Geometrical Constructions — Chapter 1 (Part One)
20 × 1-Mark Questions (Quick Recall)
Q1. What is an angle bisector?
A1. A ray that divides an angle into two equal angles (i.e. each has half the measure of the original).
Q2. What is the point of concurrence of angle bisectors of a triangle called?
A2. The incentre (denoted by \(I\)).
Q3. What is a perpendicular bisector of a segment?
A3. A line that is perpendicular to the segment and passes through its midpoint.
Q4. What is the point of concurrence of perpendicular bisectors of a triangle called?
A4. The circumcentre (denoted by \(C\)).
Q5. Are the angle bisectors concurrent inside the triangle?
A5. Yes — the incentre lies inside every triangle.
Q6. Do perpendicular bisectors always intersect inside the triangle?
A6. Not always: for acute triangles inside, for right triangles at the hypotenuse midpoint, for obtuse triangles outside.
Q7. What construction gives a point equidistant from three vertices?
A7. Intersection of perpendicular bisectors — the circumcentre is equidistant from the vertices.
Q8. What construction gives a point equidistant from three sides?
A8. Intersection of angle bisectors — the incentre is equidistant from the sides.
Q9. Write the notation for congruent segments AB and PQ.
A9. \(\overline{AB}\cong\overline{PQ}\).
Q10. If \(\angle ABC = 40^\circ\), how do we write it (measure notation)?
A10. \(m\angle ABC=40^\circ\).
Q11. Which triangle construction uses SSS?
A11. Construct triangle when three sides are given (SSS construction).
Q12. Which triangle construction uses SAS (two sides and included angle)?
A12. Construct triangle when two sides and the included angle are given (SAS).
Q13. Which construction is ASA?
A13. Two angles and the included side (ASA) — construct using rays from the base and intersection.
Q14. What is RHS construction?
A14. Right-angle, hypotenuse and one side given — unique right triangle construction (RHS).
Q15. If two circles have equal radii are they congruent?
A15. Yes — circles with equal radii are congruent.
Q16. How to verify a constructed perpendicular bisector CD of segment PS?
A16. Check CD ⟂ PS and CD passes through midpoint M of PS (i.e. \(PM=MS\)).
Q17. Where is the circumcentre of a right triangle located?
A17. At the midpoint of the hypotenuse.
Q18. Where is the incentre located relative to the triangle?
A18. Always inside the triangle (intersection of angle bisectors).
Q19. Notation: length of segment AB?
A19. \(l(AB)\) or \(\overline{AB}\) (context-dependent). Example: \(l(AB)=5\text{ cm}\).
Q20. What does "concurrent" mean in geometry?
A20. Three or more lines (or curves) meeting at a single point.
20 × 2-Mark Questions (Short Reasoning / Short Steps)
Q1. How to construct the perpendicular bisector of a segment of length \(4\text{ cm}\)?
A1. Steps:
- Draw segment \(PS\) of \(4\text{ cm}\).
- With center \(P\), radius > half (e.g., 2.5 cm), draw arcs above and below.
- With center \(S\), same radius draw arcs; join the intersection points to form line \(CD\).
- \(CD\) is the perpendicular bisector — it meets \(PS\) at midpoint \(M\) and is perpendicular.
Q2. Prove that the three angle bisectors of a triangle are concurrent.
A2. Outline:
- Construct bisectors of \(\angle P,\angle Q\). They meet at \(I\).
- Point \(I\) is equidistant from sides \(PQ,QR,RP\) (drop perpendiculars IA, IB, IC).
- Because IA=IB=IC, \(I\) lies on bisector of \(\angle R\) as well. Hence all three are concurrent at incentre \(I\).
Q3. How to construct \(\triangle XYZ\) when \(XY=6\text{ cm}, YZ=4\text{ cm}, XZ=5\text{ cm}\) (SSS)?
A3. Steps:
- Draw base \(XY=6\text{ cm}\).
- With center \(X\) radius \(5\) cm draw arc; with center \(Y\) radius \(4\) cm draw arc; their intersection is \(Z\).
- Join to get \(\triangle XYZ\).
Q4. How to draw \(\triangle PQR\) when \(PQ=5.5\text{ cm},\ \angle P=50^\circ,\ PR=5\text{ cm}\) (SAS)?
