Chapter 11 — Constructions

Introduction
Constructions in geometry means drawing precise figures (lines, angles, triangles, circles, etc.) using only certain standard instruments. In classical (Euclidean) constructions you normally use an unmarked ruler (or straightedge) and a compass. The goal is to create exact figures using basic rules and procedures that rely on geometric properties (like equality of radii, intersection of arcs, perpendicularity and bisected angles).

Why constructions matter
They build understanding of geometric properties, give rigorous ways to produce figures, and are applied in drafting, engineering, architecture and many practical crafts. In Class 10 you use constructions as tools to solve problems such as bisecting an angle, drawing perpendiculars, and constructing triangles from given data.

Basic principles
Most constructions follow a small set of moves: (1) choose a base or reference line, (2) use a compass to copy distances or draw arcs, (3) locate intersection points of arcs/lines, (4) join points with the straightedge. Intersections of two arcs, or an arc and a line, are used as new defined points. Constructions are justified by basic theorems (e.g., perpendicular bisector points are equidistant from segment endpoints).

Instruments and how to use them

• Unmarked ruler / straightedge: Used to draw (extend) straight lines and join points. Do not use it to measure lengths when working with classical constructions (unless a scale is allowed).
• Compass: Used to draw circles and arcs and to transfer distances. Keep the compass point steady and the pencil sharp for accurate arcs. A fixed radius draws all points at that distance from the center.
• Divider: Two pointed ends; used to transfer lengths without a pencil mark (or to compare distances).
• Protractor: Used to measure or draw specific angles (when constructions allow angle measurement). Useful in practical drafting and verifying results.
• Set squares (45° and 30°–60°): For drawing common angles and perpendicular/parallel lines quickly in practical work.
• Pencil, eraser, sharpener: Use a sharp HB pencil for fine lines; erase construction arcs only after final drawing is confirmed.
• Scale (metric): For measuring lengths when allowed by the problem.

Practical tips

• Keep the compass width steady while transferring or drawing arcs; mark intersections clearly.
• Draw light construction arcs first; darken final lines after confirming correctness.
• Label points as you create them (A, B, C, P, Q…) to follow steps and justify results.
• Work on a stable surface; small errors in compass placement can shift results noticeably.

What are Basic Constructions? Basic constructions are the fundamental geometric constructions made with a compass and an unmarked straightedge (ruler). They form the building blocks for more complex constructions. In Class 10, typical basic constructions include:

1. Perpendicular bisector of a line segment
2. Angle bisector
3. Perpendicular to a line through a given point (point on the line and point not on the line)
4. Construction of triangles using given data (SSS, SAS, RHS/Right-angle-Hypotenuse-Side, ASA/AAS)

Why they work (key ideas): Constructions use properties of circles and symmetry. For example, if two circles with equal radii are drawn with centers at the ends of a segment, their intersection points are equidistant from the segment ends — this gives the perpendicular bisector. Similarly, arcs drawn from the sides of an angle help locate points equidistant from both sides, which yields the angle bisector.

Step-by-step (core constructions):

1. Perpendicular bisector of segment AB — Place compass at A, draw an arc above and below the segment with radius > AB/2. Repeat from B with same radius. Join the two intersection points of the arcs; this line is the perpendicular bisector and meets AB at its midpoint.
2. Bisect angle ∠XAY — With center A, draw an arc that cuts both AX and AY at P and Q. With centers P and Q and radius greater than half PQ, draw two arcs that intersect at R. Join AR. AR is the angle bisector (it divides ∠XAY into two equal angles).
3. Perpendicular to a line l through an external point P — With center P, draw an arc cutting l at two points A and B. With centers A and B and equal radius, draw arcs that meet at Q and R on the side of P. Join Q and R; the line QR is the perpendicular to l and passes through P (or join P to midpoint of arc intersections appropriately).
4. Perpendicular to line l through a point P on l — With center P, draw arcs that cut l at A and B. With centers A and B and equal radius greater than AB/2, draw arcs above and below l; join their intersections to get the perpendicular through P.
5. Triangle constructions (overview) — Use circle intersections to transfer side lengths and construct vertices: for SSS, draw base AB, then draw circles centered at A and B with radii equal to the other two given sides; their intersection gives the third vertex. For SAS, draw base and use an angle plus circle intersection. For RHS (right triangle), mark hypotenuse and draw a circle or use perpendicular construction with Pythagorean check.

Practical tips: always use a sharp pencil, set the compass firmly, keep the same radius when required, and label intersection points clearly. After construction, you can verify results using properties (e.g., equal distances or equal angles) or with a protractor/ruler as a check.

Overview

SSS (Side–Side–Side) is a standard construction method: when the three side lengths of a triangle are given, you can construct the triangle using a ruler and compass. If the three lengths satisfy the triangle inequality, the triangle is unique up to congruence.

