Overview
Chapter: Constructions (Class 9, NCERT) Introduction: This chapter introduces classical Euclidean constructions using only an unmarked straightedge (ruler) and a compass. It teaches standard procedures to draw required lines, angles and triangles accurately and to justify those constructions using basic geometry results. Importance: Constructions develop precision, spatial reasoning and logical thinking. They reinforce congruence criteria, loci ideas (sum/difference of distances) and provide tools used throughout geometry. These techniques are frequently tested in exams and form a foundation for higher‑level geometry. Key themes: basic constructions (angle bisector, perpendicular and perpendicular bisector), division of a segment, and constructing triangles from given data (SSS, SAS, ASA, RHS and special cases involving sum/difference of two sides). The chapter also emphasizes step‑wise procedures and short justifications based on congruence or locus arguments. What the student will learn: students will gain the ability to carry out standard compass‑and‑ruler constructions (bisect angles and segments, draw perpendiculars), divide a line segment into equal parts, construct…
Learning Objectives
- Define basic terms used in constructions such as point, line segment, ray, perpendicular bisector and angle bisector.
- Explain the principles and properties (like congruence and perpendicularity) underlying classical compass-and-straightedge constructions.
- Construct the perpendicular bisector of a given line segment using compass and straightedge and locate its midpoint.
- Construct the bisector of a given angle using compass and straightedge and verify that it divides the angle into two equal parts.
- Construct a perpendicular to a given line at a point on the line using appropriate construction steps.
- Construct a perpendicular to a given line from a point not on the line using compass-and-straightedge methods.
- Construct a triangle when its three sides are given (SSS construction) and justify the construction.
- Construct a triangle when two sides and the included angle are given (SAS construction) and justify the construction.
Topics in this chapter
5 topics · tap a topic title to jump straight to it.
Introduction and Tools
What is Construction?
In geometry, a construction is a precise way to create geometric figures using only specified tools (traditionally a straightedge and compass). Construction differs from freehand drawing because it follows fixed steps and logical reasoning so the resulting figure satisfies the given conditions exactly.
Why learn constructions?
- Develops logical thinking and stepwise problem solving.
- Helps understand properties of geometric objects (angles, bisectors, perpendiculars, etc.).
- Useful in real-life technical tasks: drafting, carpentry, map-making, engineering and design.
Basic rules (Euclidean style)
- You may use an unmarked straightedge (ruler without scale) and a compass.
- You may draw straight lines through two existing points and draw circles with a given center and radius.
- Measurements (reading distances on a scale) are not allowed in pure constructions; instead copy lengths with a compass.
Main tools and their use
- Compass: To draw circles/arcs and to transfer distances (copy a length). Set the compass width to a length and reproduce it elsewhere.
- Straightedge (ruler without scale): To draw a straight line through any two points. It does not measure distances.
- Protractor: Used in practical drawing to measure or construct a specific angle (in pure constructions we avoid measuring angles directly).
- Divider: Like a compass but used only to transfer or compare distances (no drawing).
- Set squares: Help draw common angles (30°, 45°, 60°, 90°) and parallels in practical drawing classes.
- Other accessories: Pencil, eraser, sharpener — keep the pencil sharp for accuracy.
Standard format of a construction solution
- Given — list the data provided in the problem (points, lengths, angles).
- Required — state what you must construct.
- Construction — give step-by-step geometric operations (with compass/straightedge) to build the figure.
- Proof/Explanation — justify why the construction meets the requirements (use geometric properties and theorems).
Common elementary constructions (introduced here)
Examples you will learn to perform precisely: bisect a line segment, bisect an angle, construct a perpendicular bisector of a segment, and draw a perpendicular to a line from a point on or off the line. These are building blocks for more complex constructions (e.g., constructing triangles).
Precautions & tips
- Keep compass width fixed when copying a length.
- Make arcs large enough so they intersect clearly; avoid tiny arcs that cause errors.
- Work lightly with the pencil for construction marks; darken only final lines.
- Label all important intersection points (A, B, C, etc.) to refer to them in the proof.