A4. Steps:
- Draw \(PQ=5.5\text{ cm}\).
- At \(P\) draw ray making \(50^\circ\) with \(PQ\).
- With center \(P\) radius \(5\) cm cut the ray to locate \(R\); join \(Q\) and \(R\).
Q5. How many circumcentres lie inside/on/outside for acute/right/obtuse triangles?
A5. Acute: inside. Right: on hypotenuse midpoint. Obtuse: outside the triangle.
Q6. Given triangle sides \(5.5,4.2,3.5\) cm — which construction method and why?
A6. Use SSS construction because all three sides are given — draw base then intersect arcs as in SSS steps.
Q7. Define congruence of angles with an example.
A7. Angles are congruent if they have equal measures. Example: \(\angle LMN=40^\circ\) and \(\angle XYZ=40^\circ\Rightarrow \angle LMN\cong\angle XYZ.\)
Q8. Construct a right triangle with hypotenuse \(5\) cm and side \(3\) cm (RHS). Outline steps.
A8. Steps:
- Draw base \(MN=3\) cm.
- At \(M\) draw ray perpendicular to \(MN\).
- With center \(N\) radius \(5\) cm cut the ray at \(L\). Join \(L\) to \(M\) and \(N\).
Q9. When constructing triangle given two angles and included side (ASA), why can two triangles be formed?
A9. Usually ASA gives unique triangle. Two triangles can occur only if rays are drawn on different sides of base (mirror image) — but they are congruent (mirror images).
Q10. Which construction should be used if a shop is equidistant from three houses?
A10. Draw perpendicular bisectors of the triangle formed by the three houses — their intersection (circumcentre) is equidistant from the three houses.
Q11. How to verify perpendicular bisector CD of PS (what to check numerically)?
A11. Verify \(PM=MS\) (midpoint) and \(\angle (CD,PS)=90^\circ\).
Q12. Give the construction steps to make an isosceles triangle with base \(5\) cm and equal sides \(3.5\) cm.
A12. Draw base \(=5\) cm, with each end as centre draw arcs radius \(3.5\) cm; intersection gives apex; join endpoints to apex.
Q13. How to construct an equilateral triangle of side \(6.5\) cm?
A13. SSS with all sides \(6.5\) cm: draw base \(6.5\) cm; arcs from endpoints radius \(6.5\) cm; intersection point is third vertex.
Q14. If two segments are congruent and one is congruent to a third, what property holds?
A14. Congruence is transitive: if \(\overline{AB}\cong\overline{PQ}\) and \(\overline{PQ}\cong\overline{MN}\), then \(\overline{AB}\cong\overline{MN}\).
Q15. Construct \(\triangle\) with \(XY=6, XZ=5, YZ=4\). If arcs intersect at two points what does it mean?
A15. It means two possible triangles symmetric about base \(XY\) — choose either intersection (mirror images).
Q16. How to draw triangle when two sides and included angle are given but angle is obtuse?
A16. Same SAS method: draw base, construct the obtuse angle (use protractor), then mark arc for third vertex.
Q17. What does \(l(AB)=5.3\text{ cm}\) mean?
A17. Length of segment \(AB\) equals \(5.3\) centimetres.
Q18. How will you construct perpendicular from incenter \(I\) to side \(PQ\)?
A18. Using \(I\) as center, draw a ray perpendicular to \(PQ\) or drop perpendicular by constructing right angle at foot point \(A\); use compass to verify distances IA=IB=IC.
Q19. Explain why the perpendicular bisectors are concurrent.
A19. Each perpendicular bisector contains points equidistant from two vertices; intersection point is equidistant from all three vertices, hence lies on all three bisectors.
Q20. Give one quick test to check two angles are congruent using paper tracing.
A20. Trace one angle on transparent paper and place over the other; if rays and vertex align, angles are congruent.
20 × 3-Mark Questions (Longer Reasoning / Full Construction Steps)
Q1. Construct \(\triangle ABC\) with \(AB=5.5\text{ cm}, BC=4.2\text{ cm}, AC=3.5\text{ cm}\). Provide steps and a reason why only one triangle is possible.