Tools required: ruler (or straightedge), compass, pencil, eraser.

Condition

Let the sides be a, b, c. A triangle with these side lengths exists iff each length is positive and the triangle inequalities hold: a + b > c, b + c > a, and c + a > b.

Step-by-step construction (construct triangle ABC given AB, BC, CA)

1. Choose one side to draw first. Draw segment BC with length equal to the given BC.
2. With center at B and radius equal to the given length AB, draw an arc (a circle or part of it).
3. With center at C and radius equal to the given length AC, draw another arc.
4. The intersection point(s) of the two arcs gives the possible position(s) for A. (If arcs meet at two points, they give congruent triangles reflected across BC; if they meet at one point they are tangent; if they do not meet, no triangle is possible.)
5. Mark one intersection as A and join A to B and A to C. The triangle ABC is the required triangle.

Why this works / uniqueness

The locus of points at a fixed distance AB from B is a circle centered at B; similarly, points at distance AC from C lie on a circle centered at C. Their intersection is the set of points at both distances simultaneously — exactly the possible positions for A. Because the distances to B and C are fixed, and BC is fixed, any two triangles constructed this way are congruent by SSS, so the triangle is unique up to rigid motion (and reflection).

Common pitfalls

• Not checking triangle inequality before construction.
• Using imprecise compass settings — ensure radii are transferred accurately.
• Confusing which given length corresponds to which side; label clearly before starting.

Connections

From the three sides you can compute angles (using the cosine rule) and area (using Heron's formula). You can also construct circumcircle and incircle after drawing the triangle.

Meaning. In the SAS case you are given two sides of a triangle and the included angle between them (for example, lengths AB and AC and the angle ∠A). These data determine a unique triangle (up to rigid motion) by the SAS congruence rule.

Why unique. If two triangles have two pairs of equal corresponding sides and the included angle equal, the triangles are congruent (SAS). Hence only one triangle can be constructed from the given data.

Tools required. Ruler (or a scale), compass and either a protractor or the classical compass-and-straightedge method to copy an angle.

Classical construction (given sides b = AC, c = AB and included angle A = ∠CAB):

1. Draw a ray and mark point A on it. Along the ray draw AB of length c with B on the ray (or simply draw segment AB = c).
2. At A construct the given angle ∠A with one side along AB. (If using compass-and-straightedge to copy an angle: draw an arc from the angle's vertex to cut both sides, copy the same arc at A and reproduce the chord distance to locate the second ray.)
3. On the second ray of angle A, with centre A and radius b, draw an arc and mark its intersection as point C (so AC = b).
4. Join B and C. Triangle ABC is the required triangle.

Proof idea (why this works). Steps place two sides of required lengths meeting at A with the given included angle; any triangle with those two sides and that included angle has the same three corresponding elements, so by SAS congruence it is congruent to the constructed triangle.

Notes. If the given sides and included angle are specified at a different vertex (for example sides AB and BC with included angle B), use that vertex as the base and repeat the same procedure. Always ensure the angle is the one included between the two given sides.

What ASA means: ASA stands for "Angle–Side–Angle". It means two angles and the included side (the side between those two angles) of a triangle are given. Using ASA we can construct a unique triangle because two angles with the included side determine the third angle and the shape uniquely (ASA congruence).

Why it is unique: If angles A and B and the included side AB are given, the third angle C = 180° − (A + B). By ASA congruence, there is exactly one triangle (up to rigid motion) with those measures.

Required tools: ruler (or straightedge), compass and a protractor (a protractor can be used for quick angle drawing, but constructions with compass-and-straightedge only are standard).

Construction procedure (basic, using protractor):

1. Draw the given side AB of the given length.
2. At point A, construct the given angle ∠A using a protractor; draw the ray from A in the interior direction.
3. At point B, construct the given angle ∠B using a protractor; draw the ray from B.
4. The intersection point of the two rays is the third vertex C. Join C to A and C to B to complete triangle ABC.

Construction with compass and straightedge (copying an angle) (when a protractor is not used):

1. Draw base AB of the given length.
2. To copy angle α at A (given as an angle elsewhere): with centre at the angle vertex (of the given angle) draw an arc that meets both arms; measure the distance between those two arc intersections (by marking the arc intersections with the compass). Now at A draw an arc of the same radius cutting the desired ray. With the compass set to the distance between the two intersections, cut an arc from that new intersection; draw the ray through A and that new cut. Repeat similarly at B to copy the second given angle β.
3. The two rays meet at C. Join AC and BC to finish the triangle.

Notes:

• Compute the third angle first: C = 180° − (A + B). If A + B ≥ 180°, construction is impossible.
• If you need the other sides numerically, use the law of sines after finding angle C.

