}\) = × \(\frac{33+5}{15}\) = \(\frac{38}{15}\)

Try These (Page No. 32)

Question 1.
What will be
i) \(\frac{-3}{5}\) × 7?
Solution:
\(\frac{-3}{5}\) × 7 = \(\frac{-3 \times 7}{5}\) = \(\frac{-21}{5}\)

ii) \(\frac{-6}{5}\) × (-2)?
Solution:
\(\frac{-6}{5}\) × (-2) = \(\frac{-6 \times -2}{5}\) = \(\frac{12}{5}\)

Try These (Page No. 32)

Question 1.
Find
i) \(\frac{-3}{4}\) × \(\frac{1}{7}\)
Solution:
\(\frac{-3}{4}\) × \(\frac{1}{7}\) = \(\frac{-3 \times 1}{4 \times 7}\) = \(\frac{-3}{28}\)

ii) \(\frac{2}{3}\) × \(\frac{-5}{9}\)
Solution:
\(\frac{2}{3}\) × \(\frac{-5}{9}\) = \(\frac{2 \times -5}{3 \times 9}\) = \(\frac{-10}{27}\)

Try These (Page No. 34)

Question 1.
What will be the reciprocal of \(\frac{-6}{11}\)? and \(\frac{-8}{5}\)?
Solution:
Reciprocal of \(\frac{-6}{11}\) is \(\frac{11}{-6}\) = \(\frac{-11}{6}\)
Reciprocal of \(\frac{-8}{5}\) is \(\frac{5}{-8}\) = \(\frac{-5}{8}\)

Try These (Page No. 36)

Question 1.
Find
i) \(\frac{2}{3}\) × \(\frac{-7}{8}\)
Solution:
\(\frac{2}{3}\) × \(\frac{-7}{8}\) = \(\frac{2}{3}\) × \(\frac{-7}{2 \times 4}\) = \(\frac{1}{3}\) × \(\frac{-7}{4}\) = \(\frac{-7}{12}\)

ii) \(\frac{-6}{7}\) × \(\frac{5}{7}\)
Solution:
\(\frac{-6}{7}\) × \(\frac{5}{7}\) = \(\frac{-6 \times 5}{7 \times 7}\) = \(\frac{-30}{49}\)

Well-designed AP 7th Class Maths Guide Chapter 8 Rational Numbers Exercise 8.2 offers step-by-step explanations to help students understand problem-solving strategies.

Rational Numbers Class 7 Exercise 8.2 Solutions – 7th Class Maths 8.2 Exercise Solutions

Question 1.
i) \(\frac{5}{4}\) + \(\left(\frac{-11}{4}\right)\)
Solution:
\(\frac{5}{4}\) + \(\left(\frac{-11}{4}\right)\) = \(\frac{5}{4}\) = –\(\frac{11}{4}\)
= \(\frac{5-11}{4}\) = \(\frac{-6}{4}\) = \(\frac{-3}{2}\)

ii) \(\frac{5}{3}\) + \(\frac{3}{5}\)
Solution:
\(\frac{5}{3}\) × \(\frac{5}{5}\) + \(\frac{3}{5}\) × \(\frac{3}{3}\) = \(\frac{25}{15}\) + \(\frac{9}{15}\) = \(\frac{25+9}{15}\) = \(\frac{34}{15}\)

iii) \(\frac{-9}{10}\) + \(\frac{22}{15}\)
Solution:
LCM of

10, 15 = 5 × 2 × 3 = 30
= \(\frac{-9}{10}\) × \(\frac{3}{3}\) + \(\frac{22}{15}\) × \(\frac{2}{2}\)
= \(\frac{-27}{30}\) + \(\frac{44}{30}\) = \(\frac{-27+44}{30}\) = \(\frac{17}{30}\)

iv) \(\frac{-3}{-11}\) + \(\frac{5}{9}\)
Solution:
\(\frac{-3}{-11}\) + \(\frac{5}{9}\) = \(\frac{3}{11}\) + \(\frac{5}{9}\)
LCM of 9, 11 = 99
= \(\frac{-3}{-11}\) × \(\frac{9}{9}\) + \(\frac{5}{9}\) × \(\frac{11}{11}\) = \(\frac{27}{99}\) + \(\frac{55}{99}\) = \(\frac{82}{99}\)

v) \(\frac{-8}{19}\) + \(\left(\frac{-2}{57}\right)\)
Solution:
LCM of 19, 57 is 57
= \(\frac{-8}{19}\) × \(\frac{3}{3}\) + \(\left(\frac{-2}{57}\right)\)
= \(\frac{-24}{57}\) – \(\frac{2}{7}\) = \(\frac{-24-2}{57}\) = \(\frac{-26}{57}\)

vi) \(\frac{-2}{3}\) + 0
Solution:
\(\frac{-2}{3}\) + 0 = \(\frac{-2}{3}\)

vii) -2\(\frac{1}{3}\) + 4\(\frac{3}{5}\)
Solution:
-2\(\frac{1}{3}\) + 4\(\frac{3}{5}\) \(\frac{-7}{3}\) + \(\frac{23}{5}\)
LCM of 3, 5 is 15
= \(\frac{-7}{3}\) × \(\frac{5}{5}\) + \(\frac{23}{5}\) × \(\frac{3}{3}\) = \(\frac{-35}{15}\) + \(\frac{69}{15}\) = \(\frac{-35+69}{15}\) = \(\frac{34}{15}\) = 2\(\frac{4}{15}\)

Question 2.
Find
i) \(\frac{7}{24}\) – \(\frac{17}{36}\)
Solution:

