Exercise 13.2 — Triangle Constructions
Construction of triangles in special cases.
About Exercise 13.2
Exercise 13.2 is from Chapter 13 — Geometrical Constructions of the Class 9 Mathematics textbook, prescribed by the CBSE, Telangana, and Andhra Pradesh boards. This exercise covers two major construction types that appear regularly in board examinations:
- Triangle constructions when the perimeter condition involves the sum or difference of two sides — situations where three sides are not directly given.
- Circle segment constructions — constructing a circular arc over a chord such that the inscribed angle equals a given value (90°, 45°, or 120°).
The key tool used in almost every question here is the perpendicular bisector. Understanding why the perpendicular bisector is used — not just how to draw it — is what separates a student who scores full marks from one who gets partial credit.
Question 1 — Construct △ABC: BC = 7 cm, ∠B = 75°, AB + AC = 12 cm
Why this method works
We mark point D on ray BX such that BD = AB + AC = 12 cm. Then D is 12 cm from B along BX. After drawing the perpendicular bisector of CD, any point on that bisector is equidistant from C and D. The intersection with BD gives point A, where AC = AD (because A is on the perp. bisector of CD). So AB + AC = AB + AD = BD = 12 cm. ✓
Steps of Construction
- Draw line segment BC of length 7 cm.
- At B, draw ray BX such that ∠XBC = 75°.
- With B as centre and radius 12 cm, draw an arc on BX. Mark the intersection as D. (So BD = AB + AC = 12 cm.)
- Join C and D.
- Draw the perpendicular bisector of CD. (Use compass arcs from C and D to find the bisector.)
- Mark the point where the perpendicular bisector intersects BD as point A.
- Join A and C.
- △ABC is the required triangle. ∠B = 75°, BC = 7 cm, and AB + AC = AB + AD = BD = 12 cm. ✓
Question 2 — Construct △PQR: QR = 8 cm, ∠Q = 60°, PQ − PR = 3.5 cm
Why this method works (difference case)
This time the difference PQ − PR = 3.5 cm is given. We mark S on QX so that QS = PQ − PR = 3.5 cm. The perpendicular bisector of RS meets QX at P. Since P is on the perp. bisector of RS, we have PR = PS. So PQ − PR = PQ − PS = QS = 3.5 cm. ✓
Important angle note: The ray QX is drawn at half of ∠Q = 30° (not 60°) on the opposite side of QR because the difference case requires S to be below the triangle.
Steps of Construction
- Draw line segment QR of length 8 cm.
- At Q, draw ray QX such that ∠XQR = 30° (which is half of ∠Q = 60°, drawn below QR towards the exterior).
- With Q as centre and radius 3.5 cm, draw an arc on QX. Mark the intersection as S. (QS = PQ − PR = 3.5 cm.)
- Join R and S.
- Draw the perpendicular bisector of RS.
- Mark the point where the perpendicular bisector meets QX as point P.
- Join P and R.
- △PQR is formed. PQ − PR = PQ − PS = QS = 3.5 cm, and ∠Q = 60°, QR = 8 cm. ✓
Question 3 — Construct △XYZ: ∠Y = 30°, ∠Z = 60°, XY + YZ + ZX = 10 cm
Why this method works
Draw AB = 10 cm (the full perimeter). Draw ray AC at ½∠Y = 15° from A, and ray BD at ½∠Z = 30° from B. These meet at X. The perpendicular bisectors of AX and BX cut AB at Y and Z respectively. Since Y is on the perp. bisector of AX, XY = AY. Similarly XZ = BZ. So XY + YZ + ZX = AY + YZ + BZ = AB = 10 cm. ✓
Steps of Construction
- Draw line segment AB = 10 cm (equal to the full perimeter XY + YZ + ZX).
- At A, draw ray AC such that ∠CAB = ½ × ∠Y = ½ × 30° = 15°.
- At B, draw ray BD such that ∠DBA = ½ × ∠Z = ½ × 60° = 30°.
- Mark the intersection of rays AC and BD as point X.
- Draw the perpendicular bisector of AX. Mark where it crosses AB as point Y.
- Draw the perpendicular bisector of BX. Mark where it crosses AB as point Z.
- Join X, Y and X, Z.
- Verify that ∠Y = 30° and ∠Z = 60°.
- △XYZ is formed with the given measurements.
Question 4 — Right Triangle: Base = 7.5 cm, Hypotenuse + Other Side = 15 cm
Why this method works
This is essentially the same as Q1, but with the angle at B fixed at 90° instead of 75°. D is placed on the perpendicular ray BX at distance BD = AB + AC = 15 cm. The perpendicular bisector of CD meets BD at A, where AC = AD. So AB + AC = AB + AD = BD = 15 cm. The right angle is guaranteed because BX ⊥ BC.
Steps of Construction
- Draw ray BC = 7.5 cm.
- At B, draw ray BX perpendicular to BC (∠XBC = 90°).
- With B as centre and radius 15 cm, draw an arc on BX. Mark intersection as D. (BD = AB + AC = 15 cm.)
- Join C and D.
- Draw the perpendicular bisector of CD.
- Mark the point where the perp. bisector crosses BD as A.
- Join A and C.
- △ABC is formed: ∠B = 90°, BC = 7.5 cm, AB + AC = 15 cm. ✓
Question 5 — Construct Circle Segments on a 5 cm Chord Containing Given Angles
This question asks us to construct a circular arc over a chord AB = 5 cm such that any point P on the arc produces an inscribed angle ∠APB equal to a specified value. This uses the Inscribed Angle Theorem: the inscribed angle is half the central angle subtended by the same arc.
Why this is special: By the Thales theorem, the angle in a semicircle = 90°. So the required arc is a semicircle with AB as its diameter!
