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rd sharma class 9 solutions offer detailed explanations for every topic, making math easy and understandable. Find step-by-step answers to problems and improve your performance in exams.

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RD Sharma Class 9 Solutions provide comprehensive solutions to help students tackle math problems with ease. These solutions are designed to offer step-by-step guidance, allowing students to understand complex concepts more clearly. With expert-backed explanations, students can approach their studies with confidence.

RD Sharma Class 9 Solutions – Questions and Answers

Sample Questions and Answers:

Find the value of $x$ in the equation $2 x + 5 = 15$ .
Answer: $x = 5$ .
What is the area of a triangle with base 4 cm and height 6 cm?
Answer: Area = $\frac{1}{2} \times 4 \times 6 = 12 \, \text{cm}^2$ .
Simplify: $3 a - 4 a + 5 b$ .
Answer: $- a + 5 b$ .
Solve for $y$ : $3 y + 7 = 22$ .
Answer: $y = 5$ .
What is the perimeter of a rectangle with length 8 cm and width 3 cm?
Answer: Perimeter = $\times (8 + 3) = 22 \, \text{cm}$ .
Solve: $2 x - 7 = 9$ .
Answer: $x = 8$ .

Books for RD Sharma Class 9 Solutions

RD Sharma Mathematics Class 9 – RD Sharma (Dhanpat Rai Publications)
Topics: Algebra, Geometry, Trigonometry, Statistics. Focus on practice problems with step-by-step solutions.
Mathematics Class 9 – R.S. Aggarwal (S. Chand Publishing)
Topics: Algebra, Probability, Coordinate Geometry, and more. Contains detailed explanations of mathematical problems.
Mathematics for Class 9 – R.D. Sharma (Dhanpat Rai)
Topics: Quadratic equations, Geometry, Mensuration. Ideal for comprehensive practice with solutions.
ICSE Mathematics Class 9 – M.L. Aggarwal (Arihant Publications)
Topics: Coordinate Geometry, Algebra. Perfect for ICSE students, with solutions explained.
Comprehensive Mathematics for Class 9 – S.K. Gupta (Rama Brothers)
Topics: Algebra, Trigonometry, Linear Equations. Provides explanations for key problems and practice solutions.
Mathematics Class 9 – M.L. Aggarwal (Arihant Publications)
Topics: Probability, Algebra, Arithmetic Progression. Strong focus on comprehensive understanding.
New Mathematics for Class 9 – R.D. Sharma (Rama Publications)
Topics: Coordinate Geometry, Mensuration, Statistics. Helps students understand core concepts and provides exercises for practice.
Understanding Mathematics Class 9 – K.K. Gupta (S. Chand Publishing)
Topics: Algebra, Geometry. Includes sample problems, solved examples, and practice questions.
Mathematics Class 9 – K. C. Sinha (Laxmi Publications)
Topics: Polynomials, Linear Equations. Covers key mathematical topics with explained solutions.
Comprehensive Mathematics Class 9 – Dr. S.K. Gupta (S. Chand Publishing)
Topics: Geometry, Algebra, Trigonometry. Offers solutions for comprehensive practice.
Higher Algebra for Class 9 – R. S. Agarwal (S. Chand Publishing)
Topics: Polynomials, Algebraic Expressions. In-depth solutions for Algebra concepts.
Foundation of Mathematics for Class 9 – R.D. Sharma (Dhanpat Rai)
Topics: Arithmetic Progression, Geometry. It contains solved examples and exercises.
Mathematics Workbook Class 9 – S. K. Gupta (Rama Brothers)
Topics: Equations, Trigonometry, Geometry. Workbook with practice problems.
Maths Class 9 - A Complete Guide – R.D. Sharma (Dhanpat Rai Publications)
Topics: Coordinate Geometry, Algebra. Offers problem-solving strategies and practice questions.
Mathematics for Class 9 – R.D. Sharma (Dhanpat Rai Publications)
Topics: Probability, Trigonometry, Linear Equations. Includes solved problems with answers for easy understanding.
Class 9 Mathematics Workbook – A. K. Singh (S. Chand Publishing)
Topics: Quadratic Equations, Geometry. Workbook focusing on exercises and practice questions.
Class 9 Mathematics Solutions – M. L. Aggarwal (Arihant Publications)
Topics: Algebra, Geometry. Provides a thorough understanding through solved examples.
Integrated Mathematics Class 9 – Dr. K.C. Sinha (Rama Brothers)
Topics: Trigonometry, Algebra, Mensuration. Emphasizes step-by-step solutions and practice problems.
Mathematics for Class 9 - A Complete Guide – R.D. Sharma (Dhanpat Rai Publications)
Topics: Coordinate Geometry, Algebra. Easy-to-follow solutions with practice exercises.
Maths Lab Manual for Class 9 – R. D. Sharma (Dhanpat Rai Publications)
Topics: Lab-based activities for class 9 mathematics. Hands-on learning with practical solutions.
Question 1:
Solve the equation:
$2 x + 5 = 15$