- Bisecting a line segment AB: Place compass at A and B with radius > AB/2, draw arcs above and below the segment. From A and B, draw two pairs of arcs so they intersect at two points; join the intersection points to get the perpendicular bisector which meets AB at its midpoint.
- Bisecting an angle ∠XAY: Draw an arc centred at A intersecting both arms at P and Q. With the same radius draw arcs from P and Q so they intersect at R. Line AR is the angle bisector, dividing ∠XAY into two equal angles.
- Drawing a perpendicular from a point P on line l: With centre P draw equal arcs that cut the line l at two points, say M and N. From M and N draw arcs with same radius that meet above the line at Q. Join P and Q to get a perpendicular to l at P.
- Copying a given length AB to start at point C: Place compass at A and set width to AB. Without changing compass width move to C and draw an arc; the point where the arc meets the chosen direction gives the copied length from C.
- Practical: A carpenter uses a compass-like divider to transfer exact lengths from a plan to wood, and a set-square to get a perfect right angle when assembling frames.
- Perpendicular bisector property: Any point on the perpendicular bisector of segment AB is equidistant from A and B (PA = PB).
- Angle bisector theorem (used in proofs): An internal angle bisector of ∠A in triangle ABC divides the opposite side BC in the ratio of the adjacent sides: (BD/DC) = (AB/AC), where AD is the bisector.
- Triangle inequality (useful check): For any triangle ABC, AB + BC > CA, BC + CA > AB, and CA + AB > BC.
- Locus definition (used conceptually): The perpendicular bisector is the locus of points equidistant from two fixed points; an angle bisector is the locus of points equidistant from the sides of the angle.
Basic Constructions
What are Basic Constructions?
Basic constructions are the fundamental compass-and-straightedge procedures used to make accurate geometric figures without measurement. Using only an unmarked ruler (straightedge) and a compass, you can copy segments, bisect segments and angles, and draw perpendiculars and perpendicular bisectors. These constructions are the building blocks for more advanced constructions (like constructing triangles).
Tools and conventions
- Compass — to draw arcs and transfer lengths.
- Straightedge — to draw straight lines between two points.
- We do not measure numerically with a scale; we use geometric intersection properties.
Common basic constructions (with short method)
- Copy a line segment AB to start at point C
- Place compass on A and B; set it to length AB.
- With center C and same compass opening, draw an arc to mark point D so that CD = AB.
- Draw the segment CD.
- Bisect a segment AB (find midpoint M)
- With radius > AB/2, draw arcs with centers A and B; they intersect at two points.
- Join those intersection points; their line meets AB at the midpoint M.
- Bisect an angle ∠X (angle bisector)
- With center at vertex X, draw an arc cutting both sides at P and Q.
- With centers P and Q and equal radius > PX, draw two arcs that intersect at R.
- Draw XR. XR bisects the angle into two equal angles.
- Construct a perpendicular to line l through a point P on l
- With center P, draw an arc cutting l at A and B (equal distance each side).
- With centers A and B and radius > AB/2 draw arcs intersecting at Q and R above and below l.
- Join Q and R; this line is perpendicular to l and passes through P.
- Construct a perpendicular to line l through a point P not on l
- With center P draw an arc that meets l at A and B.
- Bisect segment AB (as above) to get midpoint M.
- Join P and M; PM is perpendicular to l.
- Perpendicular bisector of segment AB
- Draw equal-radius arcs from A and B to get two intersection points.
- Join intersections; that line is the perpendicular bisector and meets AB at its midpoint and is perpendicular to AB.
Why these work (intuition)
Arcs from equal radii create points equidistant from given centers; intersections of such loci give points satisfying required equal-distance or equal-angle conditions. For example, points on the perpendicular bisector are equidistant from A and B; the angle bisector consists of points equidistant from the sides of the angle.
Where these are used
They serve in triangle constructions, dividing segments into equal parts, designing symmetric shapes, engineering layouts, carpentry joinery, map and land subdivision and optical alignment.
- Carpentry: Marking the midpoint of a board by constructing the perpendicular bisector of its end points to place a dowel or hinge centrally.