LCM of 24, 36
= 2 × 2 × 3 × 2 × 3 = 72
\(\frac{7}{24}\) × \(\frac{3}{3}\) – \(\frac{17}{36}\) × \(\frac{2}{2}\)
\(\frac{21}{72}\) – \(\frac{34}{72}\)
\(\frac{21-34}{72}\) = \(\frac{-13}{72}\)

ii) \(\frac{5}{63}\) – \(\left(\frac{-6}{21}\right)\)
Solution:
\(\frac{5}{63}\) – \(\left(\frac{-6}{21}\right)\) = \(\frac{5}{63}\) + \(\frac{6}{21}\)

LCM of 21, 63
= 3 × 7 × 3 = 63
= \(\frac{5}{63}\) + \(\frac{6}{21}\) × \(\frac{3}{3}\) = \(\frac{5}{63}\) + \(\frac{18}{63}\) = \(\frac{5+18}{63}\) = \(\frac{23}{63}\)

iii) \(\frac{-6}{13}\) – \(\left(\frac{-7}{15}\right)\)
Solution:
\(\frac{-6}{13}\) – \(\left(\frac{-7}{15}\right)\)
\(\frac{-6}{13}\) + \(\frac{7}{15}\)
LCM of 13,15 is 13 × 15 = 195
\(\frac{-6}{13}\) × \(\frac{15}{15}\) + \(\frac{7}{15}\) × \(\frac{13}{13}\)
= \(\frac{-90}{195}\) + \(\frac{91}{195}\) = \(\frac{-90+91}{195}\) = \(\frac{1}{195}\)

iv) \(\frac{-3}{8}\) – \(\frac{7}{11}\)
Solution:
LCM of 8, 11 = 8 × 11 = 88
\(\frac{-3}{8}\) × \(\frac{11}{11}\) – \(\frac{7}{11}\) × \(\frac{8}{8}\)
= \(\frac{-33}{88}\) – \(\frac{56}{88}\) = \(\frac{-33-56}{88}\) = \(\frac{-89}{88}\)

v) -2\(\frac{1}{9}\) – 6
Solution:
-2\(\frac{1}{9}\) – 6 = \(\frac{-19}{9}\) – \(\frac{6}{1}\)
LCM of 1, 9 = 9
\(\frac{-19}{9}\) – \(\frac{6}{1}\) × \(\frac{9}{9}\)
= \(\frac{-19}{9}\) – \(\frac{54}{9}\) = \(\frac{-73}{9}\) = -8\(\frac{1}{9}\)

Question 3.
Find the product.
i) \(\frac{9}{2}\) × \(\left(\frac{-7}{4}\right)\)
Solution:
\(\frac{9}{2}\) × \(\left(\frac{-7}{4}\right)\) = \(\frac{9}{2}\) × \(\frac{-7}{2 \times 2}\) = \(\frac{-63}{8}\)

ii) \(\frac{3}{10}\) × (-9)
Solution:
\(\frac{3}{10}\) × (-9) = \(\frac{3 \times-9}{10}\) = \(\frac{-27}{10}\)

iii) \(\frac{-6}{5}\) × \(\frac{9}{11}\)
Solution:
\(\frac{-6}{5}\) × \(\frac{9}{11}\) = \(\frac{-6 \times 9}{5 \times 11}\) = \(\frac{-54}{55}\)

iv) \(\frac{3}{7}\) × \(\left(\frac{-2}{5}\right)\)
Solution:
\(\frac{3}{7}\) × \(\left(\frac{-2}{5}\right)\) = \(\frac{3 \times-2}{7 \times 5}\) = \(\frac{-6}{35}\)

v) \(\frac{3}{11}\) × \(\frac{2}{5}\)
Solution:
\(\frac{3}{11}\) × \(\frac{2}{5}\) = \(\frac{3 \times 2}{11 \times 5}\) × \(\frac{6}{55}\)

vi) \(\frac{3}{-5}\) × \(\frac{-5}{3}\)
Solution:
\(\frac{3}{-5}\) × \(\frac{-5}{3}\) = \(\frac{1}{1}\) × \(\frac{1}{1}\) = 1

Question 4.
Find the value of:
i) (-4) ÷ \(\frac{2}{3}\)
Solution:
(-4) ÷ \(\frac{2}{3}\) = -4 × \(\frac{3}{2}\) = -2 × 3 = -6

ii) \(\frac{-3}{5}\) ÷ 2
Solution:
\(\frac{-3}{5}\) ÷ 2 = \(\frac{-3}{5}\) × \(\frac{1}{2}\) = \(\frac{-3}{10}\)

iii) \(\frac{-4}{5}\) ÷ (-3)
Solution:
\(\frac{-4}{5}\) × \(\frac{-1}{3}\) = × \(\frac{4}{15}\)

iv) \(\frac{-1}{8}\) ÷ \(\frac{3}{4}\)
Solution:
\(\frac{-1}{8}\) ÷ \(\frac{3}{4}\) = \(\frac{-1}{8}\) × \(\frac{4}{3}\) = \(\frac{-1 \times 1}{2 \times 3}\) = \(\frac{-1}{6}\)

v) \(\frac{-2}{13}\) ÷ \(\frac{1}{7}\)
Solution:
\(\frac{-2}{13}\) ÷ \(\frac{1}{7}\) = \(\frac{-2}{13}\) × \(\frac{7}{-1}\) = \(\frac{-2 \times 7}{13 \times 1}\) = \(\frac{-14}{13}\)

vi) \(\frac{-7}{12}\) ÷ \(\left(\frac{-2}{13}\right)\)
Solution:
\(\frac{-7}{12}\) ÷ \(\left(\frac{-2}{13}\right)\) = \(\frac{-7}{12}\) × \(\frac{13}{-2}\) = \(\frac{-91}{-24}\) = \(\frac{91}{24}\)

vii) \(\frac{3}{13}\) ÷ \(\left(\frac{-4}{65}\right)\)
Solution:
\(\frac{3}{13}\) ÷ \(\frac{-4}{65}\) = \(\frac{3}{13}\) × \(\frac{65}{-4}\) = \(\frac{3 \times 5}{1 \times-4}\) = \(\frac{15}{-4}\)

Well-designed AP 7th Class Maths Guide Chapter 8 Rational Numbers Exercise 8.1 offers step-by-step explanations to help students understand problem-solving strategies.

