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State Board · Class 9 · Mathematics Part 2

Lesson 4: Constructions of Triangles

5 concepts · 40 practice questions with answers · Free to read

This lesson is also in: मराठी

Basics of triangle constructions

From daily life

A tailor can cut a triangular flag only if told enough: for example the base, one corner angle and how long the other two sides are together. Triangle constructions work from such "just enough" facts.

Simple definition

To construct a triangle, three suitable conditions are needed. Here A is always found as the intersection of a ray and the perpendicular bisector (लंबदुभाजक) of a segment.

Rule 1: Known constructions and the theorem used — Earlier we constructed a triangle from three sides; from the base and two adjacent angles; from two sides and the included angle; and a right-angled triangle from the hypotenuse and one side. Perpendicular bisector theorem: every point on the perpendicular bisector of a segment is equidistant from its end points, and every point equidistant from the end points lies on it. A point on two different lines is their intersection.

Three steps: given data, mark D, A is on the perpendicular bisectorIdea behind all three constructionsGiven: base,one angle, sum ordifference or perimeterMark a point Don a ray so thatBD = that lengthA lies on theperpendicular bisectorof seg CD

In the figure : In the figure every construction uses one idea: A lies on the perpendicular bisector of the segment joining C to the marked point.

  • Easy : Easy: Any point on the perpendicular bisector of seg CD is the same distance from C and D.
  • Medium : Medium: If AC = AD, then A lies on the perpendicular bisector of seg CD.
  • Exam-type : Exam: Sides 3 cm, 4 cm, 8 cm cannot form a triangle since 3 + 4 = 7 < 8.

Rule 2: The three new constructions — (I) base, an angle adjacent to the base, and the sum of the other two sides. (II) base, an angle adjacent to the base, and the difference of the other two sides. (III) perimeter and the two angles on the base. In each, a helper point is marked so that A falls on a perpendicular bisector.

  • Easy : Easy: Sum of two sides → Construction I.
  • Medium : Medium: Difference of two sides → Construction II.
  • Exam-type : Exam: Perimeter and base angles → Construction III.
Remember

A is the meeting point of a ray and a perpendicular bisector.

Common mistake

Using sides that break the triangle rule. Check that the sum of two sides exceeds the third.

Shortcut

Memory clue: "किरण + लंबदुभाजक = A".

Spot the difference

GivenConstruction
Base, angle, sum of sidesI
Base, angle, difference of sidesII
Perimeter, base anglesIII

Practice questions

Easy
  1. 1How many conditions are needed to construct a triangle?
  2. 2What is A in these constructions?
  3. 3State the perpendicular bisector theorem.
Medium
  1. 4Which construction uses the sum of two sides?
  2. 5Which uses the perimeter?
  3. 6Can a triangle have sides 5, 6, 12?
Challenging
  1. 7A point is on two lines. What is it called?
  2. 8Why is the perpendicular bisector useful in these constructions?
Show answers (try them yourself first)
  1. 1Three.
  2. 2Intersection of a ray and a perpendicular bisector.
  3. 3Points on it are equidistant from the end points, and the converse.
  4. 4Construction I.
  5. 5Construction III.
  6. 6No, 5 + 6 = 11 < 12.
  7. 7Their intersection.
  8. 8It gives all points equal in distance from two marked points, which fixes A.

Construction I: base, adjacent angle and sum of two sides

From daily life

Suppose a rope of 10 m is tied from B to A and then A to C, while B and C are 6 m apart. A can be anywhere on a certain path. Fixing the angle at B and the rope length pins A to one spot.

Simple definition

Construction I: given BC, ∠B and AB + AC, construct ΔABC.

Rule 1: Steps for Construction I — Example: BC = 6 cm, ∠B = 45°, AB + AC = 10 cm. (1) Draw seg BC = 6 cm. (2) Draw ray BP with ∠PBC = 45°. (3) On ray BP mark D with BD = 10 cm. (4) Join D to C. (5) Draw the perpendicular bisector of seg DC; it meets ray BP at A. (6) Join A to C.