Why this works
When ∠APB = 90°, the arc APB is a semicircle. The centre O is the midpoint of AB (= 2.5 cm). The radius OA = OB = 2.5 cm. Any point P on the semicircle above AB will give ∠APB = 90° by Thales' theorem (angle in a semicircle).
Steps of Construction
- Draw line segment AB = 5 cm.
- Draw the perpendicular bisector of AB. It meets AB at midpoint O.
- With O as centre and OB as radius (= 2.5 cm), draw a circle (or the upper semicircle).
- Plot any point P on the upper semicircle.
- Join PA and PB.
- Observe that ∠APB = 90° (Thales' theorem — angle in a semicircle).
- The upper semicircular segment APB is the required segment containing a 90° angle.
Finding O: Draw rays AX and BY each at 45° to AB from A and B respectively. They meet at O (the centre).
Why this works
Since the inscribed angle ∠APB = 45°, the central angle ∠AOB = 2 × 45° = 90°. This means O forms a right isosceles triangle OAB, with ∠OAB = ∠OBA = 45°. So we find O by drawing 45° rays from A and B — their intersection is the centre. P must be taken on the major arc (the larger arc, above AB).
Steps of Construction
- Draw line segment AB = 5 cm.
- Draw ray AX such that ∠XAB = 45°, and ray BY such that ∠YBA = 45° (on the same side of AB). They intersect at O.
- With O as centre and OB as radius, draw a circle.
- Plot any point P on the major arc (the larger arc on the same side as O).
- Join PA and PB.
- Observe that ∠APB = 45°.
- The major arc APB is the required segment containing a 45° angle.
Finding the central angle: The inscribed angle from the major arc = 180° − 120° = 60°. So central ∠AOB = 2 × 60° = 120°... but since ∠APB is from the minor arc, the rays from A and B are drawn at 180° − 120° = 60°, giving each base angle = 30°.
Why P is on the minor arc
When the required angle is obtuse (> 90°), the point P must lie on the minor arc. The corresponding inscribed angle from the major arc would be 180° − 120° = 60°, giving central angle 2 × 60° = 120°. We construct by drawing 30° rays at A and B (so ∠OAB = ∠OBA = 30°) to find O. P is then placed on the minor arc.
Steps of Construction
- Draw line segment AB = 5 cm.
- Draw ray AX such that ∠XAB = 30°, and ray BY such that ∠YBA = 30° (on the same side). They intersect at O.
- With O as centre and OB as radius, draw a circle.
- Plot any point P on the minor arc (the smaller arc on the opposite side from O).
- Join PA and PB.
- Observe that ∠APB = 120°.
- The minor arc APB is the required segment containing a 120° angle.
Quick Reference — All 5 Questions Summarised
| Q | Given | Key Tool | Core Principle |
|---|---|---|---|
| 1 | BC=7 cm, ∠B=75°, AB+AC=12 cm | Perp. bisector of CD | A on perp. bisector → AC = AD → AB+AC = BD = 12 cm |
| 2 | QR=8 cm, ∠Q=60°, PQ−PR=3.5 cm | Perp. bisector of RS | P on perp. bisector → PR = PS → PQ−PR = QS = 3.5 cm |
| 3 | ∠Y=30°, ∠Z=60°, perimeter=10 cm | Two perp. bisectors of AX and BX | Rays at ½∠Y and ½∠Z find X; perp. bisectors find Y and Z on AB |
| 4 | ∠B=90°, BC=7.5 cm, AB+AC=15 cm | Perp. bisector of CD | Same as Q1 with ∠B = 90° (Thales' theorem also applies) |
| 5(i) | Chord = 5 cm, angle = 90° | Semicircle (midpoint of AB) | Angle in semicircle = 90° (Thales). O is midpoint of AB. |
| 5(ii) | Chord = 5 cm, angle = 45° | Rays at 45° from A and B | Central angle = 90°; P on major arc; inscribed = 45° |
| 5(iii) | Chord = 5 cm, angle = 120° | Rays at 30° from A and B | P on minor arc; base angles 30° → inscribed angle = 120° |
Common Mistakes to Avoid in Exercise 13.2
- Q1 vs Q2 — wrong side for auxiliary point: In Q1 (sum case), D is marked on the ray BX at distance = sum. In Q2 (difference case), S is marked at distance = difference on the ray below QR. Mixing these up gives a completely wrong triangle.
- Q2 — wrong angle for the ray: The ray QX should be at half of ∠Q = 30°, not at 60°. Many students use the full angle instead of half.
- Q3 — forgetting to halve the angles: The rays at A and B must be at half the respective angles (½∠Y and ½∠Z). Drawing them at the full angles gives a wrong intersection point.
- Q5 — choosing wrong arc (minor vs major): For angles less than 90° (like 45°), P goes on the major arc. For angles greater than 90° (like 120°), P goes on the minor arc. Getting this wrong reverses the answer.
- Q5 — wrong base angle calculation for centre: For angle θ on the major arc, each base angle = θ. For angle θ on the minor arc, each base angle = 180° − θ, divided by 2. Always derive the base angle from the required inscribed angle before starting.
What Exercise 13.2 Prepares You For
The perpendicular bisector technique from Q1–Q4 appears again whenever triangles are constructed in later classes given incomplete information (perimeter, partial sides, etc.). The angle-bisector approach in Q3 connects to the Triangles chapter, particularly the angle sum property and the incircle.
The circle segment constructions in Q5 build directly on the Inscribed Angle Theorem from Chapter 12 — Circles. If you found Q5 conceptually challenging, reviewing the angle subtended by a chord at the centre (Exercise 12.2) and the inscribed angle theorem will help significantly.