- Road design: Creating a perpendicular from a point to a planned road line to locate a service access or property boundary.
- Architecture: Bisecting angles when designing symmetric roof trusses or interior layouts so two parts are equal.
- Land surveying: Copying a measured distance (segment) from a known reference point to mark plot corners without numeric measuring.
- Art and pattern design: Repeatedly bisecting angles to create symmetric motifs (e.g., star patterns).
- Angle Bisector Theorem: In triangle ABC, if AD is the bisector of angle A meeting BC at D, then BD/DC = AB/AC.
- Perpendicular bisector property: Any point on the perpendicular bisector of segment AB is equidistant from A and B (PA = PB).
- Congruence/constructibility rules often used to justify constructions: SSS, SAS, ASA, RHS (Right-angle-Hypotenuse-Side).
- For perpendicular construction from external point P: If M is midpoint of AB where the arc from P meets l at A and B, then PM is perpendicular to l (PM^2 = PA·PB + (difference used as geometric relation)).
Triangle Constructions
What is Triangle Construction?
Triangle construction is the process of creating a triangle using only a ruler (straightedge) and a compass from a given minimal set of data (sides, angles, or combination). Constructions use geometric principles: intersection of straight lines and circles, the triangle inequality, and angle-sum properties.
Tools: ruler (or straightedge), compass, protractor (for checking/learning), pencil and eraser.
Key ideas:
- Triangle inequality: the sum of any two sides is greater than the third.
- Sum of interior angles = 180°.
- Construction principle: a point at a fixed distance from a given point lies on a circle (use arcs). A point seen under a fixed angle from a segment lies on an arc of a circle (angle locus).
Standard construction cases (Class 9)
- SSS (three sides given a, b, c)
Steps: Draw one side (say BC = a). With centre B radius = c draw an arc. With centre C radius = b draw another arc. Their intersection is A. Join A to B and C to complete triangle ABC. - SAS (two sides and included angle given)
Steps: Draw the included angle at one vertex or draw the base (one side). From one endpoint, draw an arc equal to the given adjacent side; from the other endpoint, draw an arc equal to the other adjacent side. The intersection gives the third vertex. Alternatively: draw base BC, at B fold/construct angle ABC, along the ray mark BA of the given length, then from C draw circle radius = CA; intersection is A. - ASA (two angles and included side given)
Steps: Draw the given side BC. At B construct the given angle ABC, at C construct the given angle BCA. The two rays from B and C meet at A. Join A to B and C. (Check angles sum < 180°.) - RHS (right triangle: hypotenuse and one side given)
Steps: If hypotenuse AC and a leg AB are given and angle B is 90°, draw AC. With center A and radius AB draw an arc; with center C and radius equal to the other known leg (or use perpendicular construction) find B on the circle such that ∠ABC = 90° (i.e., B lies on the circle with AC as diameter). Use the Thales theorem: any angle subtended by a diameter is a right angle.
Practical tips: check triangle inequality before construction; make light pencil marks for arcs and darken final edges; verify angles and lengths after construction.
Why it works: intersections of circles (SSS, SAS) or of rays (ASA) provide unique points satisfying distance/angle constraints. For RHS, Thales' circle ensures a right angle.
- Surveying/triangulation: locating a point by measuring angles or distances from fixed stations (constructing triangles helps find positions).
- Roof truss and supports: designing triangular frames where side lengths and angles must be precise for load-bearing structures.
- Navigation: bearing-based position fixing uses triangles formed by lines of sight and known baseline points.
- Graphics and CAD: building triangular meshes and precise components in mechanical parts uses geometric constructions.
- Sum of interior angles: ∠A + ∠B + ∠C = 180°
- Triangle inequality: for any triangle with sides a, b, c → a + b > c, b + c > a, c + a > b
- Pythagoras (right triangle): if sides a, b are legs and c is hypotenuse → a² + b² = c²
- Area (useful when combining construction & measurement): area = (1/2)·base·height; also area = (1/2)·ab·sin C
Constructions Related to Triangle Centers
Overview: A triangle has several important centers — points defined by special concurrent lines. The four classical centers are the circumcenter, incenter, centroid and orthocenter. Each center is constructed with a ruler and compass and has simple geometric properties useful in solving problems and in real-life design.