A1. Steps:
- Draw base \(AB=5.5\) cm.
- With center \(A\), radius \(AC=3.5\) cm draw an arc.
- With center \(B\), radius \(BC=4.2\) cm draw another arc; intersection gives \(C\).
- Join \(AC\) and \(BC\). Unique because three side lengths determine a unique triangle (SSS) up to reflection; reflection is congruent.
Q2. Construct triangle given \(PQ=7\text{ cm},\ PR=5\text{ cm},\ \angle P=60^\circ\) (SAS variant). Show diagram steps.
A2. Steps:
- Draw \(PQ=7\) cm as base.
- At \(P\) draw ray making \(60^\circ\) with base using a protractor or angle construction.
- With center \(P\) radius \(5\) cm cut the ray to find \(R\). Join \(QR\).
Q3. Construct a triangle with \(YX=6\) cm, \(\angle ZXY=30^\circ\), \(\angle XYZ=100^\circ\) (ASA). Provide the steps and explain why the third angle is determinable.
A3. Steps:
- Draw base \(YX=6\) cm.
- At \(X\) draw ray forming \(30^\circ\) on one side of base.
- At \(Y\) draw ray forming \(100^\circ\) on the same side of base.
- The intersection of the two rays gives \(Z\). Third angle is \(180^\circ-30^\circ-100^\circ=50^\circ\) by triangle angle-sum property.
Q4. Given right triangle data hypotenuse \(5\text{ cm}\), one side \(4\text{ cm}\) (RHS). Construct and show how to verify.
A4. Steps:
- Draw base \(MN=4\) cm (the given side).
- At \(M\) draw a perpendicular ray to base.
- With center \(N\) radius \(5\) cm cut the perpendicular to locate \(L\); \(\triangle LMN\) is right with hypotenuse \(LN=5\).
- Verify using Pythagoras: check \(LM^2 + MN^2 = LN^2\) (numerically: \(3^2+4^2=5^2\) if LM is 3 cm in this specific example). If lengths match, construction is correct.
Q5. Explain construction and proof that incentre is equidistant from triangle sides.
A5. Explanation:
- Incentre \(I\) is intersection of angle bisectors. Drop perpendiculars \(IA,IB,IC\) to sides \(QR,RP,PQ\).
- Because \(I\) lies on bisector of \(\angle P\), \(IA\) is perpendicular to \(QR\) and symmetric; triangle congruence (RHS or angle-side) shows \(IA=IB=IC\).
- Thus incentre is center of the inscribed circle (incircle) touching all three sides.
Q6. Construct a triangle when three vertices are equidistant from a point — which is this point and how to construct it?
A6. That point is the circumcentre \(C\). Construct perpendicular bisectors of two sides; their intersection is \(C\). From \(C\) measure distance to any vertex to get circumradius.
Q7. Draw triangle with \(MA=5.2\text{ cm},\ \angle A=80^\circ,\ AT=6\text{ cm}\). Give construction steps.
A7. Steps:
- Draw segment \(MA=5.2\) cm.
- At \(A\) construct \(80^\circ\) ray.
- With center \(A\) radius \(AT=6\) cm cut the ray at \(T\). Join \(M\) to \(T\).
Q8. Why can two triangles be formed when an arc intersects a ray twice (example earlier with arc from C and ray from B)?
A8. Because the arc may cut the ray in two positions depending on which side of the base the vertex lies, producing two distinct (mirror) triangles satisfying the same data.
Q9. Construct an equilateral triangle and show why all angles are \(60^\circ\).
A9. Steps:
- Draw base \(AB\) of given side.
- With centers \(A\) and \(B\) radius \(AB\) draw arcs; intersection is \(C\).
- By construction \(AC=BC=AB\) → triangle is equilateral ⇒ all angles are \(60^\circ\) by symmetry or angle-sum divided equally.
Q10. Give steps to construct a scalene triangle when you choose side lengths yourself.
A10. Steps (example):
- Choose three different lengths (e.g., \(6,4,5\) cm) satisfying triangle inequality.
- Draw base equal to largest length; use arcs from endpoints with other two lengths to find vertex.