Rational Numbers Class 7 Exercise 8.1 Solutions – 7th Class Maths 8.1 Exercise Solutions

Question 1.
List five rational numbers between:
i) -1 and 0
Solution:
\(\frac { -1 }{ 2 }\), \(\frac { -1 }{ 3 }\), \(\frac { -2 }{ 5 }\), \(\frac { -2 }{ 3 }\), \(\frac { -2 }{ 7 }\)

ii) -2 and -1
Solution:
\(\frac { -3 }{ 2 }\), \(\frac { -5 }{ 3 }\), \(\frac { -8 }{ 5 }\), \(\frac { -10 }{ 7 }\), \(\frac { -9 }{ 5 }\)

iii) \(\frac { -4 }{ 5 }\) and \(\frac { -2 }{ 3 }\)
Solution:
\(\frac { -4 }{ 5 }\) and \(\frac { -2 }{ 3 }\) converting into like fractions, we get
\(\frac { -4 }{ 5 }\) = \(\frac{(-4 \times 9)}{(5 \times 9)}\) = \(\frac { -36 }{ 45 }\)
\(\frac { -2 }{ 3 }\) = \(\frac{(-2 \times 13)}{(3 \times 15)}\) = \(\frac { -30 }{ 45 }\)
Five rational numbers between \(\frac { -4 }{ 5 }\) and \(\frac { -2 }{ 3 }\) are,
\(\frac { -36 }{ 45 }\) < \(\frac { -35 }{ 45 }\) < \(\frac { -34 }{ 45 }\) < \(\frac { -33 }{ 45 }\) < \(\frac { -31 }{ 45 }\) < \(\frac { -30 }{ 45 }\)
\(\frac { -4 }{ 5 }\) < \(\frac { -35 }{ 45 }\) < \(\frac { -34 }{ 45 }\) < \(\frac { -33 }{ 45 }\) < \(\frac { -32 }{ 45 }\) < \(\frac { -31 }{ 45 }\) < \(\frac { -2 }{ 3 }\)
Thus, the five rational numbers between \(\frac{-4}{5}\) and \(\frac{-2}{3}\) are,
\(\frac { -7 }{ 9 }\), \(\frac { -34 }{ 45 }\), \(\frac { -11 }{ 15 }\), \(\frac { -32 }{ 45 }\), \(\frac { -31 }{ 45 }\)

iv) –\(\frac { 1 }{ 2 }\) and \(\frac { 2 }{ 3 }\)
Solution:
\(\frac { -1 }{ 2 }\) × \(\frac { 3 }{ 3 }\) and \(\frac { 2 }{ 3 }\) × \(\frac { 2 }{ 2 }\) = \(\frac { -3 }{ 6 }\) and \(\frac { 4 }{ 6 }\)
The five rational numbers are
\(\frac { 3 }{ 6 }\), \(\frac { 2 }{ 6 }\), \(\frac { 1 }{ 6 }\), 0, \(\frac { -1 }{ 6 }\) i.e., \(\frac { 1 }{ 2 }\), \(\frac { 1 }{ 3 }\), \(\frac { 1 }{ 6 }\), 0, \(\frac { -1 }{ 6 }\)

Question 2.
Write four more rational numbers in each of the following patterns.
i) \(\frac { -3 }{ 5 }\), \(\frac { -6 }{ 10 }\), \(\frac { -9 }{ 15 }\), \(\frac { -12 }{ 20 }\), ……………..
Solution:
\(\frac { -3 }{ 5 }\), \(\frac { -6 }{ 10 }\), \(\frac { -9 }{ 15 }\), \(\frac { -12 }{ 20 }\)
\(\frac { -12 – 3 }{ 20 + 5 }\) = \(\frac { -15 }{ 25 }\) ⇒ \(\frac{-15-3}{25+5}\) = \(\frac{-18}{30}\)
\(\frac{-18-3}{30+5}\) = \(\frac { -21 }{ 35 }\) ⇒ \(\frac{-21-3}{35+5}\) = \(\frac { -24 }{ 40 }\)
∴ \(\frac { -3 }{ 5 }\), \(\frac { -6 }{ 10 }\), \(\frac { -9 }{ 15 }\), \(\frac { -12 }{ 20 }\), \(\frac { -15 }{ 25 }\), \(\frac { -18 }{ 30 }\), \(\frac { -21 }{ 35 }\), \(\frac { -24 }{ 40 }\)

ii) \(\frac { -1 }{ 4 }\), \(\frac { -2 }{ 8 }\), \(\frac { -3 }{ 12 }\), ……………..
Solution:
\(\frac { -1 }{ 4 }\), \(\frac { -2 }{ 8 }\), \(\frac { -3 }{ 12 }\), ……………..
\(\frac{-3-1}{12+4}\) = \(\frac { -4 }{ 16 }\)
\(\frac{-4-1}{16+4}\) = \(\frac{-5}{20}\)
\(\frac{-5-1}{20+4}\) = \(\frac{-6}{24}\)
\(\frac{-6-1}{24+4}\) = \(\frac{-7}{28}\)
∴ \(\frac{-1}{4}\), \(\frac{-2}{8}\), \(\frac{-3}{12}\), \(\frac{-4}{16}\), \(\frac{-5}{20}\), \(\frac{-6}{24}\), \(\frac{-7}{28}\)

