NCERT Solutions for Class 9 Maths Chapter 9 Exercise 9.2 Solutions

Content Curator

NCERT Solutions for Class 9 Maths Chapter 9 Areas of Parallelograms and Triangles Exercise 9.2 Solutions are based on parallelograms with the same base and parallel lines.

Download PDF: NCERT Solutions for Class 9 Maths Chapter 9 Exercise 9.2 Solutions

Check out NCERT Solutions for Class 9 Maths Chapter 9 Exercise 9.2 Solutions

Exercise Solutions of Class 9 Maths Chapter 9 Areas Of Parallelograms And Triangles

Also check other Exercise Solutions of Class 9 Maths Chapter 9 Areas Of Parallelograms And Triangles

Also Check:

Important Chapter Related Topics
Area of Triangle	Perimeter of a Parallelogram
Area of a Trapezoid Formula	Trapezoids

Also check:

Math Study Guides
Maths Important Formulas	Maths MCQs	Comparison Articles in Maths
Difference Between Topics in Maths	Math Study Material	Algebra
Trigonometry	Number Systems	Mensuration
Math Formula	NCERT Solutions for Class 9 Maths	Geometry

CBSE X Related Questions

1.
If 3 cot A = 4, check whether \(\frac{(1-\text{tan}^2 A)}{(1+\text{tan}^2 A)}\) = cos² A – sin²A or not

View Solution

2.
Find the sums given below :
\(7 + 10\frac 12+ 14 + ....... + 84\)
\(34 + 32 + 30 + ....... + 10\)
\(–5 + (–8) + (–11) + ....... + (–230)\)

View Solution

3.
Form the pair of linear equations for the following problems and find their solution by substitution method.
(i) The difference between two numbers is 26 and one number is three times the other. Find them.
(ii) The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them.
(iii) The coach of a cricket team buys 7 bats and 6 balls for Rs 3800. Later, she buys 3 bats and 5 balls for Rs 1750. Find the cost of each bat and each ball.
(iv) The taxi charges in a city consist of a fixed charge together with the charge for the distance covered. For a distance of 10 km, the charge paid is Rs 105 and for a journey of 15 km, the charge paid is Rs 155. What are the fixed charges and the charge per km? How much does a person have to pay for travelling a distance of 25 km.
(v) A fraction becomes\(\frac{ 9}{11}\), if 2 is added to both the numerator and the denominator. If, 3 is added to both the numerator and the denominator it becomes \(\frac{5}{6}\). Find the fraction.
(vi) Five years hence, the age of Jacob will be three times that of his son. Five years ago, Jacob’s age was seven times that of his son. What are their present ages?

View Solution

4.
Prove the following identities, where the angles involved are acute angles for which the expressions are defined:\(\frac{(\text{1 + tan² A})}{(\text{1 + cot² A})} = (\frac{\text{1 - tan A }}{\text{ 1 - cot A}})^²= \text{tan² A}\)

View Solution

5.
The lengths of 40 leaves of a plant are measured correct to the nearest millimetre, and the data obtained is represented in the following table :
Length (in mm)
Number of leaves
118 - 126
3
127 - 135
5
136 - 144
9
145 - 153
12
154 - 162
5
163 - 171
4
172 - 180
2
Find the median length of the leaves.
(Hint : The data needs to be converted to continuous classes for finding the median, since the formula assumes continuous classes. The classes then change to 117.5 - 126.5, 126.5 - 135.5, . . ., 171.5 - 180.5.)

Length (in mm)	Number of leaves
118 - 126	3
127 - 135	5
136 - 144	9
145 - 153	12
154 - 162	5
163 - 171	4
172 - 180	2

View Solution

6.
Solve the following pair of linear equations by the substitution method.
(i) x + y = 14
x – y = 4
(ii) s – t = 3
  \(\frac{s}{3} + \frac{t}{2}\) =6
(iii) 3x – y = 3
9x – 3y = 9
(iv) 0.2x + 0.3y = 1.3
0.4x + 0.5y = 2.3
(v)\(\sqrt2x\) + \(\sqrt3y\)=0
  \(\sqrt3x\) - \(\sqrt8y\) = 0
(vi) \(\frac{3x}{2} - \frac{5y}{3}\) =-2,
  \(\frac{ x}{3} + \frac{y}{2}\) = \(\frac{ 13}{6}\)

View Solution

NCERT Solutions for Class 9 Maths Chapter 9 Exercise 9.2 Solutions

Exercise Solutions of Class 9 Maths Chapter 9 Areas Of Parallelograms And Triangles

CBSE X Related Questions

1.
If 3 cot A = 4, check whether \(\frac{(1-\text{tan}^2 A)}{(1+\text{tan}^2 A)}\) = cos² A – sin²A or not

2.
Find the sums given below :
\(7 + 10\frac 12+ 14 + ....... + 84\)
\(34 + 32 + 30 + ....... + 10\)
\(–5 + (–8) + (–11) + ....... + (–230)\)

4.
Prove the following identities, where the angles involved are acute angles for which the expressions are defined:\(\frac{(\text{1 + tan² A})}{(\text{1 + cot² A})} = (\frac{\text{1 - tan A }}{\text{ 1 - cot A}})^²= \text{tan² A}\)

Similar Mathematics Concepts

Comments

SUBSCRIBE TO OUR NEWS LETTER

NCERT Solutions for Class 9 Maths Chapter 9 Exercise 9.2 Solutions

Exercise Solutions of Class 9 Maths Chapter 9 Areas Of Parallelograms And Triangles

CBSE X Related Questions

1. If 3 cot A = 4, check whether \(\frac{(1-\text{tan}^2 A)}{(1+\text{tan}^2 A)}\) = cos2 A – sin2 A or not

2. Find the sums given below :\(7 + 10\frac 12+ 14 + ....... + 84\)\(34 + 32 + 30 + ....... + 10\)\(–5 + (–8) + (–11) + ....... + (–230)\)

4. Prove the following identities, where the angles involved are acute angles for which the expressions are defined:\(\frac{(\text{1 + tan² A})}{(\text{1 + cot² A})} = (\frac{\text{1 - tan A }}{\text{ 1 - cot A}})^²= \text{tan² A}\)

Similar Mathematics Concepts

Comments

SUBSCRIBE TO OUR NEWS LETTER

1.
If 3 cot A = 4, check whether \(\frac{(1-\text{tan}^2 A)}{(1+\text{tan}^2 A)}\) = cos² A – sin²A or not

2.
Find the sums given below :
\(7 + 10\frac 12+ 14 + ....... + 84\)
\(34 + 32 + 30 + ....... + 10\)
\(–5 + (–8) + (–11) + ....... + (–230)\)

4.
Prove the following identities, where the angles involved are acute angles for which the expressions are defined:\(\frac{(\text{1 + tan² A})}{(\text{1 + cot² A})} = (\frac{\text{1 - tan A }}{\text{ 1 - cot A}})^²= \text{tan² A}\)