Construction of a triangle from the base, an adjacent angle and the sum of the other two sides: D is marked on the ray, the perpendicular bisector of CD meets the ray at A60°BCADBC = 5BD = 8AD = ACA is on the perpendicularbisector of CD (dashed)BC = 5, ∠B = 60°, AB + AC = 8

In the figure : In the figure construction I: BC = 6 cm, ∠B = 45°, AB + AC = 10 cm.

  • Easy : Easy: Draw BD = AB + AC along the ray from B.
  • Medium : Medium: Reason: BA + AD = BA + AC = 10, so AD = AC.
  • Exam-type : Exam: Since AD = AC, A is on the perpendicular bisector of seg DC.

Rule 2: Using the perimeter and other cases — If the perimeter and base are given, AB + AC = perimeter − base. For BC = 4 cm and perimeter 11 cm, AB + AC = 7 cm. The sum must be more than the base (7 > 4). If the given angle is ∠C, draw the ray from C instead.

Live figure

Construction I: base, adjacent angle and sum of two sides

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  • Easy : Easy: Perimeter 12, BC = 5 gives AB + AC = 7.
  • Medium : Medium: BC = 6 and AB + AC = 5 is impossible, since 5 < 6.
  • Exam-type : Exam: Construct ΔABC with BC = 5 cm, ∠C = 60°, AB + AC = 9 cm: draw the ray from C and mark CD = 9 cm.
Remember

D is on the ray with BD = AB + AC. A is on the bisector of DC.

Common mistake

Taking BD as the base. BD equals the sum of the two sides.

Shortcut

Memory clue: "बेरीज = BD, A = DC चा लंबदुभाजक".

Spot the difference

StepAction
1BC = 6 cm
2∠PBC = 45°
3BD = 10 cm on ray BP
4Bisector of DC meets ray at A

Practice questions

Easy
  1. 1Which three things are given in Construction I?
  2. 2How long is BD?
  3. 3Why does A lie on the bisector of DC?
Medium
  1. 4Perimeter 15 cm and base 6 cm. Find AB + AC.
  2. 5Can we build a triangle with BC = 8 cm and AB + AC = 7 cm?
  3. 6If ∠C is given, from which point is the ray drawn?
Challenging
  1. 7BC = 4.5 cm, perimeter 12 cm. Find AB + AC.
  2. 8Prove AD = AC in Construction I.
Show answers (try them yourself first)
  1. 1BC, ∠B and AB + AC.
  2. 2AB + AC.
  3. 3Because AD = AC.
  4. 49 cm.
  5. 5No, the sum of two sides must exceed the third.
  6. 6From C.
  7. 77.5 cm.
  8. 8BD = BA + AD = AB + AC, and BD is made equal to AB + AC, so AD = AC.

Construction II: base, adjacent angle and difference of two sides

From daily life

Two runners start from the same gate. One runs 2 km farther than the other. The difference decides where they stand relative to a mark. In triangles, the difference of two sides can also fix a corner.

Simple definition

Construction II: given BC, ∠B and AB − AC (or AC − AB), construct ΔABC.

Rule 1: When AB > AC — Example: BC = 5 cm, ∠B = 50°, AB − AC = 2 cm. Draw seg BC. Draw ray BL with ∠LBC = 50°. On ray BL mark D with BD = 2 cm. Join D to C. Draw the perpendicular bisector of seg DC; it meets ray BL at A. Reason: B − D − A and BD = AB − AD = 2, so AD = AC.

Construction from the base, an adjacent angle and the difference of two sides: BD equals the difference and A lies on the perpendicular bisector of CD50°BCADBC = 6BD = 2AD = ACD on BA, BD = AB − ACBC = 6, ∠B = 50°, AB − AC = 2

In the figure : In the figure construction II: BC = 5 cm, ∠B = 50°, AB − AC = 2 cm.

  • Easy : Easy: BD = the difference, here 2 cm.
  • Medium : Medium: D lies on ray BL, between B and A.
  • Exam-type : Exam: The difference 2 cm must be less than BC = 5 cm.

Rule 2: When AC > AB — Example: BC = 6 cm, ∠B = 40°, AC − AB = 2 cm. Take D on the ray opposite to ray BL with BD = 2 cm. Then AD = AB + BD = AB + 2 = AC, so again A is on the perpendicular bisector of seg DC.