What is RHS? RHS is a congruence criterion for right triangles: if the length of the Right angle is common (i.e. both triangles are right-angled), and the Hypotenuse and one Side (leg) of one triangle are respectively equal to the hypotenuse and a corresponding side of another triangle, then the two right triangles are congruent (RHS).

Construction problem: Construct a right triangle ABC with given hypotenuse AB of length c and a given side AC of length b, where the right angle is at C.

Key idea used: Thales' theorem — any angle subtended by a diameter of a circle is a right angle. So every point C on the circle with diameter AB satisfies ∠ACB = 90°.

1. Draw the given segment AB of length c (this will be the hypotenuse).
2. Draw the circle with diameter AB (centre at midpoint of AB, radius = c/2). Any point on this circle will make angle ACB = 90°.
3. With A as centre and radius equal to the given side AC = b, draw a circle (centre A, radius b). The required point C is at the intersection of this circle and the circle with diameter AB.
4. Pick one of the intersection points as C. Join C to A and C to B. Triangle ABC is the required right triangle (right-angled at C) with AC = b and AB = c.

Why this works (brief proof):

• Because C lies on the circle with diameter AB, ∠ACB = 90° (Thales).
• Because C lies on the circle centred at A with radius b, AC = b (by construction).
• Thus triangle ABC has the given hypotenuse AB and given side AC, and is right-angled at C.

Notes:

• There are generally two possible positions for C (one above AB and one below AB), giving two congruent mirror-image triangles.
• Construction is possible only when b < c (strictly), because for a right triangle with hypotenuse c and leg b we must have b^2 < c^2 (else no real right triangle exists).

Coordinate interpretation (useful for graphing):

A = (0,0), B = (c,0). Let AC = b and BC = a where a = sqrt(c^2 - b^2).
Then the intersection coordinates are
 C = (x, y) with x = b^2 / c,  y = ± (a * b) / c.
So there are two symmetric solutions: y positive and y negative.

What the topic means

In some triangle-construction problems you are given the base BC and either the sum AB + AC or the difference |AB − AC| of the two other sides (often together with one more datum such as an angle at A). The set of possible locations of the vertex A (for fixed B and C) is a known plane locus:

• AB + AC = constant → an ellipse with foci at B and C.
• |AB − AC| = constant → a hyperbola (two branches) with foci at B and C.

Existence conditions (triangle inequalities)

• If the given sum is s = AB + AC, then s must satisfy s > BC (otherwise no triangle).
• If the given difference is d = |AB − AC|, then d < BC (otherwise no triangle).

How this helps in constructions

Construction reduces to locating the point A as an intersection of two loci: (i) the ellipse or hyperbola defined by B, C and the given sum/difference, and (ii) the locus defined by the other given data (for example: a ray making the given angle at A, a given altitude, a median, etc.). In practical school constructions you usually:

1. draw BC (given),
2. construct the auxiliary locus that encodes AB+AC or |AB−AC| (by a geometric method or a practical string/pins method),
3. draw the geometric constraint coming from the extra given data (e.g. the ray for the given angle at A),
4. take the intersection(s) to get A and then join AB and AC.

Practical notes

• Ellipse: can be drawn approximately with the string-and-pins method (tie a string of fixed length s around two pins placed at B and C and trace points keeping the string taut).
• Hyperbola: not as easy with basic tools; alternative straightedge-and-compass approaches use auxiliary constructions that convert the difference condition into a workable intersection (for example, use translations/reflections or construct points whose distances add up to or differ by the required amount). In many textbook problems an additional angle or a segment makes the intersection reducible to a standard SAS/SSS construction.

What it means: Techniques & Auxiliary Constructions are methods used to simplify a construction problem by drawing extra (auxiliary) lines or points so that the problem reduces to one or more standard constructions (e.g., SSS, SAS, ASA, RHS, perpendicular bisector, angle bisector).

Common ideas / techniques:

• Introduce auxiliary lines: draw an altitude, median, extension of a side, or a parallel line to convert the given data into a known construction.
• Use circles (compass arcs) to transfer lengths and to find intersection points that satisfy distance conditions.
• Use perpendicular bisectors for locating points equidistant from two points (circumcenter) and angle bisectors for ratio conditions (incenter).
• Reduce problems about sum or difference of sides to intersections of circles: sum AB+AC = given translates to a point on a circle centered at A with radius AB+AC, often realized by placing lengths head-to-tail.
• Use parallel lines and similar triangles to divide a segment in a given ratio (auxiliary ray and parallel through segment division points).
• Use reflection or symmetry: reflecting a point across a line can turn a difference condition into a simple distance condition.