Circumcenter (O)
- Construction steps (ruler & compass):
- Take triangle ABC. Construct the perpendicular bisector of AB: place compass at A and B, draw equal arcs above and below AB, join their intersection points to get one perpendicular bisector.
- Similarly construct the perpendicular bisector of BC (or AC).
- The intersection point O of the two perpendicular bisectors is the circumcenter. With O as center and radius OA, draw the circumcircle through A, B, C.
- Key property: O is equidistant from A, B and C (OA = OB = OC). The circumcircle passes through all three vertices.
Incenter (I)
- Construction steps:
- From vertex A, construct the internal angle bisector of ∠A (use arcs to mark equal distances along the sides and join intersection of arcs to A).
- Construct the angle bisector of ∠B (or ∠C).
- The intersection I of the two angle bisectors is the incenter. With I as center, draw a circle tangent to any side — this is the incircle.
- Key property: I is equidistant from the three sides (distance = inradius r). The incircle is tangent to all three sides.
Centroid (G)
- Construction steps:
- Find midpoint of BC by drawing perpendicular bisector or using equal-radius arcs on B and C; join A to midpoint M of BC — this is a median.
- Similarly draw a median from B to midpoint of AC (or from C).
- The medians meet at the centroid G.
- Key property: The centroid divides each median in the ratio 2:1 counting from the vertex (AG : GM = 2 : 1). G is the centre of mass of a triangular lamina.
Orthocenter (H)
- Construction steps:
- From vertex A draw a line through A perpendicular to BC (use compass to construct a perpendicular).
- From B draw a line through B perpendicular to AC.
- The intersection of these perpendiculars is the orthocenter H (altitudes meet at H). For obtuse triangles some altitudes meet outside the triangle.
- Key property: H is the intersection of the three altitudes of the triangle.
Remarks and useful facts:
- All four centers coincide only for an equilateral triangle.
- Perpendicular bisectors (circumcenter), angle bisectors (incenter), medians (centroid) and altitudes (orthocenter) are each three concurrent lines defining the respective center.
- Advanced relation: For a non-equilateral triangle the circumcenter O, centroid G and orthocenter H are collinear on the Euler line with OG : GH = 1 : 2.
- Engineering: Centroid used to find the center of mass of triangular components (balance point for beams, plates).
- Architecture/landscape: Circumcenter used to draw a circle passing through three marked boundary points (e.g., placing a circular fountain to touch three points).
- Manufacturing/plumbing: Incenter used to place a circular tank or sprinkler that must touch three straight barriers — incircle gives the largest circle tangent to all sides.
- Navigation/GNSS concept: Triangulation and circumcircle ideas help locate a point equidistant from known landmarks (circumcenter idea used conceptually in some positioning methods).
- Centroid division: Each median is divided by centroid G in the ratio 2:1 (vertex to centroid : centroid to midpoint = 2:1).
- Angle bisector theorem: An internal angle bisector from A divides BC in the ratio AB : AC.
- Inradius relation: Area Δ = r × s, where r is inradius and s is semiperimeter (s = (a+b+c)/2). Thus r = Δ / s.
- Circumradius relations: R = a / (2 sin A) and R = (a b c) / (4 Δ), where a, b, c are side lengths and Δ is the area.
- Coordinates (useful if vertices known in coordinate plane): - Centroid G = ((x1+x2+x3)/3, (y1+y2+y3)/3). - Circumcenter can be found as intersection of perpendicular bisector equations; incenter as intersection of angle-bisector lines (or weighted average by sides).
Construction Strategies and Problem-Solving
What this topic covers: Construction strategies are step-by-step methods used with ruler (straightedge) and compass to create geometric figures accurately. Problem-solving in constructions means translating given data into a sequence of basic constructions (drawing lines, arcs, bisectors, perpendiculars, circles) and proving that the construction meets the given conditions using congruence or basic theorems.