Live figure

Construction II: base, adjacent angle and difference of two sides

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  • Easy : Easy: AB − AC = 2 means AB is longer: D on ray BL.
  • Medium : Medium: AC − AB = 2 means AC is longer: D on the opposite ray.
  • Exam-type : Exam: In both cases BD = the given difference, and A is on the bisector of seg DC.
Remember

BD = difference. Longer side is AB: D on the ray. Longer is AC: D on the opposite ray.

Common mistake

Placing D on the wrong side. Check which of AB, AC is longer.

Shortcut

Memory clue: "AB मोठी = किरणावर; AC मोठी = विरुद्ध किरणावर".

Spot the difference

CasePosition of D
AB > ACOn ray BL
AC > ABOn the opposite ray

Practice questions

Easy
  1. 1What is given in Construction II?
  2. 2How long is BD?
  3. 3If AB > AC, where is D?
Medium
  1. 4If AC > AB, where is D?
  2. 5Can the difference of two sides be 9 cm when BC = 7 cm?
  3. 6Why is AD = AC when AB > AC?
Challenging
  1. 7Why is AD = AC when AC > AB?
  2. 8Name the point whose position the perpendicular bisector of DC decides.
Show answers (try them yourself first)
  1. 1BC, ∠B and the difference of two sides.
  2. 2The given difference.
  3. 3On ray BL.
  4. 4On the opposite ray.
  5. 5No, it must be less than BC.
  6. 6AD = AB − BD = AB − (AB − AC) = AC.
  7. 7AD = AB + BD = AB + (AC − AB) = AC.
  8. 8A.

Construction III: perimeter and angles adjacent to the base

From daily life

A farmer fences a triangular plot using a 10 m rope round it. He knows only the two angles along the base. Stretching the rope along the base line, one end beyond each corner, helps find the shape.

Simple definition

Construction III: given the perimeter and the angles ∠B, ∠C on the base, construct ΔABC.

Rule 1: Idea and angles — Example: perimeter 10 cm, ∠B = 80°, ∠C = 40°. On line BC take P and Q with PB = AB and CQ = AC. Then PQ = AB + BC + CA = 10 cm. In ΔPBA, PB = BA, so ∠APB = ∠PAB. Their sum is the exterior angle ∠ABC = 80° (remote interior angles), so each is 40°. Similarly ∠CQA = ∠CAQ = 20°.

Construction from the perimeter and the base angles: P and Q are taken on line BC with PB = AB and CQ = AC, so PQ is the perimeterPBCQAPQ = AB + BC + CA = perimeter (10 cm)PB = BA and CQ = CA, so PQ equals the perimeter

In the figure : In the figure construction III: P and Q are marked on line BC so that PQ equals the perimeter.

  • Easy : Easy: The angle at P is half of ∠B: ∠B = 60° gives 30°.
  • Medium : Medium: ∠C = 50° gives 25° at Q.
  • Exam-type : Exam: ∠B = 80°, ∠C = 40° give 40° at P and 20° at Q, so ∠PAQ = 180° − 60° = 120°.

Rule 2: Steps — (1) Draw seg PQ = 10 cm. (2) Draw 40° at P and 20° at Q; the rays meet at A. (3) Draw the perpendicular bisectors of seg AP and seg AQ; they meet PQ at B and C. (4) Join A to B and A to C. ΔABC has perimeter 10 cm and angles 80° at B and 40° at C.

  • Easy : Easy: B lies on the bisector of AP, so BA = BP.
  • Medium : Medium: C lies on the bisector of AQ, so CA = CQ.
  • Exam-type : Exam: Then PQ = PB + BC + CQ = AB + BC + CA, the perimeter.
Remember

Angles at P and Q are half of ∠B and ∠C. Bisectors of AP, AQ give B and C.

Common mistake

Drawing the full angles ∠B and ∠C at P and Q. Use half of each.

Shortcut

Memory clue: "P, Q वर कोन निम्मे".