iii) \(\frac { -1 }{ 6 }\), \(\frac { 2 }{ -12 }\), \(\frac { 3 }{ -18 }\), \(\frac { 4 }{ -24 }\) …………..
Solution:
\(\frac { -1 }{ 6 }\), \(\frac { 2 }{ -12 }\), \(\frac { 3 }{ -18 }\), \(\frac { 4 }{ -24 }\)
\(\frac { 4 + 1 }{ – 24 – 6 }\) = \(\frac { 5 }{ -30 }\); \(\frac { 5 + 1 }{ -30 – 6 }\) = \(\frac { 6 }{ -36 }\)
\(\frac { 6 + 1 }{ – 36 – 6 }\) = \(\frac { 7 }{ -42 }\); \(\frac { 7 + 1 }{ – 42 – 6 }\) = \(\frac { 8 }{ -48 }\)
∴ \(\frac { -1 }{ 6 }\), \(\frac { 2 }{ -12 }\), \(\frac { 3 }{ -18 }\), \(\frac { 4 }{ -24 }\), \(\frac { 5 }{ -30 }\), \(\frac { 6 }{ -36 }\), \(\frac { 7 }{ -42 }\), \(\frac { 8 }{ -48 }\)

iv) \(\frac { -2 }{ 3 }\), \(\frac { 2 }{ -3 }\), \(\frac { 4 }{ -6 }\), \(\frac { 6 }{ -9 }\), …………
Solution:
\(\frac { -2 }{ 3 }\), \(\frac { 2 }{ -3 }\), \(\frac { 4 }{ -6 }\), \(\frac { 6 }{ -9 }\), …………
\(\frac { 6 + 2 }{ – 9 – 3 }\) = \(\frac { 8 }{ -12 }\)
\(\frac { 8 + 2 }{ – 12 – 3 }\) = \(\frac { 10 }{ -15 }\)
\(\frac { 10 + 2 }{ – 15 – 3 }\) = \(\frac { 12 }{ -18 }\)
\(\frac { 12 + 2 }{ – 18 – 3 }\) = \(\frac { 14 }{ -21 }\)
∴ \(\frac { -2 }{ 3 }\), \(\frac { 2 }{ -3 }\), \(\frac { 4 }{ -6 }\), \(\frac { 6 }{ -9 }\), \(\frac { 8 }{ -21 }\), \(\frac { 10 }{ -15 }\), \(\frac { 12 }{ -18 }\), \(\frac { 14 }{ -21 }\)

Question 3.
Give four rational numbers equivalent to:
i) \(\frac { -2 }{ 7 }\)
Solution:
\(\frac { -2 }{ 7 }\) × \(\frac { 2 }{ 2 }\) = \(\frac { -4 }{ 14 }\)
\(\frac { -2 }{ 7 }\) × \(\frac { 3 }{ 3 }\) = \(\frac { -6 }{ 21 }\)
\(\frac { -2 }{ 7 }\) × \(\frac { 4 }{ 4 }\) = \(\frac { -8 }{ 28 }\)
\(\frac { -2 }{ 7 }\) × \(\frac { 5 }{ 5 }\) = \(\frac { -10 }{ 35 }\)
∴ The four rational numbers equivalent to \(\frac { -2 }{ 7 }\) are \(\frac { -4 }{ 14 }\), \(\frac { -6 }{ 21 }\), \(\frac { -8 }{ 28 }\), \(\frac { -10 }{ 35 }\).

ii) \(\frac { 5 }{ -3 }\)
Solution:
\(\frac { 5 }{ -3 }\) = \(\frac { -5 }{ 3 }\)
\(\frac { -5 }{ 3 }\) × \(\frac { 2 }{ 2 }\) = \(\frac { -10 }{ 6 }\)
\(\frac { -5 }{ 3 }\) × \(\frac { 3 }{ 3 }\) = \(\frac { -15 }{ 9 }\)
\(\frac { -5 }{ 3 }\) × \(\frac { 4 }{ 4 }\) = \(\frac { -20 }{ 12 }\)
\(\frac { -5 }{ 3 }\) × \(\frac { 5 }{ 5 }\) = \(\frac { -25 }{ 15 }\)
∴ The four rational numbers equivalent to \(\frac { -5 }{ 3 }\) are \(\frac { -10 }{ 6 }\), \(\frac { -15 }{ 9 }\), \(\frac { -20 }{ 12 }\), \(\frac { -25 }{ 15 }\)

iii) \(\frac { 4 }{ 9 }\)
Solution:
\(\frac { 4 }{ 9 }\) × \(\frac { 2 }{ 2 }\) = \(\frac { 8 }{ 18 }\)
\(\frac { 4 }{ 9 }\) × \(\frac { 3 }{ 3 }\) = \(\frac { 12 }{ 27 }\)
\(\frac { 4 }{ 9 }\) × \(\frac { 4 }{ 4 }\) = \(\frac { 16 }{ 36 }\)
\(\frac { 4 }{ 9 }\) × \(\frac { 5 }{ 5 }\) = \(\frac { 20 }{ 45 }\)
∴ The four rational numbers equivalent to \(\frac { 4 }{ 9 }\) are \(\frac { 8 }{ 18 }\), \(\frac { -12 }{ 27 }\), \(\frac { 16 }{ 36 }\), \(\frac { 20 }{ 45 }\).