General problem-solving strategy (stepwise):
- Read and understand: Identify exactly what is given and what must be constructed (sides, angles, medians, perpendiculars, ratios).
- Check feasibility: Use triangle inequality, angle-sum (180°), or given numeric conditions to ensure construction is possible.
- Choose the main method: Match the data to a standard construction type (SSS, SAS, ASA, RHS) or a locus-based method (intersections of lines/circles).
- Plan auxiliary constructions: Decide if you need angle bisectors, perpendiculars, parallel lines, or circles to locate a point (e.g., apex of triangle as intersection of two circles).
- Execute with care: Draw a clear base, set compass widths precisely, mark intersections, and only draw required lines.
- Prove and verify: Use congruence (SSS, SAS, ASA, RHS) or properties (angle bisector theorem, perpendicular bisector property) to justify the construction and verify lengths/angles.
Common basic constructions used as building blocks:
- Bisect an angle (to get equal angles)
- Construct perpendicular bisector of a segment (to find midpoint or locus of equidistant points)
- Draw perpendicular from a point to a line (point on or off the line)
- Divide a segment in a given ratio (using parallel/ray method)
- Construct triangle from given elements using SSS, SAS, ASA, or RHS rules
- Use circle intersections: intersection of two circles gives points at given distances from two centers (used to locate a triangle's third vertex)
Justification techniques: After construction, show that the constructed figure satisfies the given conditions by proving congruence between constructed parts (use SSS, SAS, ASA, RHS) or by invoking known properties (e.g., perpendicular bisector points are equidistant from endpoints).
Practical tips: Always draw a clear base line (place a side on the x-axis mentally), label points, mark compass radii, and lightly draw auxiliary lines if needed. If the intersection does not occur, re-check feasibility (triangle inequality or angle sizes).
- Constructing a triangle ABC when sides AB = 6 cm, BC = 5 cm and CA = 7 cm (SSS): Draw base AB = 6 cm, then with centers A and B draw circles of radii 7 cm and 5 cm respectively. Their intersection gives C. Join C to A and B and verify using SSS congruence.
- Bisecting an angle: To bisect ∠X, draw equal arcs from X cutting the sides at P and Q, then from P and Q draw equal arcs that meet at R. XR is the angle bisector. Real-life: placing two equal-angled supports in roof framing.
- Construct perpendicular from an external point P to a line l: With center P draw a circle meeting l at A and B. Construct perpendicular bisector of AB; its intersection with the circle gives foot of perpendicular. Real-life: drop a perpendicular from a survey point to a road to measure shortest distance.
- Divide a road segment AB in ratio 3:2 for utility-placement: From A draw a ray, mark 5 equal segments on it, connect the 5th point to B and draw parallels through the 3rd mark to intersect AB — that point divides AB internally in 3:2.
- Triangle inequality: For any triangle with sides a, b, c: a + b > c, b + c > a, and c + a > b (used to check feasibility).
- Sum of interior angles of triangle: ∠A + ∠B + ∠C = 180° (useful to compute a missing angle before construction).
- Pythagoras theorem (right triangles): If right-angled at C, then AB² = AC² + BC² (used in RHS constructions or verification).
- Angle-bisector theorem: If AD is internal bisector of ∠A in triangle ABC meeting BC at D, then BD/DC = AB/AC (helps to justify constructions that use bisectors).
- Section (internal division) idea (constructive, not numeric formula): To divide AB in ratio m:n, draw a ray from A, mark (m + n) equal parts, connect the last point to B, and draw a parallel through the m-th mark to meet AB. (Coordinate form: if A(x1,y1), B(x2,y2) then point dividing internally in m:n is ((nx1 + mx2)/(m+n), (ny1 + my2)/(m+n)).)
Key Concepts
- Compass
- A drafting tool with two arms used to draw arcs and circles and transfer distances.
- Straightedge
- An unmarked ruler used to draw straight lines between two points without measuring distance.