Spot the difference

GivenAt PAt Q
∠B = 80°, ∠C = 40°40°20°
∠B = 60°, ∠C = 50°30°25°
∠B = 90°, ∠C = 70°45°35°

Practice questions

Easy
  1. 1What is given in Construction III?
  2. 2What does PQ equal?
  3. 3∠B = 70°. What angle is drawn at P?
Medium
  1. 4∠C = 100°. What angle is drawn at Q?
  2. 5How are B and C found on PQ?
  3. 6Why is ∠APB = ∠PAB?
Challenging
  1. 7For ∠B = 60°, ∠C = 40°, find ∠PAQ.
  2. 8Which theorem shows ∠APB + ∠PAB = ∠ABC?
Show answers (try them yourself first)
  1. 1Perimeter and the two base angles.
  2. 2The perimeter.
  3. 335°.
  4. 450°.
  5. 5By the perpendicular bisectors of AP and AQ.
  6. 6In ΔPBA, PB = BA.
  7. 7180° − (30° + 20°) = 130°.
  8. 8Theorem of remote interior angles.

Choosing and applying the constructions

From daily life

A craftsperson first reads the order: "perimeter with ratio", "sum of sides", "difference of sides". Choosing the right method is half the work.

Simple definition

Choose the method from the given data. Always check that the sum of any two sides is greater than the third.

Rule 1: Choosing the construction — Sum of two sides → I. Difference of two sides → II. Perimeter with base angles → III. If the perimeter and the ratio of sides are given, find the sides first. Perimeter 18 cm in the ratio 2 : 3 : 4: one part = 18 ÷ 9 = 2 cm, so the sides are 4, 6, 8 cm, and we construct from three sides.

  • Easy : Easy: Perimeter 20 cm, ratio 1 : 2 : 2 gives 4, 8, 8 cm.
  • Medium : Medium: Perimeter 24 cm, ratio 3 : 4 : 5 gives 6, 8, 10 cm.
  • Exam-type : Exam: Check 4 + 6 = 10 > 8, so the triangle with sides 4, 6, 8 exists.

Rule 2: Applying the constructions — For the perimeter construction use half of each base angle: ∠M = 70°, ∠N = 50° give 35° and 25°. If XY + XZ = 9.5 cm and YZ = 5 cm, the perimeter is 14.5 cm. A triangle with BC = 6 cm and AB + AC = 5 cm is impossible. You can also try these constructions in Geogebra.

  • Easy : Easy: ∠M = 80° gives 40°.
  • Medium : Medium: XY + XZ = 8, YZ = 4: perimeter 12.
  • Exam-type : Exam: Is there a triangle with BC = 7, AB − AC = 8? No, 8 > 7.
Remember

Sum → I. Difference → II. Perimeter → III. Check the triangle rule.

Common mistake

Drawing full angles in construction III. Draw half of each.

Shortcut

Memory clue: "बेरीज I, फरक II, परिमिती III".

Spot the difference

DataMethod
Base, angle, AB + ACI
Base, angle, AB − ACII
Perimeter, base anglesIII
Perimeter, ratioFind sides first

Practice questions

Easy
  1. 1Which construction for a sum of sides?
  2. 2Which construction for a difference of sides?
  3. 3Which for perimeter and base angles?
Medium
  1. 4Perimeter 15 cm, ratio 1 : 2 : 2. Find the sides.
  2. 5∠M = 50°, ∠N = 90°. Angles to draw?
  3. 6XY + XZ = 11 cm, YZ = 4.2 cm. Perimeter?
Challenging
  1. 7Is AB + AC = 4 cm with BC = 5 cm possible?
  2. 8Perimeter 24 cm, ratio 2 : 3 : 7. Does the triangle exist?
Show answers (try them yourself first)
  1. 1I.
  2. 2II.
  3. 3III.
  4. 43, 6, 6 cm.
  5. 525° and 45°.
  6. 615.2 cm.
  7. 7No, the sum of two sides must exceed the third.
  8. 8Sides 4, 6, 14; 4 + 6 = 10 < 14, so no.

Quick revision

  • In these constructions A is the intersection of a ray and the perpendicular bisector of a segment.
  • Sum of two sides: mark D with BD = AB + AC; difference of two sides: BD = the difference, on the opposite ray if AC > AB.
  • Perimeter construction: PQ = perimeter, with angles at P and Q equal to half of the base angles.