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NCERT Solutions for Class 9 Maths Chapter 11 Constructions

NCERT Solutions for Class 9 Maths Chapter 11 Constructions covers the basic geometric constructions using a ruler and compass, including bisecting angles, constructing specific angles, and constructing triangles under given conditions.

Overview of Topics & Subtopics (Click to Jump)

1. Basic Constructions

Short Note:

• To bisect a given angle: draw arcs from the vertex cutting both arms, then arcs from those points to intersect; join the vertex to the intersection point.

• To construct the perpendicular bisector of a line segment: draw arcs of equal radius (more than half the segment) from both endpoints on either side; join the intersection points.

• To construct an angle of 60°: draw an arc from a point on a ray, then with the same radius draw an arc from that intersection; join the vertex to the new point.

• Angles of 30°, 90°, 45°, and 15° can be constructed by bisecting 60°, combining constructions, or bisecting a right angle.

2. Construction of Triangles

Short Note:

• Given base, a base angle, and sum of the other two sides: extend the base beyond the base angle by the given sum, join, then draw the perpendicular bisector to locate the third vertex.

• Given base, a base angle, and difference of the other two sides: similar method, marking off the difference along or beyond the base line and using a perpendicular bisector.

• Given perimeter and the two base angles: mark the full perimeter on a line, construct half the base angles at the ends, then use perpendicular bisectors to find the third vertex.

NCERT Textbook Exercise Questions

Construct an angle of 90° at the initial point of a given ray and justify the construction.
Answer: First construct a 60° angle, then bisect the adjacent 120° angle formed by extending the construction, or alternatively construct 60° and 30° adjacent to it, summing to 90°. Justification follows from the equilateral-triangle construction ensuring each angle is exactly 60°, and bisection giving exact halves.
Construct a triangle ABC in which BC = 7 cm, ∠B = 75° and AB + AC = 13 cm.
Answer: Draw BC = 7 cm. At B construct ∠B = 75°. On this ray mark D such that BD = AB + AC = 13 cm; join DC. Draw the perpendicular bisector of DC meeting BD at A. Join AC to complete ΔABC.
Construct a triangle ABC in which BC = 8 cm, ∠B = 45° and AB − AC = 3.5 cm.
Answer: Draw BC = 8 cm and ∠B = 45° at B. Since AB > AC, mark D on the ray BA such that BD = AB − AC = 3.5 cm; join DC. Draw the perpendicular bisector of DC; it meets the ray at A. Join AC to complete the triangle.
Construct a triangle XYZ with perimeter 11 cm and base angles 60° and 45°.
Answer: Draw a line segment PQ = 11 cm (the perimeter). At P construct 60° and at Q construct 45°, then bisect both these angles. The bisectors meet at X. Draw perpendicular bisectors of PX and QX meeting PQ at Y and Z respectively; join XY and XZ to complete ΔXYZ.
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More for this chapter

Class 9 › Mathematics › Constructions
Chapter notesRead the chapterPractice MCQsCheck what you rememberChapter test10 questions on one pageOnline testTest 1, 2 and 3, one question at a timePYQPrevious year questionsStudy materialPDF notes
Topics in this chapter: Basic ConstructionsConstruction of Triangles