Answer 1:
$2 x = 15 - 5$
$2 x = 10$
$\frac{10}{2}$
$x = 5$

Question 2:
Find the area of a triangle with base 4 cm and height 6 cm.

Answer 2:
Area = $12×Base×Height\frac{1}{2} \times \text{Base} \times \text{Height}$
Area = $12×4×6\frac{1}{2} \times 4 \times 6$
Area = 12 cm²

Question 3:
Simplify:
$3 a - 4 a + 5 b$

Answer 3:

$3 a - 4 a + 5 b = - a + 5 b$

Question 4:
Solve for $y$ :
$3 y + 7 = 22$

Answer 4:
$3 y = 22 - 7$
$3 y = 15$
$\frac{15}{3}$
$y = 5$

Question 5:

What is the perimeter of a rectangle with length 8 cm and width 3 cm?

Answer 5:
Perimeter = $\times (\text{Length} + \text{Width})$
Perimeter = $\times (8 + 3)$
Perimeter = 22 cm

Question 6:
Solve:
$2 x - 7 = 9$

Answer 6:
$2 x = 9 + 7$
$2 x = 16$
$\frac{16}{2}$
$x = 8.$

Question 7:
Find the value of $x$ in the equation:
$3 x + 5 = 17$

Answer 7:
$3 x = 17 - 5$
$3 x = 12$
$\frac{12}{3}$
$x = 4$

Question 8:
What is the area of a rectangle with length 12 cm and width 5 cm?

Answer 8:
Area = $Length×Width\text{Length} \times \text{Width}$
Area = $12 \times 5$
Area = 60 cm²

Question 9:
Simplify the expression:
$4 x - 3 y + 2 xy.$

Answer 9:
$4 x + 2 x - 3 y + y = 6 x - 2 y$
Question 10:
Solve the quadratic equation:
$x^2 - 5x + 6 = 0$

Answer 10:
Factor the equation:
$(x - 2) (x - 3) = 0$
So, $x = 2$ or $x = 3$

Question 11:
Find the surface area of a cube with side 6 cm.

Answer 11:
Surface area = $\times \text{side}^2$
Surface area = $\times 6^2 = 6 \times 36 = 216 \, \text{cm}^2$

Question 12:
Find the value of $x$ if:
$5 x - 3 = 2 x + 6$

Answer 12:
$5 x - 2 x = 6 + 3$
$3 x = 9$
$x = 3$

Question 13:
Find the area of a circle with radius 7 cm.

Answer 13:
Area = $πr2\pi r^2$
Area = $3.14 \times 7^2$
Area = $3.14 \times 49 = 153.86 \, \text{cm}^2$

Question 14:
Simplify the expression:
$7 x - 5 x + 2 y$

Answer 14:
$7 x - 5 x + 2 y = 2 x + 2 y$

Question 15:
Solve the equation:
$x + 2 = 8$

Answer 15:
$x = 8 - 2$
$x = 6$

Question 16:
Find the value of $x$ in the equation:
$3 x + 4 = 16$

Answer 16:
$3 x = 16 - 4$
$3 x = 12$
$\frac{12}{3}$
$x = 4$

Question 17:
What is the perimeter of a square with side length 9 cm?

Answer 17:
Perimeter = $\times \text{side length}$
Perimeter = $\times 9 = 36 \, \text{cm}$

Question 18:
Simplify:
$2 x + 3 y - x + 4 y$

Answer 18:
$2 x - x + 3 y + 4 y = x + 7 y$

Question 19:
Solve the equation:
$4 x - 7 = 2 x + 5$

Answer 19:
$4 x - 2 x = 5 + 7$
$2 x = 12$
$\frac{12}{2}$
$x = 6$

Question 20:
Find the value of $y$ in the equation:
$3 y - 5 = 16$

Answer 20:
$3 y = 16 + 5$
$3 y = 21$
$\frac{21}{3}$
$y = 7$
Question 21:
Solve the quadratic equation:
$x^2 - 9 = 0$

Answer 21:
$x^2 = 9$
$x = 3$ or $x = - 3$

Question 22:
Find the area of a trapezium with parallel sides 8 cm and 10 cm, and height 6 cm.