Question 4.
Draw the number line and represent the following rational numbers on it:
Solution:

Question 5.
The points P, Q, R, S, T, U, A and B on the number line are such that, TR = RS = SU and AP = PQ = QB. Name the rational numbers represented by P, Q, R and S.

Solution:
P represents \(\frac { 7 }{ 3 }\)
Q represents \(\frac { 8 }{ 3 }\)
R represents \(\frac { -4 }{ 3 }\)
S represents \(\frac { -5 }{ 3 }\)

Question 6.
Which of the following pairs represent the same rational number?
i) \(\frac { -7 }{ 21 }\) and \(\frac { 3 }{ 9 }\)
Solution:
\(\frac { -7 }{ 21 }\) and \(\frac { 3 }{ 9 }\)
Both are not same as one is negative and other is positive rational number.

ii) \(\frac { -16 }{ 20 }\) and \(\frac { 20 }{ -25 }\)
Solution:
\(\frac { -16 }{ 20 }\) and \(\frac { 20 }{ -25 }\) \(\frac { -16 }{ 20 }\)
= \(\frac { – 4 × 4 }{ 5 × 4 }\) = \(\frac { -4 }{ 5 }\)
\(\frac { 20 }{ -25 }\) = \(\frac { -20 }{ 25 }\)
= \(\frac { – 4 × 5 }{ 5 × 5 }\) = \(\frac { -4 }{ 5 }\)
∴ Both are same \(\frac { -16 }{ 20 }\) = \(\frac { 20 }{ -25 }\)

iii) \(\frac { -2 }{ -3 }\) and \(\frac { 2 }{ 3 }\)
Solution:
\(\frac { -2 }{ -3 }\) and \(\frac { 2 }{ 3 }\)
∴ Both are same \(\frac { -2 }{ -3 }\) and \(\frac { 2 }{ 3 }\)

iv) \(\frac { -3 }{ 5 }\) and \(\frac { -12 }{ 20 }\)
Solution:
\(\frac { -3 }{ 5 }\) and \(\frac { -12 }{ 20 }\)
\(\frac { -12 }{ 20 }\) = \(\frac{-4 \times 3}{5 \times 4}\) = \(\frac { -3 }{ 5 }\)
Both are same

v) \(\frac { 8 }{ -5 }\) and \(\frac { -24 }{ 15 }\)
Solution:
\(\frac { 8 }{ -5 }\) and \(\frac { -24 }{ 15 }\)
\(\frac { -24 }{ 15 }\) = \(\frac{-8 \times 3}{5 \times 3}\) = \(\frac { -8 }{ 5 }\)
\(\frac { 8 }{ -5 }\) = \(\frac { -8 }{ 5 }\) Both are same

vi) \(\frac { 1 }{ 3 }\) and \(\frac { -1 }{ 9 }\)
Solution:
\(\frac { 1 }{ 3 }\) and \(\frac { -1 }{ 9 }\)
Both are not same as one is negative and other is positive rational number.

vii) \(\frac { -5 }{ -9 }\) and \(\frac { 5 }{ -9 }\)
Solution:
\(\frac { -5 }{ -9 }\) and \(\frac { 5 }{ -9 }\)
\(\frac { -5 }{ -9 }\) ≠ \(\frac { 5 }{ -9 }\)
Both are not same as one is negative and other is positive rational number.

Question 7.
Rewrite the following rational numbers in the simplest form :
i) \(\frac { -8 }{ 6 }\)
Solution:
\(\frac { -8 }{ 6 }\) = \(\frac{-2 \times 2 \times 2}{2 \times 3}\) = \(\frac{-4}{3}\)

ii) \(\frac { 25 }{ 45 }\)
Solution:
\(\frac { 25 }{ 45 }\) = \(\frac{5 \times 5}{9 \times 5}\) = \(\frac { 5 }{ 9 }\)

iii) \(\frac { -44 }{ 72 }\)
Solution:
\(\frac { -44 }{ 72 }\) = \(\frac{-2 \times 2 \times 11}{2 \times 2 \times 3 \times 3}\) = \(\frac { -11 }{ 18 }\)

iv) \(\frac { -8 }{ 10 }\)
Solution:
\(\frac { -8 }{ 10 }\) = \(\frac{-2 \times 4}{2 \times 5}\) = \(\frac { -4 }{ 5 }\)

Question 8.
Fill in the boxes with the correct symbol out of >, <, and =

Negative rational number is always less than a positive rational number.



The number ‘ 0 ‘ is always greater than a negative rational number. (Or)
‘0’ lies to the right of \(\frac{-7}{6}\) on number line, so ‘ 0 ‘ is greater than (>) \(\frac{-7}{6}\).

Question 9.
Which is greater in each of the following :
i) \(\frac{2}{3}\), \(\frac{5}{2}\)
Solution:
\(\frac{2}{3}\), \(\frac{5}{2}\)
\(\frac{2}{3}\) × \(\frac{2}{2}\), \(\frac{5}{2}\) × \(\frac{3}{3}\)
\(\frac{4}{6}\), \(\frac{15}{6}\)
\(\frac{15}{6}\) > \(\frac{4}{6}\)
\(\frac{5}{2}\) > \(\frac{2}{3}\)

ii) \(\frac{-5}{6}\), \(\frac{-4}{3}\)
Solution:
\(\frac{-5}{6}\), \(\frac{-4}{3}\)
\(\frac{-4}{3}\) × \(\frac{2}{2}\) = \(\frac{-8}{6}\)
\(\frac{-5}{6}\) > \(\frac{-8}{6}\)
∴ \(\frac{-5}{6}\) > \(\frac{-4}{3}\)