Q11. Construct triangle with \(FU=5\) cm, \(UN=4.6\) cm, \(\angle U=110^\circ\). Steps and comment on obtuse angle handling.
A11. Steps:
- Draw side \(UN=4.6\) cm.
- At \(U\) construct \(110^\circ\) ray (obtuse drawn outward using protractor/angle copy).
- With center \(U\) radius \(5\) cm cut the ray to locate \(F\). Join \(F\) to \(N\).
Q12. Construct triangle with \(RS=5.5\) cm, \(RP=4.2\) cm, \(\angle R=90^\circ\). Steps.
A12. Steps:
- At \(R\) construct right angle; along one leg place \(RP=4.2\) cm, along other leg place \(RS=5.5\) cm, join endpoints to form triangle.
Q13. Construct triangle when two angles and included side are given but angles sum more than \(180^\circ\). What happens?
A13. Impossible — two given angles must sum to less than \(180^\circ\) so the third angle has positive measure. If sum ≥ \(180^\circ\), triangle cannot exist.
Q14. Show how to draw perpendicular from a point on one side of triangle to the opposite side.
A14. Use standard perpendicular construction: with equal radii draw two arcs from the point cutting the line on both sides; join arc intersections to get perpendicular.
Q15. How to find point equidistant from three sides using angle bisectors practically?
A15. Construct bisectors of two angles of triangle; their intersection \(I\) is incentre equidistant from all three sides. Measure perpendicular to any side to obtain inradius.
Q16. Using construction tools, how would you check if \(\triangle\) is isosceles?
A16. Measure two sides with ruler/divider; if two equal, triangle is isosceles. Alternatively, draw perpendicular bisector of base — if it passes through opposite vertex, triangle is isosceles.
Q17. Construct triangle with \(AT=6.4\) cm, \(\angle A=45^\circ, \angle T=105^\circ\) (ASA). Steps and compute third angle.
A17. Steps:
- Draw \(AT=6.4\) cm.
- At \(A\) draw \(45^\circ\) ray; at \(T\) draw \(105^\circ\) ray on same side.
- Intersection gives other vertex. Third angle \(=180-45-105=30^\circ\).
Q18. Explain why the circumcentre can be outside the triangle (give diagram logic).
A18. In obtuse triangle, perpendicular bisectors of sides intersect at a point outside because centre equidistant from vertices lies beyond the obtuse vertex to maintain equal distances.
Q19. What is the construction to draw a triangle with sides \(PQ=4.5\), \(PR=11.7\), and right angle at \(Q\)?
A19. Steps:
- Draw \(PQ=4.5\) cm.
- At \(Q\) construct perpendicular ray; along this mark point so that \(PR\) (distance from \(P\) to that point) becomes \(11.7\) cm by trial arcs — more robust is draw circle center \(P\) radius \(11.7\) cm and intersect perpendicular ray.
Q20. Why must triangle inequality hold when selecting sides? State inequality and give example.
A20. Triangle inequality: for sides \(a,b,c\): \(a+b>c\), \(b+c>a\), \(c+a>b\). Example: 2,2,5 fail since \(2+2\ngtr5\) so cannot form triangle.
Textbook Exercise Questions — Practice Sets (All extracted & solved)
Practice Set 1 — Perpendicular bisectors & Angle bisectors
PS1-Q1. Draw line segments of lengths \(5.3\text{ cm},\ 6.7\text{ cm},\ 3.8\text{ cm}\) and draw their perpendicular bisectors. Where do the points of concurrence lie?
PS1-A1. Solutions (one each):
- For each segment draw arcs from endpoints with radius > half length; join arc intersections to get perpendicular bisector.
- For a single isolated segment, there is no “concurrence” of multiple bisectors unless segments are sides of same triangle. If the segments are treated as sides of a triangle, the perpendicular bisectors meet at the circumcentre which may be inside/outside/on depending on triangle type.
Remark: If you construct a triangle with those three as sides (check triangle inequality), you can find circumcentre by intersecting two perpendicular bisectors.
PS1-Q2. Draw angles \(105^\circ, 55^\circ, 90^\circ\) and their bisectors.
PS1-A2. Steps:
- Draw each angle using protractor.