General strategy (stepwise):

1. Read and label given elements precisely (points, lengths, angles).
2. Ask: which standard construction does this resemble? If none, which auxiliary line would convert it into one?
3. Draw the auxiliary line(s) — altitude, extension, median, parallel, or extra point. Use circle intersections where lengths must be equal.
4. Complete the reduced standard construction(s) and verify with the given conditions.
5. Erase auxiliary marks mentally (they are not part of the final figure) but keep them for checking accuracy.

Tools and accuracy: use a compass for arcs and circles, a ruler (unmarked) for straight lines, and a protractor only when an angle measure is given directly. Auxiliary constructions help reduce work and improve precision.

What the terms mean

Proof of a construction shows that the steps actually produce a figure satisfying the given conditions. A proof usually reduces to showing two constructed triangles are congruent or that a constructed point lies on a required locus (line, circle, angle bisector).

Validity / Existence asks whether a figure with the given data exists at all. In constructions this is tested by seeing whether the geometric loci used in the method intersect. If the loci have no common point, no figure exists.

Uniqueness asks whether the figure is the only possible one. Uniqueness is shown when the construction leads to a unique intersection of loci or when a congruence rule forces only one solution. Non-uniqueness happens when loci meet in more than one point (commonly two).

Common geometric reasoning used

• Use of loci: A point at a fixed distance from A is a circle; points at fixed difference/sum of distances form arcs/Apollonius loci; points equidistant from two points lie on the perpendicular bisector; points making a fixed angle with two rays lie on an arc of a circle.
• Congruence criteria (SSS, SAS, ASA, RHS) prove that the constructed triangles satisfy given lengths/angles and that the construction produces the required triangle.
• Triangle inequality (sum of any two sides > third) is a necessary condition for existence when side lengths are specified.

Typical proof structure for a construction

1. Describe the loci used and why the unknown point must lie on them.
2. Show that intersection(s) of those loci correspond exactly to the required point(s).
3. Use congruence or distance/angle equalities to verify all given conditions are satisfied.
4. Discuss the number of intersections to conclude uniqueness or show multiple/no solutions.

Important special case to remember: SSA – the ambiguous case

If two sides and a non-included angle are given (side–side–angle, SSA), you may get two distinct triangles, one triangle, or none depending on lengths (this is the classic ambiguous case). Geometrically this corresponds to a circle intersecting a line in 0, 1 or 2 points.

Worked examples and exercises in the Constructions chapter teach how to make accurate figures with ruler and compass and how to reason about each construction. The usual approach is: (1) read the requirement and list given data, (2) identify which standard construction or theorem applies (SSS, SAS, ASA, RHS, perpendicular bisector, angle bisector, division in given ratio, tangent from a point, etc.), (3) plan a construction sequence using arcs and straight lines, (4) perform the construction carefully with clear labels, and (5) justify each step using geometric facts.

Key practical tips: use a sharp pencil, draw light construction arcs first, mark intersection points clearly, measure only when required (not as the main construction tool), and always name points (A, B, C, D...) so you can refer to them in the justification. Typical justifications rely on congruence criteria, properties of perpendicular bisectors and angle bisectors, and proportionality (Thales). Exercises train procedural skill and geometric reasoning, and most problems reduce to a few standard patterns (construct triangle by SSS, SAS, ASA, RHS; divide a segment in m:n; draw tangent from an external point; construct triangle given base and sum/difference of other sides, etc.).

In Chapter "Constructions" the theme "Applications & Connections" shows how classical ruler-and-compass constructions are used to solve geometric problems and how these constructions link to other parts of the Class 10 syllabus (coordinate geometry, trigonometry, mensuration and real-life design). The idea is not only to follow construction steps but to understand why a construction works and how its properties (perpendicular bisector, angle bisector, tangent, locus) connect to algebraic formulas and practical tasks.

Key points:

• Common constructions: triangle construction (SSS, SAS, ASA, RHS), perpendicular bisector, angle bisector, common tangent from an external point, construction of incircle and circumcircle.
• Connections to other topics: perpendicular bisector = set of points equidistant from two points (link to locus and coordinates); angle bisector = set of points equidistant from the sides (link to trigonometry and distances); tangent length from external point uses right-triangle (Pythagoras) and the power of a point idea.
• Practical uses: design and surveying (accurate angles and distances), carpentry and engineering (triangulation, making bisected angles for joints), map work and plotting (dividing lines in given ratios), optics and mechanical linkages (constructing normals and tangents).

When solving application problems, standard steps are: (1) translate the word problem into geometric conditions (e.g., point equidistant from A and B), (2) identify which basic construction enforces that condition (e.g., perpendicular bisector), (3) carry out the construction and optionally use algebra/trigonometry to compute lengths/angles if numeric answers are required.