Construction Ex 11.2 Class 9 NCERT: Constructing Triangles

Welcome! In this chapter on Constructions, we move beyond basic angles and lines. Exercise 11.2 presents some fascinating challenges where you'll construct triangles with specific, tricky conditions. Have you ever wondered how to draw a triangle if you only know its base, one angle, and the sum of the other two sides? Or what if you're given the difference? It seems complex, but it's all based on a beautiful geometric property involving perpendicular bisectors.

This guide will break down each type of construction from the construction ex 11 2 class 9 ncert syllabus. We'll explore the logic behind each step, so you don't just memorize them but truly understand them. By the end, you'll be able to confidently construct any triangle when given its base, a base angle, and the sum or difference of the other two sides, or when given its perimeter and two base angles. Let's get our compass and ruler ready!

Understanding the Logic Behind Triangle Constructions in Ex 11.2

In this exercise, we aren't given all three sides or two angles and a side directly. Instead, we're given combined information. The key to solving these constructions lies in the properties of a perpendicular bisector. Remember: any point on the perpendicular bisector of a line segment is equidistant from the endpoints of that segment. We use this property to find the third vertex of the triangle.

  1. Given Base, Base Angle, and SUM of other two sides (e.g., BC, ∠B, AB + AC): We first draw a ray BX from B making the given angle. We cut a line segment BD = AB + AC on this ray. Now, we have a point D, but we need point A. If we join DC, vertex A must lie on BD. Also, we know AC must be equal to AD. For this to be true, A must lie on the perpendicular bisector of the segment DC. The intersection of this perpendicular bisector with BD gives us the required vertex A.
  1. Given Base, Base Angle, and DIFFERENCE of other two sides (e.g., BC, ∠B, AB - AC): This has two cases. If AB > AC, we cut a segment BD = AB - AC on the ray BX. We join DC. The vertex A must lie on the perpendicular bisector of DC. If AC > AB, the difference is AC - AB. We extend the ray BX downwards and cut BD = AC - AB on the extended line. Again, joining DC and drawing its perpendicular bisector gives us vertex A.

Step-by-Step: Construct a Triangle with Base, Base Angle, and Sum of Sides

  1. Step 1: Draw the Base and Angle — Draw the base BC of the given length. At point B, construct a ray BX making an angle equal to the given base angle (e.g., ∠XBC = 60°).
  2. Step 2: Mark the Sum of the Other Two Sides — From the ray BX, cut a line segment BD equal to the given sum of the other two sides (AB + AC). Use a compass to measure and mark this length.
  3. Step 3: Join and Find the Midpoint — Join the points D and C to form the line segment DC.
  4. Step 4: Construct the Perpendicular Bisector — Construct the perpendicular bisector of the line segment DC. To do this, place the compass at D and draw arcs above and below DC with a radius more than half of DC. Repeat with the same radius from point C, intersecting the first arcs. Join the intersection points of the arcs.
  5. Step 5: Locate the Third Vertex and Complete the Triangle — Let the perpendicular bisector intersect the line segment BD at a point A. Join A to C. The triangle ABC is the required triangle. (Justification: Since A lies on the perpendicular bisector of DC, AD = AC. Also, BD = BA + AD. Therefore, BD = BA + AC).

Tips for Acing Construction Questions

Accuracy and neatness are your best friends in this chapter. Here's how to avoid losing marks:

  • Use a Sharp Pencil and Good Instruments: A blunt pencil leads to thick lines and inaccurate measurements. Ensure your compass is tight and your ruler has clear markings.
  • Show Construction Arcs: Do not erase the arcs you use to bisect angles or lines. They are proof of your method and examiners look for them.
  • Distinguish the 'Difference' Cases: Pay close attention when given the difference of sides. If the side adjacent to the given angle is longer (e.g., AB > AC, given ∠B), you cut the difference on the main ray. If the other side is longer (AC > AB), you must extend the ray downwards and cut the difference there. This is a very common point of confusion.
  • Write the Steps of Construction: Unless the question says otherwise, always write down the steps of construction clearly after drawing the figure. This carries marks.

Practice Questions with Solutions

  • Q: Construct a triangle ABC in which BC = 7 cm, ∠B = 75° and AB + AC = 13 cm. A: Step 1: Draw BC = 7 cm. At B, construct ∠XBC = 75°. Make sure ray BX is long. Step 2: From ray BX, cut off BD = 13 cm. Join CD. Step 3: Draw the perpendicular bisector of CD. Let it intersect BD at A. Join AC. Final answer: ΔABC is the required triangle.
  • Q: Construct a triangle ABC in which BC = 8 cm, ∠B = 45° and AB - AC = 3.5 cm. A: Step 1: Draw BC = 8 cm. At B, construct ∠XBC = 45°. Make sure ray BX is long. Step 2: From ray BX, cut off BD = 3.5 cm. Join CD. Step 3: Draw the perpendicular bisector of CD. Let it intersect BX at A. Join AC. Final answer: ΔABC is the required triangle.
  • Q: Construct a triangle ABC in which BC = 6 cm, ∠B = 60° and AC - AB = 2 cm. A: Step 1: Draw BC = 6 cm. At B, construct ∠XBC = 60°. Extend ray XB downwards to form ray BY. Step 2: From ray BY (extended part), cut off BD = 2 cm. Join CD. Step 3: Draw the perpendicular bisector of CD. Let it intersect BX at A. Join AC. Final answer: ΔABC is the required triangle.
  • Q: Construct a triangle ABC in which ∠B = 60°, ∠C = 45° and AB + BC + CA = 11 cm. A: Step 1: Draw a line segment PQ = 11 cm. Step 2: At P, construct ∠XPQ = 60°/2 = 30°. At Q, construct ∠YQP = 45°/2 = 22.5°. Step 3: Let PX and QY intersect at A. Step 4: Draw the perpendicular bisector of AP, intersecting PQ at B. Draw the perpendicular bisector of AQ, intersecting PQ at C. Join AB and AC. Final answer: ΔABC is the required triangle.

Frequently Asked Questions

Why do we use a perpendicular bisector in these constructions?

The perpendicular bisector is key because any point on it is equidistant from the two endpoints of the segment it bisects. We use this property to find the third vertex 'A' such that it satisfies the condition like AD = AC, which allows us to correctly form the triangle.

What is the main difference when constructing a triangle with the sum of sides versus the difference of sides?

For the 'sum' (AB + AC), you mark the total length on the ray of the angle. For the 'difference' (AB - AC), you mark the smaller length. The crucial difference is that if the side opposite the angle is longer (AC > AB), you must extend the ray in the opposite direction to mark the difference.

Is it mandatory to write the justification for each construction in an exam?

In most CBSE exams, you are required to write the 'Steps of Construction'. Justification, which explains why the method works, is usually not required unless the question specifically asks for it. However, understanding the justification helps you remember the steps correctly.