- For each angle copy half-measure bisector: e.g., bisector of \(105^\circ\) is \(52.5^\circ\) from one arm, of \(55^\circ\) is \(27.5^\circ\), of \(90^\circ\) is \(45^\circ\).
- Use compass-and-straightedge: mark equal arcs on arms and join intersections to vertex to produce bisector.
PS1-Q3. Draw an obtuse-angled triangle and a right-angled triangle. Find points of concurrence of angle bisectors. Where do the points lie?
PS1-A3. Solution:
- Construct the two triangles.
- Draw all three angle bisectors; they meet at the incentre \(I\) (always inside the triangle) — for both obtuse and right triangles the incentre remains inside.
PS1-Q4. Draw a right-angled triangle. Draw perpendicular bisectors of its sides. Where does the point of concurrence lie?
PS1-A4. Solution:
- For a right triangle, the perpendicular bisectors concur at the midpoint of the hypotenuse; this is the circumcentre and lies on the hypotenuse.
PS1-Q5. Maithili, Shaila and Ajay live in three different places. A toy shop is equidistant from the three houses. Which geometrical construction should be used?
PS1-A5. Use perpendicular bisectors of the triangle formed by the three houses. Their intersection (circumcentre) is equidistant from the three vertices and is the shop's location.
Practice Set 2 — SSS constructions
PS2-Q1(a). Draw \(\triangle ABC\) with \(AB=5.5\text{ cm}, BC=4.2\text{ cm}, AC=3.5\text{ cm}\).
PS2-A1(a). Steps:
- Draw base \(AB=5.5\text{ cm}\).
- With centre \(A\), radius \(AC=3.5\) cm draw arc.
- With centre \(B\), radius \(BC=4.2\) cm draw arc; intersection is \(C\). Join \(AC,BC\).
PS2-Q1(b). Draw \(\triangle STU\) with \(ST=7\text{ cm}, TU=4\text{ cm}, SU=5\text{ cm}\).
PS2-A1(b). Steps same as SSS: draw base \(ST\), arcs with radii \(SU\) and \(TU\) to get \(U\).
PS2-Q2. Draw an isosceles triangle with base \(5\) cm and other equal sides \(3.5\) cm.
PS2-A2. Draw base \(=5\) cm; arcs radius \(3.5\) from endpoints; intersection apex; join to get isosceles triangle.
PS2-Q3. Draw an equilateral triangle with side \(6.5\) cm.
PS2-A3. SSS with all sides \(6.5\) cm: draw base then arcs radius \(6.5\) from endpoints to find third vertex.
PS2-Q4. Draw one equilateral, one isosceles and one scalene triangle (student chooses sides).
PS2-A4. Follow SSS/isosceles/equilateral constructions; ensure chosen scalene triple satisfies triangle inequality.
Practice Set 3 — SAS / RHS constructions
PS3-Q1. Draw \(\triangle MAT\) with \(MA=5.2\text{ cm}, \angle A=80^\circ, AT=6\text{ cm}\).
PS3-A1. Steps:
- Draw \(MA=5.2\text{ cm}\).
- At \(A\) construct \(80^\circ\) ray.
- With center \(A\) radius \(6\) cm cut the ray to get \(T\). Join \(M\) to \(T\).
PS3-Q2. Draw \(\triangle NTS\) with \(\angle T=40^\circ\) and \(NT=TS=5\text{ cm}\) (isosceles with vertex at T).
PS3-A2. Steps:
- Draw base \(NS\) by first locating \(N\) and constructing isosceles around \(T\): draw \(NT=5\) and \(TS=5\) from \(T\). Practically draw \(NT=5\) and place \(T\) so \(\angle T\) becomes \(40^\circ\) between the equal sides.
- Construct by first drawing a ray for one equal side then marking other equal side with angle constraint.
Note: alternative simpler method: construct isosceles by drawing base and perpendicular bisector to locate apex at distance satisfying given angle.
PS3-Q3. Draw \(\triangle FUN\) with \(FU=5\text{ cm}, UN=4.6\text{ cm}, \angle U=110^\circ\).
PS3-A3. Use SAS method: draw \(UN\), at \(U\) form \(110^\circ\) ray, with center \(U\) radius \(5\) cm locate \(F\).