iii) \(\frac{-3}{4}\), \(\frac{2}{-3}\)
Solution:
\(\frac{-3}{4}\), \(\frac{2}{-3}\)
\(\frac{2}{-3}\) × \(\frac{3}{3}\) = \(\frac{-9}{12}\)
\(\frac{-9}{12}\) < \(\frac{-8}{12}\)
\(\frac{2}{-3}\) > \(\frac{-3}{4}\)

iv) \(\frac{-1}{4}\), \(\frac{1}{4}\)
Solution:
\(\frac{-1}{4}\), \(\frac{1}{4}\)

v) -3\(\frac{2}{7}\), -3\(\frac{4}{5}\)
Solution:
-3\(\frac{2}{7}\), -3\(\frac{4}{5}\)
\(\frac{-23}{7}\), \(\frac{-19}{5}\)
\(\frac{-23}{7}\) × \(\frac{5}{5}\)
\(\frac{-19}{5}\) × \(\frac{7}{7}\)
\(\frac{-115}{35}\) × \(\frac{-133}{35}\)
\(\frac{-115}{35}\) > \(\frac{-133}{35}\)
∴ -3\(\frac{2}{7}\) > -3\(\frac{4}{5}\)

Question 10.
Write the following rational numbers in ascending order :
i) \(\frac{-3}{5}\), \(\frac{-2}{5}\), \(\frac{-1}{5}\)
Solution:
\(\frac{-3}{5}\), \(\frac{-2}{5}\), \(\frac{-1}{5}\)
\(\frac{-3}{5}\) < \(\frac{-2}{5}\) < \(\frac{-1}{5}\)

ii) \(\frac{-1}{3}\), \(\frac{-2}{9}\), \(\frac{-4}{3}\)
Solution:
\(\frac{-1}{3}\), \(\frac{-2}{9}\), \(\frac{-4}{3}\)
\(\frac{-1}{3}\) × \(\frac{3}{3}\), \(\frac{-2}{9}\), \(\frac{-4}{3}\) × \(\frac{3}{3}\)
\(\frac{-3}{9}\), \(\frac{-2}{9}\), \(\frac{-12}{9}\)
\(\frac{-12}{9}\) < \(\frac{-3}{9}\) < \(\frac{-2}{9}\)
\(\frac{-4}{3}\) < \(\frac{-1}{3}\) < \(\frac{-2}{9}\)

iii) \(\frac{-3}{7}\), \(\frac{-3}{2}\), \(\frac{-3}{4}\)
Solution:
\(\frac{-3}{7}\), \(\frac{-3}{2}\), \(\frac{-3}{4}\)
LCM of 2, 4, 7 is 28
\(\frac{-3}{7}\) × \(\frac{4}{4}\), \(\frac{-3}{2}\) × \(\frac{14}{14}\), \(\frac{-3}{4}\) × \(\frac{7}{7}\)
\(\frac{-12}{28}\), \(\frac{-42}{28}\), \(\frac{-21}{28}\)
\(\frac{-42}{28}\) < \(\frac{-21}{28}\) < \(\frac{-12}{28}\)
\(\frac{-3}{2}\) < \(\frac{-3}{4}\) < \(\frac{-3}{7}\)

Well-designed AP 7th Class Maths Guide Chapter 6 The Triangle and its Properties InText Questions offers step-by-step explanations to help students understand problem-solving strategies.

AP 7th Class Maths 6th Chapter The Triangle and its Properties InText Questions

Try These (Page No. 176)

Question 1.
Write the six elements (i.e., the 3 sides and the 3 angles) of △ABC.
Solution:

In △ABC
Three sides are AB, BC and AC
Three angles are \(\angle \mathrm{A}\), \(\angle \mathrm{B}\) and \(\angle \mathrm{C}\)

Question 2.
Write the :
i) Side opposite to the vertex Q of △PQR
Solution:

In △PQR, the side opposite to \(\angle \mathrm{Q}\) is PR .

ii) Angle opposite to the side LM of △LMN
Solution:

In △LMN, the angle opposite to the side \(\angle \mathrm{M}\) is \(\angle \mathrm{N}\).

iii) Vertex opposite to the side RT of △RST

Solution:
In △RST, the vertex opposite to the side RT is \(\angle \mathrm{S}\).

Question 3.
Look at figures and classify each of the triangles according to its
a) Sides
b) Angles

Solution:
i) Two sides are equal 8 cm and 8 cm
∴ △ABC is an isosceles triangle.

ii) In △PQR, \(\angle \mathrm{R}\) = 90°
∴ △PQR is a Right angled triangle

iii) In △MNL, \(\angle \mathrm{MNL}\) > 90°
i.e., \(\angle \mathrm{N}\) is an obtuse angle
∴ MNL is an obtuse angled triangle.

iv) In △RST, RS = ST = RT = 5.2 cm
All three sides are equal
∴ RST is an equilateral triangle

v) In △ABC, \(\angle \mathrm{B}\) is an obtuse angle
∴ \(\angle \mathrm{B}\) > 90°
∴ △ABC is an obtuse angled triangle

vi) In △PQR, \(\angle \mathrm{Q}\) = 90°
∴ △PQR is a right angled triangle

Think discuss and write (Page No: 178)

Question 1.
How many medians can a triangle have ?
Solution:
A triangle can have 3 medians.

Question 2.
Does a median lie wholly in the interior of the triangle? (If you think that this is not true, draw a figure to show such a case).
Solution:
Yes, the median of a triangle lies inside of the triangle as it is to be drawn from a vertex to the midpoint of its opposite side.