Answer 22:
Area = $12×(a+b)×h\frac{1}{2} \times (a + b) \times h$
Area = $12×(8+10)×6\frac{1}{2} \times (8 + 10) \times 6$
Area = $cm2\frac{1}{2} \times 18 \times 6 = 54 \, \text{cm}^2$

Question 23:
Solve:
$2 x + 4 = 12$

Answer 23:
$2 x = 12 - 4$
$2 x = 8$
$\frac{8}{2}$
$x = 4$

Question 24:
What is the surface area of a sphere with radius 5 cm?

Answer 24:
Surface area = $\pi r^2$
Surface area = $\times 3.14 \times 5^2$
Surface area = $\times 3.14 \times 25 = 314 \, \text{cm}^2$

Question 25:
Simplify:
$4 y - 2 y + 6$

Answer 25:
$4 y - 2 y + 6 = 2 y + 6$

Question 26:
Solve the equation:
$6 x + 2 = 20$

Answer 26:
$6 x = 20 - 2$
$6 x = 18$
$\frac{18}{6}$
$x = 3$

Question 27:
Find the value of $x$ in the equation:
$7 x - 3 = 25$

Answer 27:
$7 x = 25 + 3$
$7 x = 28$
$\frac{28}{7}$
$x = 4$

Question 28:
What is the area of a square with side length 5 cm?

Answer 28:
Area = $side2\text{side}^2$
Area = $cm25^2 = 25 \, \text{cm}^2$

Question 29:
Simplify:
$5 x - 3 x + 2 y$

Answer 29:
$5 x - 3 x + 2 y = 2 x + 2 y$

Question 30:
Find the perimeter of a rectangle with length 10 cm and width 6 cm.

Answer 30:
Perimeter = $\times (\text{Length} + \text{Width})$
Perimeter = $\times (10 + 6) = 32 \, \text{cm}$

.RD Sharma Class 9 Solutions are an essential resource for students aiming to excel in their mathematics studies. Mathematics at this level includes various chapters such as algebra, geometry, mensuration, and statistics, and RD Sharma's well-structured books provide clear, concise, and detailed solutions to every problem.

The primary advantage of using RD Sharma Class 9 Solutions is that they simplify difficult topics for students, making the learning process more accessible. The solutions are not just limited to answering questions but also include step-by-step explanations that guide students on how to approach different types of problems.

One of the unique features of RD Sharma solutions is the variety of problems provided for practice. This ensures that students have ample opportunities to hone their skills and grasp concepts thoroughly. Whether students are working through basic algebraic expressions or solving complex geometry problems, RD Sharma provides a diverse set of problems that cover every aspect of the Class 9 curriculum.

Moreover, the detailed explanations ensure that students can follow the problem-solving methods without confusion. Each solution is clearly written with the necessary mathematical steps, making it easier for students to understand the methodology behind solving the problem. This is crucial for building a strong mathematical foundation, which will benefit students in higher classes as well.

In addition to the clear solutions, RD Sharma also focuses on reinforcing concepts with regular practice. This approach allows students to get better with each attempt. Regular practice through RD Sharma Class 9 Solutions helps in retaining formulas, rules, and methods that are vital for solving mathematics problems efficiently.

Furthermore, these solutions act as a self-learning tool, which means students can study independently and resolve doubts as they go along. This independence fosters confidence in students and allows them to approach their math homework and exams with a positive attitude.

The layout and structure of RD Sharma Class 9 Solutions are also designed with the needs of students in mind. The problems are presented in an organized manner, making it easy for students to identify their areas of difficulty and focus on improving them.

In conclusion, RD Sharma Class 9 Solutions offer a comprehensive approach to learning mathematics, providing students with clear explanations, plenty of practice, and the confidence to tackle even the toughest math problems. Whether used alongside classroom learning or as a primary study resource, these solutions are an excellent tool for mastering mathematics at the Class 9 level.