Class 9 Mathematics Notes Chapter 11 (Constructions) – Mathematics Book

Alright class, let's focus on Chapter 11: Constructions. This is a practical geometry chapter, but understanding the 'why' behind each step is crucial, especially for competitive exams where concepts are tested. We'll be using only an unmarked ruler and compasses for these constructions.

Chapter 11: Constructions - Detailed Notes

Core Idea: To construct geometric figures like angle bisectors, perpendicular bisectors, specific angles, and triangles with given conditions using only a straight edge (unmarked ruler) and compasses. Justification (proving why the construction works) is based on geometric principles, mainly congruence of triangles.

1. Basic Constructions

a) Construction 11.1: To construct the bisector of a given angle.
* Given: An angle, say ∠ABC.
* Steps:
1. With B as the centre and any convenient radius, draw an arc intersecting rays BA and BC at points E and D respectively.
2. With D as the centre and a radius greater than half of DE, draw an arc.
3. With E as the centre and the same radius as in step 2, draw another arc intersecting the previous arc at point F.
4. Draw the ray BF. This ray BF is the required bisector of ∠ABC.
* Justification: Join DF and EF. In ΔBEF and ΔBDF:
* BE = BD (Arcs of the same radius)
* EF = DF (Arcs of the same radius)
* BF = BF (Common side)
* Therefore, ΔBEF ≅ ΔBDF (SSS congruence).
* Hence, ∠EBF = ∠DBF (CPCT - Corresponding Parts of Congruent Triangles). So, BF bisects ∠ABC.

b) Construction 11.2: To construct the perpendicular bisector of a given line segment.
* Given: A line segment AB.
* Steps:
1. With A as the centre and a radius greater than half of AB, draw arcs on both sides of AB.
2. With B as the centre and the same radius as in step 1, draw arcs intersecting the previous arcs at points P and Q respectively.
3. Join PQ. Let PQ intersect AB at point M.
4. The line PMQ is the required perpendicular bisector of AB. (M is the midpoint, and ∠PMA = 90°).
* Justification: Join AP, AQ, BP, BQ.
* In ΔPAQ and ΔPBQ: AP = BP (Arcs of equal radius), AQ = BQ (Arcs of equal radius), PQ = PQ (Common). So, ΔPAQ ≅ ΔPBQ (SSS). Thus, ∠APM = ∠BPM (CPCT).
* Now in ΔAPM and ΔBPM: AP = BP (Arcs of equal radius), ∠APM = ∠BPM (Proved above), PM = PM (Common). So, ΔAPM ≅ ΔBPM (SAS). Thus, AM = BM (CPCT) and ∠AMP = ∠BMP (CPCT).
* Since ∠AMP + ∠BMP = 180° (Linear Pair) and ∠AMP = ∠BMP, we have 2∠AMP = 180°, so ∠AMP = 90°.
* Therefore, PMQ is the perpendicular bisector of AB.

c) Construction 11.3: To construct an angle of 60° at the initial point of a given ray.
* Given: A ray AB with initial point A.
* Steps:
1. With A as the centre and any convenient radius, draw an arc intersecting AB at point D.
2. With D as the centre and the same radius as in step 1, draw an arc intersecting the first arc at point E.
3. Draw the ray AC passing through E.
4. ∠CAB is the required angle of 60°.
* Justification: Join DE. By construction, AD = AE = DE (Radii of same arcs). Therefore, ΔADE is an equilateral triangle. Hence, ∠EAD = 60°.

d) Constructing Other Standard Angles (Using Ruler and Compass):
* 90°: Construct 60°. Then construct another 60° adjacent to it (making 120°). Bisect the second 60° angle (between 60° and 120° mark). 60° + 30° = 90°. Alternatively, construct the perpendicular bisector to a line at a point on it.
* 45°: Construct 90° and then bisect it. (90°/2 = 45°).
* 30°: Construct 60° and then bisect it. (60°/2 = 30°).
* 22.5° (or 22 ½°): Construct 45° and then bisect it. (45°/2 = 22.5°).
* 75°: Construct 60° and 90°. Bisect the angle between the 60° and 90° rays. (60° + (90°-60°)/2 = 60° + 15° = 75°).
* 105°: Construct 90° and 120°. Bisect the angle between the 90° and 120° rays. (90° + (120°-90°)/2 = 90° + 15° = 105°).
* 135°: Construct 90° and 180° (straight line). Bisect the angle between them (90° + (180°-90°)/2 = 90° + 45° = 135°). Or, construct 90° and 45° adjacent to it on a straight line.