PS3-Q4. \(\triangle PRS\) with \(RS=5.5\), \(RP=4.2\), \(\angle R=90^\circ\). Construct and show steps.
PS3-A4. At \(R\) erect perpendicular axes; place \(RP\) and \(RS\) on legs to get right triangle; join endpoints.
Practice Set 4 — ASA constructions
PS4-Q1. Construct \(\triangle SAT\) with \(AT=6.4\text{ cm}, \angle A=45^\circ, \angle T=105^\circ\).
PS4-A1. Draw base \(AT\); at \(A\) draw \(45^\circ\) ray, at \(T\) draw \(105^\circ\) ray on same side; intersection is third vertex.
PS4-Q2. Construct \(\triangle MNP\) with \(NP=5.2\text{ cm},\ \angle N=70^\circ,\ \angle P=40^\circ\).
PS4-A2. Same ASA steps; third angle = \(180-70-40=70^\circ\) so triangle will be isosceles with equal base angles.
PS4-Q3. Construct \(\triangle EFG\) with \(EG=6\text{ cm}, \angle F=65^\circ, \angle G=45^\circ\).
PS4-A3. Draw base \(EG\). Construct given angles at \(E\) and \(G\) and take intersection as vertex \(F\).
PS4-Q4. Construct \(\triangle XYZ\) with \(XY=7.3\text{ cm},\ \angle X=34^\circ,\ \angle Y=95^\circ\).
PS4-A4. ASA steps; third angle \(=180-34-95=51^\circ\).
Practice Set 5 — Right triangle / RHS constructions
PS5-Q1. Construct \(\triangle MAN\) with \(\angle MAN=90^\circ,\ AN=8\text{ cm},\ MN=10\text{ cm}\).
PS5-A1. Steps:
- Draw base \(MN=10\) cm.
- At \(M\) (or at the appropriate vertex) construct right angle; along one leg measure \(AN=8\) if labeling corresponds; place vertices accordingly to satisfy given sides and right angle.
Note: assign vertices carefully according to given notation to ensure correct side placements.
PS5-Q2. Right triangle with hypotenuse \(SU=5\) and side \(ST=4\): construct \(\triangle STU\).
PS5-A2. Draw base \(ST=4\). At \(T\) erect perpendicular; with center \(S\) radius \(5\) intersect perpendicular to get \(U\).
PS5-Q3. Construct \(\triangle ABC\) with \(AC=7.5\text{ cm},\ \angle ABC=90^\circ,\ BC=5.5\text{ cm}\).
PS5-A3. Draw \(BC=5.5\) cm; at \(B\) erect perpendicular; with center \(C\) radius \(7.5\) cut perpendicular to find \(A\).
PS5-Q4. Construct \(\triangle PQR\) with \(PQ=4.5\) cm, \(PR=11.7\) cm, and \(\angle PQR=90^\circ\). Steps.
PS5-A4. Draw \(PQ=4.5\). At \(Q\) construct perpendicular; draw circle center \(P\) radius \(11.7\) cm; intersection with perpendicular gives \(R\).
Practice Set 6 & 7 — Congruence of segments, angles and circles
PS6-Q1. From a figure, write pairs of congruent segments (example given in book).
PS6-A1. Use divider to compare lengths; pairs with equal measures are congruent. Example answers depend on figure labels: (i) AB ≅ DC, (ii) AE ≅ BH, etc. — verify using divider.
PS6-Q2. On a line with equally spaced points \(Q,P,Z,Y,X,W,A,B,C\), fill blanks like \(\overline{AB}\cong\ ?\).
PS6-A2. If spacing between adjacent points is equal, then \(\overline{AB}\cong\overline{BC}\), \(\overline{AP}\cong\overline{PZ}\) etc. (match using positions).
PS7-Q1. Using given angle figures identify congruent angles.
PS7-A1. Angles with same numerical measures are congruent (e.g. \(45^\circ\) with \(45^\circ\), \(90^\circ\) with \(90^\circ\), etc). Use tracing method for verification.
PS7-Q2. Which circles among radii \(1\), \(2\), \(1\), \(1.3\) cm are congruent?
PS7-A2. Circles with same radius 1 cm are congruent (figs (a) and (c) in book). Others differ.