- Geometric construction
- A precise drawing made using only a compass and straightedge following logical steps.
- Angle bisector
- A ray that divides an angle into two equal angles.
- Perpendicular bisector
- A line that is perpendicular to a segment and passes through its midpoint, dividing it into equal parts.
- Perpendicular to a line (from a point on the line)
- A line through a point on a given line that meets the given line at a right angle (90°).
- Perpendicular to a line (from an external point)
- A line from a point not on the given line that meets the line at a right angle, constructed using arcs.
- SSS (Side-Side-Side) construction
- Constructing a triangle when the lengths of all three sides are given using three arcs to locate vertices.
- SAS (Side-Angle-Side) construction
- Constructing a triangle when two sides and the included angle are given by drawing one side, angle, and an arc for the other side.
- ASA (Angle-Side-Angle) construction
- Constructing a triangle when two angles and the included side are given by constructing angles at ends of the side and joining rays.
- RHS (Right angle-Hypotenuse-Side) construction
- Constructing a right triangle when the hypotenuse and one side are given: build the hypotenuse and draw a circle to locate the right-angle vertex.
- Division of a line segment into n equal parts
- A method using parallel lines and proportional segments to split a given segment into n equal pieces.
- Locus
- The set of all points satisfying a given condition or a rule.
- Arc
- A continuous part of the circumference of a circle, usually drawn with a compass during constructions.
- Chord
- A line segment joining two points on a circle.
- Radius
- A line segment from the center of a circle to any point on the circle; its length is the circle's radius.
- Tangent (to a circle)
- A line that touches a circle at exactly one point and is perpendicular to the radius at the point of contact.
- Auxiliary line
- An extra line drawn to simplify a construction or proof, often used to create congruent triangles or parallel lines.
- Congruent triangles
- Two triangles that are identical in shape and size, used to justify correctness of constructions via congruence criteria.
End-of-Chapter Trial Paper & Test Questions
Topic-wise questions to test your understanding of every concept in this chapter.
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Which tools are permitted in Euclidean geometric constructions? / यूक्लिडियन ज्यामितीय निर्माण में कौन से उपकरणों की अनुमति है? (a) Ruler (with markings) and protractor / मापदंड (निशानों के साथ) और चाँदा (b) Compass and unmarked straightedge / परकार और अचिह्नित सीधा पट्टी (c) Set squares and divider / सेट स्क्वायर और डिवाइडर (d) Calculator and graph paper / कैलकुलेटर और ग्राफ पेपर
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(b) Compass and unmarked straightedge / परकार और अचिह्नित सीधा पट्टी — Classical Euclidean constructions use only a compass (to draw arcs and copy lengths) and an unmarked ruler (to draw straight lines). / शास्त्रीय यूक्लिडियन निर्माण केवल परकार (चाप खींचने और लंबाई कॉपी करने के लिए) और अचिह्नित पट्टी (सीधी रेखाएँ खींचने के लिए) का उपयोग करते हैं।
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To construct the perpendicular bisector of segment AB, you draw arcs from both A and B with the same radius. The radius must be: / रेखाखंड AB का लंब समद्विभाजक बनाने के लिए, आप A और B दोनों से एक ही त्रिज्या से चाप खींचते हैं। त्रिज्या होनी चाहिए: (a) Equal to AB / AB के बराबर (b) Less than AB/2 / AB/2 से कम (c) Greater than AB/2 / AB/2 से अधिक (d) Exactly AB/2 / ठीक AB/2
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(c) Greater than AB/2 / AB/2 से अधिक — The arcs must intersect above and below AB; for this the radius must be more than half the segment length. / चापों को AB के ऊपर और नीचे प्रतिच्छेद करना चाहिए; इसके लिए त्रिज्या रेखाखंड की आधी लंबाई से अधिक होनी चाहिए।