2. Construction of Triangles

a) Construction 11.4: To construct a triangle given its base, a base angle and the sum of the other two sides.
* Given: Base BC, base angle ∠B, and sum AB + AC = L (a given length).
* Steps:
1. Draw the base BC.
2. At point B, construct ∠XBC equal to the given angle ∠B.
3. From the ray BX, cut off a line segment BD equal to the given sum L (AB + AC).
4. Join DC.
5. Construct the perpendicular bisector of DC. Let it intersect BD at point A.
6. Join AC.
7. ΔABC is the required triangle.
* Justification: Point A lies on the perpendicular bisector of DC. Therefore, AD = AC. Now, BD = BA + AD (Point A lies on BD). Since AD = AC, we have BD = BA + AC. Also, ∠B is constructed as required, and BC is the given base.

b) Construction 11.5: To construct a triangle given its base, a base angle and the difference of the other two sides.
* Given: Base BC, base angle ∠B, and difference AB - AC = D (Case 1: AB > AC) OR AC - AB = D (Case 2: AC > AB).
* Steps (Case 1: AB > AC):
1. Draw the base BC.
2. At point B, construct ∠XBC equal to the given angle ∠B.
3. From the ray BX, cut off a line segment BD equal to the given difference D (AB - AC).
4. Join DC.
5. Construct the perpendicular bisector of DC. Let it intersect the ray BX at point A.
6. Join AC.
7. ΔABC is the required triangle.
* Steps (Case 2: AC > AB):
1. Draw the base BC.
2. At point B, construct ∠XBC equal to the given angle ∠B. Extend XB downwards to form line XBX'.
3. From the ray BX' (opposite to BX), cut off a line segment BD equal to the given difference D (AC - AB).
4. Join DC.
5. Construct the perpendicular bisector of DC. Let it intersect the ray BX at point A.
6. Join AC.
7. ΔABC is the required triangle.
* Justification (Both cases): Point A lies on the perpendicular bisector of DC. Therefore, AD = AC.
* Case 1: BD = AB - AD (Since A lies on BX beyond D). As AD = AC, BD = AB - AC.
* Case 2: BD = AD - AB (Since D lies on the extension of BX). As AD = AC, BD = AC - AB.
* In both cases, ∠B and base BC are as per construction.

c) Construction 11.6: To construct a triangle given its perimeter and its two base angles.
* Given: Perimeter P = AB + BC + CA, and two base angles ∠B and ∠C.
* Steps:
1. Draw a line segment XY equal to the perimeter P.
2. At point X, construct an angle ∠LXY equal to the given ∠B.
3. At point Y, construct an angle ∠MYX equal to the given ∠C.
4. Bisect ∠LXY and ∠MYX. Let these bisectors intersect at a point A.
5. Construct the perpendicular bisector of AX. Let it intersect XY at point B.
6. Construct the perpendicular bisector of AY. Let it intersect XY at point C.
7. Join AB and AC.
8. ΔABC is the required triangle.
* Justification:
* B lies on the perpendicular bisector of AX. So, AB = BX. Therefore, ∠BAX = ∠AXB. Since ∠AXB = ∠LXY / 2 = ∠B / 2, we have ∠BAX = ∠B / 2. Now, ∠ABC is an exterior angle for ΔABX. So, ∠ABC = ∠BAX + ∠AXB = ∠B/2 + ∠B/2 = ∠B.
* Similarly, C lies on the perpendicular bisector of AY. So, AC = CY. Therefore, ∠CAY = ∠AYC. Since ∠AYC = ∠MYX / 2 = ∠C / 2, we have ∠CAY = ∠C / 2. Now, ∠ACB is an exterior angle for ΔACY. So, ∠ACB = ∠CAY + ∠AYC = ∠C/2 + ∠C/2 = ∠C.
* Also, XY = XB + BC + CY. Since XB = AB and CY = AC, we have XY = AB + BC + AC = Perimeter P.

Key Points for Exams:

• Tools: Only unmarked ruler and compasses are allowed.
• Neatness: Draw thin lines. Clearly show all construction arcs; do not erase them.
• Labeling: Label all points, lines, and angles correctly.
• Steps: While detailed written steps might not always be required in MCQ exams, knowing the sequence is vital. For descriptive exams, write clear, concise steps.
• Justification: Understand the underlying geometric proof (usually congruence) for each construction. This helps tackle variations or complex problems.
• Accuracy: Use a sharp pencil and accurate compass settings.

