Chapter 13.2 — Exercise 13.2 — Triangle Constructions
Construction of triangles in special cases. This is Lesson 2 of 2 in Chapter 13: Geometrical Constructions.
A Triangle Defined by a Combination, Not a Single Side
None of these five problems hand over three plain, directly-drawable lengths. Each one gives a base, one angle, and either a sum or a difference of the other two sides — a combination no compass can draw directly, needing one extra construction step before the triangle itself can even begin — a genuinely new technique this exercise introduces and then reuses, with small variations, across every problem that follows.
The Core Trick: Turn a Sum Into One Drawable Segment
Given: base BC, ∠B, and AB + AC = a known total length
Draw ray BX at ∠B; mark D on BX so BD = AB + AC
Join CD; draw the perpendicular bisector of CD
Where it meets BD is exactly AThe perpendicular bisector is the key move: any point on it is equidistant from C and D, so the point A where it crosses BD satisfies AD = AC automatically. Since BD = AB + AC was drawn as one single length, and AD = AC, the remaining piece BA = BD − AD = BD − AC = AB exactly — the full sum splits itself back into the two correct original sides, without either one ever being drawn directly. Nothing about this trick is special to triangles specifically — it's really just the same "any point on a perpendicular bisector is equidistant from both endpoints" fact from Chapter 12, applied here as a construction tool rather than as a proof step inside a circle theorem.
Building a Triangle From a Side Sum
Construct △ABC with BC = 7 cm, ∠B = 75°, AB + AC = 12 cm. Draw BC = 7 cm first, then ray BX with ∠B = 75°. Mark D on BX with BD = 12 cm. Join CD, draw its perpendicular bisector, and mark A where that bisector crosses BD. Join AC directly — △ABC is now formed with exactly the required measurements.
A closely related problem asks for a right triangle instead: base 7.5 cm, with the hypotenuse and the other side summing to 15 cm. The construction is identical in every step — draw the base, draw a ray at the given angle (90° this time, at B), mark D at the full 15 cm sum, then use the same perpendicular-bisector trick to split that sum back into the two correct sides. The right angle changes nothing about the method itself; it's simply the specific angle value being used at B — a reminder that this whole construction technique is really indifferent to which angle happens to be given, whether it's 75°, 90°, or anything else entirely.
Running the Same Trick in Reverse for a Difference
Construct △PQR with QR = 8 cm, ∠Q = 60°, PQ − PR = 3.5 cm. Draw QR = 8 cm, then ray QX with ∠Q = 60°. Since PQ is the longer side here, mark S on QX with QS = PQ − PR = 3.5 cm. Join RS, draw its perpendicular bisector, and mark P where it crosses ray QX. Join PR directly — △PQR is now formed with PQ − PR = 3.5 cm exactly, since the perpendicular bisector guarantees PS = PR, making PQ = QS + SP = 3.5 + PR.
Sum case: BD = AB + AC, A found where AD = AC → AB = BD − AD
Difference case: QS = PQ − PR, P found where PS = PR → PQ = QS + SPThe two constructions are really mirror images of each other: a sum places D beyond both original vertices, so the perpendicular bisector's intersection point sits between B and D, splitting the total apart; a difference places S at the shorter side's own length along the longer side's direction, so the perpendicular bisector's intersection point sits beyond S, building the total back up instead of splitting it down.
A Third Case: Given Two Angles and a Perimeter
Construct △XYZ with ∠Y = 30°, ∠Z = 60°, and the full perimeter XY + YZ + ZX = 10 cm. Draw a straight segment AB of the full 10 cm perimeter length. At A, draw a ray making half of ∠Y (15°); at B, draw a ray making half of ∠Z (30°). Their exact intersection point is X. The perpendicular bisector of AX meets AB at Y; the perpendicular bisector of BX meets AB at Z. Joining X to both Y and Z completes △XYZ, with ∠Y and ∠Z coming out to exactly 30° and 60° as required.
This construction needs two perpendicular bisectors rather than one, since both remaining vertices, Y and Z, have to be recovered from the same single outer segment AB — each bisector peels off one of the two triangle's vertices from the auxiliary triangle ABX built first, in exactly the same "recover a hidden point" spirit already established by the sum and difference constructions covered above.
Constructing a Whole Arc, Not Just a Triangle
Construct a segment of a circle on a 5 cm chord AB, containing a specific given angle throughout. For the 90° case specifically: draw the perpendicular bisector of AB carefully to find its exact midpoint O, then draw a circle centred at O with radius OB exactly — any point P chosen on this circle gives ∠APB = 90° directly, since AB is now a diameter and every angle in a semicircle is a right angle. For 45° or 120°, the method changes: draw rays AX and BY meeting at O so that ∠A = ∠B equal a chosen value, then draw a circle centred at O with radius OB; a point P on the resulting major arc gives 45°, while a point chosen on the minor arc instead gives 120° exactly.
| Target angle | Where P is placed | Why it works |
|---|---|---|
| 90° | Anywhere on the full circle | AB is a diameter — angle in a semicircle theorem (Chapter 12) |
| 45° | On the major arc | Angle in a major segment is acute |
| 120° | On the minor arc | Angle in a minor segment is obtuse |
This entire construction leans directly on results proved back in Chapter 12 — the semicircle case is a direct, immediate application of "angle in a semicircle is 90°," and the other two cases reuse the fact that every single point in the same segment subtends an exactly equal angle, so once O is placed correctly, any point at all on the correct arc gives the required angle without needing to be positioned any more precisely than that — the whole segment of the circle, not just one single spot on it, genuinely satisfies the condition simultaneously.
From Constructions to Counting Outcomes
Every construction in this short chapter built an exact geometric figure from a handful of given measurements. Chapter 14, Probability leaves geometric construction behind entirely, turning instead to counting how many outcomes an experiment can produce and how likely each one is — a genuinely different branch of mathematics closing out this Class 9 Mathematics course.