RD Sharma Class 9 Maths Areas of Parallelograms and Triangles Solutions - Free PDF Download
FAQs on RD Sharma Solutions for Class 9 Maths Chapter 15 - Areas of Parallelograms and Triangles
1. How do the RD Sharma Solutions for Class 9 Maths Chapter 15 help in understanding complex theorems?
Vedantu's RD Sharma Solutions for Class 9 Maths Chapter 15 provide a step-by-step breakdown of each theorem's proof. Instead of just stating the result, the solutions guide you through the logical sequence, including the construction, hypothesis, and proof, making abstract concepts like 'parallelograms on the same base and between the same parallels' easy to grasp and apply in exams.
2. What is the correct method to solve problems involving a triangle and a parallelogram on the same base in Chapter 15?
The correct method, as detailed in the solutions, is to first establish the relationship: "If a triangle and a parallelogram are on the same base and between the same parallels, the area of the triangle is half the area of the parallelogram." The solutions then show how to:
- Identify the common base and the parallel lines.
- Write down the formula: Area(△) = ½ × Area(||gm).
- Substitute the known values to find the unknown area, ensuring all steps are clearly mentioned as per the CBSE marking scheme.
3. Are all exercises from the latest RD Sharma (2025-26 edition) Chapter 15 covered in these solutions?
Yes, our solutions cover all the questions from every exercise in the latest 2025-26 edition of the RD Sharma textbook for Class 9 Maths Chapter 15. Each problem is solved with detailed explanations, ensuring you have a reliable resource to verify your answers, understand different problem types, and prepare thoroughly for your exams.
4. Why is correctly identifying the 'base' and corresponding 'height' crucial for solving problems in this chapter?
Correctly identifying the base and its corresponding height is crucial because the area of a parallelogram (Base × Height) or a triangle (½ × Base × Height) depends entirely on these two specific, perpendicular measurements. A common mistake is using a slanted side as the height. The RD Sharma solutions repeatedly emphasise choosing a side as the base and then finding the perpendicular distance (the altitude) to the opposite side, which prevents incorrect calculations.
5. How do the solutions prove that a median divides a triangle into two triangles of equal area?
The solutions provide a structured proof by:
- Drawing a triangle (e.g., △ABC) with a median (e.g., AD from A to BC).
- Constructing an altitude (e.g., AN ⊥ BC). This altitude is the height for both △ABD and △ACD.
- Using the area formula: Area(△ABD) = ½ × BD × AN and Area(△ACD) = ½ × CD × AN.
- Since a median bisects the base (BD = CD), the solutions show that their areas must be equal. This step-wise method is perfect for exam preparation.
6. Can two parallelograms with completely different shapes have the same area? How do the solutions explain this?
Yes, two very different-looking parallelograms can have the same area. This is a key concept explained through the theorem: "Parallelograms on the same base and between the same parallels are equal in area." The solutions illustrate this by showing that as long as the length of the base and the perpendicular height between the parallel lines are constant, the area will remain the same, regardless of how slanted the other sides are.
7. What makes Vedantu's RD Sharma solutions for this chapter more effective than just looking at the answers?
Vedantu's solutions are more than just final answers. They focus on the 'how' and 'why' of problem-solving. Each step is justified with the relevant theorem or property from the chapter. This helps you build a strong conceptual foundation, learn the correct way to present proofs in exams, and identify potential areas where you might make mistakes, leading to better scores.
8. If a question involves multiple triangles and parallelograms within a single figure, how do the solutions simplify the approach?
For complex figures, the solutions adopt a systematic approach. They guide you to break down the complex shape into smaller, simpler components (e.g., individual triangles and parallelograms). Then, they apply the chapter's core theorems one by one to find relationships between the areas of these components, ultimately combining them to solve for the required area. This methodical process makes even the most intimidating problems manageable.