Think discuss and write (Page No : 180)

Question 1.
How many altitudes can a triangle have?
Solution:
A triangle can have 3 altitudes.

Question 2.
Draw rough sketches of altitudes from A to \(\bar{BC}\) for the following triangles (Figures) :

Solution:

AD ⊥ BC
AD is the altitude

ii)
AC is the altitude if BC is base
BC is the altitude if AC is base.

iii)
AM ⊥ BM
∴ AM is the altitude lies outside of the triangle if △ABC has base angle obtuse.

Question 3.
Will an altitude always lie in the interior of a triangle? If you think that this need not be true, draw a rough sketch to show such a case.
Solution:
In case of acute angled triangle altitude lies inside of the triangle.
But always need not be true.
In case of obtuse angled triangle the altitude lies outside of the triangle.

RS ⊥ SP
∴ RS is the altitude from outside.

Question 4.
Can you think of a triangle in which two altitudes of the triangle are two of its sides?
Solution:
Yes, in case of right angled triangle.

Question 5.
Can the altitude and median be same for a triangle?
(Hint : For Q.No. 4 and 5, investigate by drawing the altitudes for every type of triangle)
Solution:
No, it is true only in case of Isesceles and equilateral triangle.

Do This (Page No. 180)

Question 1.
Take several cut-outs of
i) an equilateral triangle
ii) an isosceles triangle and
iii) a scalene triangle.
Find their altitudes and medians. Do you find anything special about them? Discuss it with your friends.
Solution:
Students activity.

Think discuss and write (Page No. 184)

Question 1.
Exterior angles can be formed for a triangle in many ways. Three of them are shown here (Figure)

There are three more ways of getting exterior angles. Try to produce those rough sketches.
Solution:

Question 2.
Are the exterior angles formed at each vertex of a triangle equal?
Solution:
No

Question 3.
What can you say about the sum of an exterior angle of a triangle and its adjacent interior angle?
Solution:
Exterior angle of a triangle is equal to the sum of interior opposite angles

Try These (Page No : 194)

Question 1.
Two angles of a triangle are 30° and 80°. Find the third angle.
Solution:
Let the two angles 30° and 80°.
Let the find angle = x°
We know that by angle sum property of triangle
30° + 80° + x° = 180°
110° + x° = 180°
x° = 180° – 110°
x° = 70°
∴ The third angle = 70°

Question 2.
One of the angles of a triangle is 80° and the other two angles are equal. Find the measure of each of the equal angles.
Solution:
Let one of the angles of a triangle = 80° Let the measure of other two equal angles = x°
We know by angle sum property
80° + x° + x° = 180°
2x° + 80° = 180°
2x° = 180° – 80°
2x° = 100° ⇒ x° = \(\frac{100^{\circ}}{2}\)
x° = 50°
∴ The measure of equal angle = 50°

Question 3.
The three angles of a triangle are in the ratio 1 : 2 : 1. Find all the angles of the triangle. Classify the triangle in two different ways.
Solution:
Given ratio of angle of a triangle = 1 : 2 : 1
Let the first angle = x°
Second angle = 2x°;
Third angle = x°
By angle sum property in triangle
x° + 2x° + x° = 180°
4x° = 180°
x° = \(\frac{180^{\circ}}{4}\)
= \(\frac{90^{\circ}}{2}\) = 45°
∴ First angle = 45°
Second angle = 2 × 45° = 90°
Third angle = 45°
Since the angles are 45°, 45° and 90°, the triangle is an isosceles triangle.
Also one angle is right angle and two angles are equal.
Hence the triangle is called as Right angled isosceles triangle.

Think, discuss and write (Page No : 194)

Question 1.
Can you have a triangle with two right angles?
Solution:
No, if there are two right angles, the sum of two angles in a triangle will be 180°, which is wrong.

Question 2.
Can you have a triangle with two obtuse angles?
Solution:
No.

Question 3.
Can you have a triangle with two acute angles?
Solution:
Yes.

Question 4.
Can you have a triangle with all the three angles greater than 60°?
Solution:
No, if three angles greater than 60°, the sum of 3 angles will exceed 180°, which is false.

Question 5.
Can you have a triangle with all the three angles equal to 60°?
Solution:
Yes, if all angles are equal, than it will be 60° and called as an equilateral triangle.

Question 6.
Can you have a triangle with all the three angles less than 60°?
Solution:
No, if three angles are less than 60°, the sum of all 3 angles will be less than 180° which is false.

Try these (Page No: 196)

Question 1.
Find angle x in each figure:
i)
Solution:
The triangle is isosceles as two sides are equal.
Then the angles opposite to equal sides are equal.
∴ x° = 40°

ii)
Solution:
The triangle is isosceles as two sides are equal.
Then the angles opposite to equal sides are equal.
Let us draw the figure

a = 45°
∴ a° + x° + 45° = 180° (Angle sum property)
45° + x° + 45° = 180°
x° + 90° = 180°
x° = 180° – 90° ⇒ x° = 90°

iii)
Solution:
The triangle is an isosceles triangle
∴ Two sides and angles are equal
∴ x° = 50°

iv)
Solution:
Let us draw the figure

The triangle is isosceles
∴ x = a
100° + x° + a° = 180° (Angle sum property)
100° + x° + x° = 180°
2x° = 180° – 100°
2x° = 80° ⇒ x° = \(\frac{80^{\circ}}{2}\) ⇒ x° = 40°

v)
Solution:
Let us draw the figure

The triangle is isosceles
∴ x = a
∴ x + a + 90° = 180°
x° + x° + 90° = 180°
2x° = 180° – 90°
2x° = 90° ⇒ x° = \(\frac{90^{\circ}}{2}\) ⇒ x° = 45°

