Solutions for Class 9 Maths Chapter 17 Constructions(Exercise17.2)- by RD Sharma- Free PDF Download
FAQs on RD Sharma Class 9 Solutions Chapter 17 - Constructions (Exercise 17.2)
1. What are the main types of triangle constructions covered in RD Sharma Class 9 Solutions for Chapter 17, Exercise 17.2?
The solutions for this exercise primarily focus on three advanced types of triangle constructions based on specific given parameters:
- Constructing a triangle given its base, a base angle, and the sum of the other two sides.
- Constructing a triangle given its base, a base angle, and the difference of the other two sides.
- Constructing a triangle given its perimeter and both base angles.
2. How do you construct a triangle when its base, a base angle, and the sum of the other two sides are given?
To solve this type of problem from Exercise 17.2, you should follow the correct method:
- Draw the base and construct the given base angle at one of its endpoints.
- From the vertex of this angle, cut an arc equal to the sum of the other two sides along the angle's ray.
- Join the end of this arc to the other end of the base, which forms a temporary triangle.
- Construct the perpendicular bisector of this newly formed side.
- The point where the perpendicular bisector intersects the long side is the third vertex of your required triangle. Join this point to the base to complete the triangle.
3. What is the step-by-step method to construct a triangle with a known base, a base angle, and the difference between the other two sides?
The method for this construction, as per the RD Sharma solutions, involves these steps:
- Draw the base (e.g., BC) and construct the given base angle (e.g., at B).
- From the ray of the base angle, cut a line segment equal to the given difference of the sides (e.g., AB - AC).
- Join the endpoint of this segment to the other end of the base (C).
- Now, construct the perpendicular bisector of this new line segment.
- Extend the perpendicular bisector until it intersects the initial ray of the base angle. This point of intersection is the required third vertex (A). Join AC to complete the triangle.
4. How can you construct a triangle if you are given its perimeter and two base angles?
The solutions in Exercise 17.2 explain this unique construction method:
- Draw a line segment that is equal in length to the given perimeter.
- At the endpoints of this segment, construct angles that are equal to half of each of the given base angles.
- Extend the rays of these new, smaller angles until they intersect. This intersection point is the first vertex of the required triangle.
- Construct the perpendicular bisectors of the two sides of this large, temporary triangle.
- The points where these bisectors intersect the original perimeter line segment are the other two vertices of the required triangle.
5. In triangle constructions from Exercise 17.2, why is drawing a perpendicular bisector a critical step?
The perpendicular bisector is a crucial tool because any point on it is equidistant from the two endpoints of the line segment it bisects. When constructing a triangle where the sum or difference of sides is given, we use this property to find the exact location of the third vertex. It ensures that the vertex we find satisfies the given conditions, effectively splitting an extended line into the two correct side lengths of the final triangle.
6. What is the key difference in the construction process when the sum of two sides is given versus when their difference is given?
The main difference lies in the initial setup on the ray of the base angle.
- For the sum of sides, you mark the total sum on the ray and then use a perpendicular bisector to 'fold back' the extra length, which forms the third side of the triangle.
- For the difference of sides, you mark the smaller difference on the ray and then use a perpendicular bisector to 'extend' a line to find the correct, longer length of the required side.
7. When constructing a triangle from its perimeter, why must we use half of the given base angles initially?
We use half the base angles because the construction creates a large temporary triangle, and the final triangle is formed inside it. The perpendicular bisectors create two smaller isosceles triangles at the base. Due to the properties of isosceles triangles, the angles at the base of the final triangle become equal to the angles we constructed (e.g., ∠B/2). Therefore, the full base angle (∠B) is correctly formed, satisfying the problem's condition. Using the full angle initially would result in an incorrect final triangle.
8. How can you verify if the triangle you constructed for a problem in Exercise 17.2 is accurate?
After completing your construction using only a compass and a straightedge, you can perform a quick verification to check your accuracy:
- Use a ruler to measure the sides. Check if their sum, difference, or the total perimeter matches the values given in the question.
- Use a protractor to measure the angles to ensure they match the given base angles accurately.