Multiple Choice Questions (MCQs)

1. To construct the perpendicular bisector of a line segment AB, we take a radius:
   a) Equal to AB
   b) Equal to half of AB
   c) Greater than half of AB
   d) Less than half of AB
   Answer: c) Greater than half of AB

2. To construct an angle of 60°, the key step involves drawing an arc from a point on the initial arc with:
   a) A radius different from the initial arc's radius
   b) A radius equal to the initial arc's radius
   c) A radius equal to the length of the ray
   d) A radius equal to half the initial arc's radius
   Answer: b) A radius equal to the initial arc's radius

3. Which of the following angles cannot be constructed using only a ruler and compasses?
   a) 22.5°
   b) 37.5°
   c) 40°
   d) 75°
   Answer: c) 40° (Since 40 is not obtained by starting with 60° and repeatedly bisecting or combining constructible angles like 60°, 90°, etc.)

4. In the construction of a triangle ABC, given base BC, ∠B, and the sum AB + AC, the first step after drawing BC and ∠B is to:
   a) Cut off length AB + AC on the ray forming ∠B.
   b) Bisect ∠B.
   c) Draw the perpendicular bisector of BC.
   d) Cut off length AB - AC on the ray forming ∠B.
   Answer: a) Cut off length AB + AC on the ray forming ∠B.

5. In the construction of ΔABC where BC = 5 cm, ∠B = 60°, and AB + AC = 7.5 cm, we draw BD = 7.5 cm on ray BX (where ∠XBC = 60°). What is the next major step?
   a) Join AC
   b) Draw the perpendicular bisector of BC
   c) Join DC and draw its perpendicular bisector
   d) Bisect ∠BDC
   Answer: c) Join DC and draw its perpendicular bisector

6. The justification for the construction of an angle bisector relies on which congruence criterion?
   a) SAS
   b) ASA
   c) SSS
   d) RHS
   Answer: c) SSS

7. To construct a ΔABC given perimeter P and base angles ∠B and ∠C, we first draw a line segment XY = P. Then we construct angles at X and Y equal to:
   a) ∠B and ∠C respectively
   b) ∠B/2 and ∠C/2 respectively
   c) 180°-∠B and 180°-∠C respectively
   d) 2∠B and 2∠C respectively
   Answer: a) ∠B and ∠C respectively (Although we bisect these later, the initial construction is of the angles themselves). Correction: The standard construction involves constructing ∠B and ∠C at X and Y, then bisecting them. Let's rephrase the question or check the standard method again. Ah, the standard method (11.6) asks to construct ∠B at X and ∠C at Y, then bisect them. Let's refine the MCQ or stick to the book's phrasing. Let's re-evaluate the question based on the steps in 11.6. Step 2 & 3: Construct ∠LXY = ∠B and ∠MYX = ∠C. Step 4: Bisect these angles. So the initial angles constructed are indeed ∠B and ∠C. Let's re-evaluate the options. Option (a) seems correct based on the initial step. Let's check the justification again. The justification relies on the bisected angles. Hmm, maybe the question should be about the angles whose bisectors meet at A. Let's rephrase the question for clarity.

   Revised MCQ 7: To construct a ΔABC given perimeter P and base angles ∠B and ∠C, we draw XY = P. Let the bisectors of angles constructed at X and Y meet at A. What should these angles at X and Y be initially constructed as?
   a) ∠B and ∠C
   b) ∠B/2 and ∠C/2
   c) 90° and 90°
   d) 60° and 60°
   Answer: a) ∠B and ∠C (As per steps 2 and 3 of Construction 11.6)

8. In constructing a triangle with base BC, ∠B, and difference AB - AC (where AB > AC), point A is found by intersecting the ray BX (where ∠XBC = ∠B) with:
   a) The perpendicular bisector of BC
   b) The bisector of ∠B
   c) The perpendicular bisector of DC (where D is on BX such that BD = AB - AC)
   d) A line parallel to BC
   Answer: c) The perpendicular bisector of DC (where D is on BX such that BD = AB - AC)

9. Construction of a triangle is NOT possible if:
   a) Base, base angle, and sum of other two sides are given.
   b) Base, base angle, and difference of other two sides are given.
   c) Perimeter and two base angles are given.
   d) The sum of two sides is less than or equal to the third side.
   Answer: d) The sum of two sides is less than or equal to the third side. (This violates the triangle inequality theorem, making the construction impossible regardless of the method).

10. When constructing an angle of 90°, we typically bisect the angle between:
    a) 0° and 60°
    b) 60° and 120°
    c) 30° and 60°
    d) 0° and 120°
    Answer: b) 60° and 120° (Construct 60°, then extend to 120°, the bisector gives 90°).

Remember to practice these constructions physically using a ruler and compass to solidify your understanding. Good luck with your preparation!