vi)
Solution:
Let us the draw the figure

The triangle is isosceles
∴ x° = a°
∴ x° + a° + 40° = 180° (Angle sum property)
x° + x° + 40° = 180°
2x° = 180° – 40°
2x° = 140° ⇒ x° = \(\frac{140^{\circ}}{2}\) ⇒ x° = 70°

vii)
Solution:
Let us draw the figure

a + 120° = 180° (Linear Pair)
a° = 180° – 120° ⇒ a = 60°
The triangle is isosceles
∴ x° = a°
∴ x = a = 60° ⇒ x = 60°

viii)
Solution:
Let us draw the figure

The triangle is isosceles
∴ a = x°
110° = a° + x° (Exterior angle property)
110° = x° + x°
2x° = 110° ⇒ x° = \(\frac{110^{\circ}}{2}\) ⇒ x° = 55°

ix)
Solution:
Let us draw the figure.

a = 30° (Vertically opposite angle)
The triangle is isosceles
∴ a° = x°
∴ a° = x° = 30° ⇒ x° = 30°

Question 2.
Find angles x and y in each figure.
i)
Solution:
Let us draw the figure

The triangle is isosceles
y = z
z + 120° = 180° (Linear. Pair)
z = 180° – 120°
z = 60°
∴ y = z = 60°
∴ y = 60°
120° = x + y (Exterior angle property)
120° = x° + 60°
& x=120° – 60° ⇒ x° = 60°

ii)
Solution:
Let us draw the figure

The triangle is right angled isosceles triangle
∴ a° = x°
a° + x° + 90° = 180° (Angle sum property)
x° + x° + 90° = 180°
2x° = 180° – 90°
2x° = 90° ⇒ x° = \(\frac{90^{\circ}}{2}\)⇒ x = 45°
x° + y° = 180° (Linear Pair)
45° + y° = 180°
y° = 180° – 45° ⇒ y = 135°

iii)
Solution:
Let us draw the figure

a = 92° (Vertically opposite angles)
The triangle is isosoceles
x = z
a + x + z = 180°
92° + x + x° = 180°
2x° + 92° = 180°
2x° = 180° – 92°
2x° = 88 ⇒ x° = \(\frac{88^{\circ}}{2}\) ⇒ x = 44°
z° = 44°
y = a + x (exterior angle property)
y = 92° + 44°
y = 136°

Think, discuss and write (Page No: 204)

Question 1.
Is the sum of any two angles of a triangle always greater than the third angle ?
Solution:
Yes, let the angles of triangle are
60°, 80°, 40°
Sum of two angles
60° + 80° = 140° > 40°
60° + 40° = 100° > 80°
80° + 40° = 120° >60°

Try These (Page No. 208)

Question 1.
Find the unknown length x in the following figures :

i)
Solution:
x² = 3² + 4²
x² = 9 + 16
x² = 25 ⇒ x² = 5² ⇒ x = 5

ii)
Solution:
x² = 6² + 8²
x² = 36 + 64
x² = 100 ⇒ x² = 10² ⇒ x = 10

iii)
Solution:
x² = 15² + 8²
x² = 225 + 64
x² = 289 ⇒ x² = 17² ⇒ x = 17 cm

iv)
Solution:
x² = 24² + 7²
x² = 576 + 49
x² = 625 ⇒ x² = 25² ⇒ x = 25

v)
Solution:

37² = 12² + a²
1369 = 144 + a²
a² = 1369 – 144
a² = 1225 ⇒ a² = 35² ⇒ a = 35
x = a + a = 35 + 35 = 70

vi)
Solution:

5² = 3² + y²
25 = 9 + y²
y² = 25 – 9
y² = 16 ⇒ y² = 4² ⇒ y = 4
12² + 5² = (x+y)²
144 +25 = (x+4)²
169 = (x+4)²
13² = (x+4)²
x + 4 = 13
x = 13 – 4 ⇒ x = 9

Diagonals meet at ‘ O ‘
In △BOC, ∠BOC = 90°
DB = 16 cm, AC = 30 cm
DO = OB = \(\frac{\mathrm{DB}}{2}\) = \(\frac{16}{2}\) = 8 cm
CO = AO = \(\frac{\mathrm{AC}}{2}\) = \(\frac{30}{2}\) = 15 cm
BC² = CO² + BO²
BC² = 15² + 8²
BC² = 225 + 64
BC² = 289
BC² = 17² ⇒ BC = 17 cm
∴ AB = BC = CD = AD = 17 cm
Perimeter of Rhombus = 4 × 17cm = 68 cm

Think discuss and write (Page No. 212)

Question 1.
Which is the longest side in the triangle PQR, right-angled at P?
Solution:
In △PQR, \(\angle \mathrm{P}\) = 90°
QR is the longest side

Question 2.
Which is the longest side in the triangle ABC, right-angled at B?
Solution:
In △ABC, \(\angle \mathrm{B}\) = 90°
AC is the longest side

Question 3.
Which is the longest side of a right triangle ?
Solution:
The side opposite to right angle in a right angled triangle is the largest side
∴ Hypotenuse is the largest side.

Question 4.
‘The diagonal of a rectangle produce by itself the same area as produced by its length and breadth’- This is Baud-hayan Theorem. Compare it with the Pythagoras property.
Solution:
Baudhayan Theorem is defirred with respect to a rectangle.
It contains a list of pythagorean triplets discovered algebrically, a statement of the pythagorean theorem. and a geometrical proof of the pythagorean theorem for an isosceles right triangle. Pythagorean theorem can be proved for any type of triangle.