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In an SSS construction, a triangle is constructed when: / SSS निर्माण में त्रिभुज का निर्माण तब किया जाता है जब: (a) Two sides and the included angle are given / दो भुजाएँ और समाहित कोण दिए गए हों (b) All three sides are given / तीनों भुजाएँ दी गई हों (c) Two angles and the included side are given / दो कोण और समाहित भुजा दी गई हो (d) The hypotenuse and one side of a right triangle are given / समकोण त्रिभुज का कर्ण और एक भुजा दी गई हो
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(b) All three sides are given / तीनों भुजाएँ दी गई हों — SSS stands for Side-Side-Side; arcs of the given lengths from both base endpoints meet at the third vertex. / SSS का अर्थ है भुजा-भुजा-भुजा; दोनों आधार बिंदुओं से दी गई लंबाई के चाप तीसरे शीर्ष पर मिलते हैं।
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Fill in the blank: The _____ is the point where the three perpendicular bisectors of a triangle's sides meet. / रिक्त स्थान भरें: _____ वह बिंदु है जहाँ त्रिभुज की तीनों भुजाओं के लंब समद्विभाजक मिलते हैं।
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Circumcenter / परिकेंद्र — The circumcenter is equidistant from all three vertices and is the center of the circumscribed circle (circumcircle). / परिकेंद्र तीनों शीर्षों से समान दूरी पर होता है और परिवृत्त का केंद्र होता है।
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Fill in the blank: The point where the three angle bisectors of a triangle meet is called the _____. / रिक्त स्थान भरें: त्रिभुज के तीनों कोण समद्विभाजकों का मिलन बिंदु _____ कहलाता है।
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Incenter / अंतःकेंद्र — The incenter is equidistant from all three sides of the triangle and is the center of the inscribed circle (incircle). / अंतःकेंद्र त्रिभुज की तीनों भुजाओं से समान दूरी पर होता है और अंतर्वृत्त का केंद्र होता है।
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True or False: Before constructing a triangle using the SSS method, it is necessary to verify the triangle inequality. / सत्य या असत्य: SSS विधि से त्रिभुज का निर्माण करने से पहले, त्रिभुज असमानता को सत्यापित करना आवश्यक है।
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True / सत्य — If the given side lengths do not satisfy the triangle inequality (sum of any two sides must exceed the third), no valid triangle can be constructed. / यदि दी गई भुजाओं की लंबाई त्रिभुज असमानता को संतुष्ट नहीं करती (किन्हीं दो भुजाओं का योग तीसरी से अधिक होना चाहिए), तो कोई वैध त्रिभुज नहीं बनाया जा सकता।
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Describe the steps to bisect a given angle ∠XAY using a compass and straightedge. / परकार और सीधी पट्टी का उपयोग करके दिए गए कोण ∠XAY को समद्विभाजित करने के चरणों का वर्णन कीजिए।
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1. With centre A, draw an arc cutting both arms at P and Q. / A को केंद्र मानकर चाप खींचें जो दोनों भुजाओं को P और Q पर काटे। 2. With centres P and Q (same radius > PQ/2), draw arcs intersecting at R. / P और Q को केंद्र मानकर (समान त्रिज्या > PQ/2) चाप खींचें जो R पर मिलें। 3. Join A to R; ray AR is the angle bisector dividing ∠XAY into two equal parts. / A को R से जोड़ें; किरण AR कोण ∠XAY को दो समान भागों में विभाजित करती है।
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In the RHS construction of a right triangle, Thales' theorem is used. What does Thales' theorem state? / एक समकोण त्रिभुज के RHS निर्माण में, थेल्स प्रमेय का उपयोग किया जाता है। थेल्स प्रमेय क्या कहती है?
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Thales' theorem states that any angle inscribed in a semicircle (i.e., the angle subtended by a diameter at any point on the circle) is a right angle (90°). / थेल्स प्रमेय कहती है कि अर्धवृत्त में अंकित कोण (अर्थात व्यास द्वारा वृत्त के किसी भी बिंदु पर अंतरित कोण) सदैव एक समकोण (90°) होता है। In RHS construction, the vertex with 90° must lie on the circle with the hypotenuse as diameter. / RHS निर्माण में, 90° वाला शीर्ष उस वृत्त पर होना चाहिए जिसका व्यास कर्ण है।
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