Chapter 4 — Constructions of Triangles Class 9 • Maharashtra Board • Part 2
20 Most Important 1-Mark Questions & Solutions
Q1. State perpendicular bisector theorem (part I).
Every point on the perpendicular bisector of a segment is equidistant from its endpoints.
Q2. State perpendicular bisector theorem (part II).
If a point is equidistant from the endpoints of a segment then it lies on the perpendicular bisector.
Q3. How many data items needed to construct a triangle uniquely?
Three independent conditions (e.g., SSS, SAS, ASA, RHS).
Q4. If base \(BC=6.3\) cm, \(\angle B=75^\circ\) and \(AB+AC=9\) cm — point \(D\) was taken on ray making \(75^\circ\) at \(B\) so that \(BD=9\). Which locus contains vertex \(A\)?
Intersection of ray \(BD\) and the perpendicular bisector of \(CD\).
Q5. For difference construction (AB−AC=3 cm), where is vertex A located?
Intersection of ray from \(B\) and perpendicular bisector of segment joining constructed point \(D\) and \(C\) (as in method).
Q6. If perimeter \(=11.3\) cm and \(\angle B=70^\circ,\angle C=60^\circ\), what is \(\angle APB\) when PQ built as in book?
\(\angle APB=\angle PAB=35^\circ\) (half of \(70^\circ\)).
Q7. In the sum-construction method we first draw PQ = perimeter. Why?
Because \(PQ=AB+BC+CA\) using points \(P,B\) and \(Q,C\) construction simplifies locating A then B and C.
Q8. When constructing triangle with base and two adjacent angles and perimeter known, which two perpendicular bisectors do we draw?
Perpendicular bisectors of \(AP\) and \(AQ\) (with \(P\!Q=\) perimeter) to locate \(B\) and \(C\).
Q9. In Construction I (sum), why point D is on the ray making given angle?
Because BD is set equal to \(AB+AC\); A must lie on ray BD to have BA+AC equal to BD.
Q10. Which basic ruler-compass step constructs an angle \(75^\circ\)?
Use protractor; with compass only: construct \(60^\circ\) (equilateral) and add \(15^\circ\) (bisect \(30^\circ\)), etc.
Q11. If in construction II \(BD=AB-AC\) and \(BD=3\) cm, what relation do you use to get AD = AC?
Since \(AB-AC=BD\) and \(AB=AD\) (constructed), \(AD=AC\).
Q12. For \(PQ=11.3\) and angles \(35^\circ\) at \(P\) and \(30^\circ\) at \(Q\), which triangle do we construct to locate A?
Construct \(\triangle PAQ\) with known included side \(PQ\) and adjacent angles \(35^\circ,30^\circ\) (ASA).
Q13. Which construction uses the fact "point on two lines is their intersection"?
All these methods: locate A as intersection of ray and perpendicular bisector, etc.
Q14. For difference-type construction if AC−AB = 3, which ray is used for locating D?
D is taken on the opposite ray (from B) such that BD = 3 so that AD = AC.
Q15. In sum-method step (2) we draw ray at given angle at B. Why ray not line?
Because A must be on one side of BC consistent with the given orientation; ray sets the direction where BD (sum) is marked.
Q16. Which three classical triangle construction criteria guarantee uniqueness?
SSS, SAS, ASA (also RHS for right triangles).
Q17. In Construction III, why do B and C lie on perpendicular bisectors of AP and AQ respectively?
Because \(PB=BA\) and \(CQ=CA\), so B is equidistant from \(A\) and \(P\), so on perpendicular bisector of \(AP\).
Q18. If you draw PQ=perimeter and find A at intersection of rays, what is next to find B?
Construct perpendicular bisector of \(AP\) and intersect it with line \(PQ\) to get \(B\).
Q19. What is the usual tolerance rule when marking lengths with compass for school constructions?
Use accurate ruler and compass; mark points without changing compass width; small tolerances acceptable up to 1 mm in schoolwork.
Q20. Name three types of "sum/difference/perimeter" constructions covered in this chapter.
(I) Base + adjacent angle + sum-of-other-sides, (II) Base + adjacent angle + difference-of-other-sides, (III) Base + adjacent angles + perimeter.
20 Most Important 2-Mark Questions & Solutions
Q1. Construct (conceptually) triangle \(ABC\) when \(BC=6.3\), \(\angle B=75^\circ\) and \(AB+AC=9\). Write construction steps concisely.
Steps: (1) Draw \(BC=6.3\). (2) At \(B\) draw ray making \(75^\circ\). (3) On ray mark \(D\) with \(BD=9\). (4) Draw \(DC\). (5) Construct perpendicular bisector of \(DC\). (6) Intersection of ray \(BD\) and that bisector is \(A\). Join \(AB,AC\).
Q2. Explain why intersection of ray \(BD\) and perpendicular bisector of \(DC\) gives vertex \(A\) for Construction I.
Point \(A\) must satisfy \(BA+AC=BD\). With \(AD=AC\), perpendicular bisector of \(DC\) ensures \(AD=AC\). Intersection enforces both direction and distance relation, giving unique \(A\).
Q3. For Construction II (difference), give steps when \(BC=7.5,\ \angle B=40^\circ,\ AB-AC=3\).
Draw \(BC=7.5\). At \(B\) draw ray making \(40^\circ\). On that ray mark \(D\) with \(BD=3\). Construct perp bisector of \(DC\). Intersection with ray is \(A\). Join to get triangle.
Q4. For Construction II variant where \(AC-AB=3,\ BC=7\), what changes?
Take \(D\) on opposite ray from \(B\) with \(BD=3\) so that \(AD=AC\). Then perpendicular bisector of \(DC\) intersect ray gives \(A\).
Q5. In Construction III, if \(\angle B=70^\circ,\ \angle C=60^\circ,\) perimeter \(=11.3\), show how to find \(\angle P\) and \(\angle Q\) at points on PQ.
Split: \(\angle APB=\angle PAB=35^\circ\) (half of \(70^\circ\)). Similarly \(\angle AQC=\angle QAC=30^\circ\) (half of \(60^\circ\)). These are used to construct triangle \(PAQ\).
Q6. Why in Construction III we use perpendicular bisectors of \(AP\) and \(AQ\) to locate \(B\) and \(C\)?
Because \(PB=BA\) so \(B\) is equidistant from \(A\) and \(P\); hence \(B\) lies on perp bisector of \(AP\). Similarly for \(C\).
Q7. Solve: Construct triangle with \(YZ=4.2\), \(\angle Y=40^\circ\), \(XY+XZ=8.5\) — what length do you mark on ray?
Mark \(D\) on ray at \(Y\) with \(YD=XY+XZ=8.5\). Then perpendicular bisector method like Construction I: intersection is \(X\).
Q8. Give concise reason why these constructions produce unique triangle.
Loci intersection: ray fixes direction, perpendicular bisector fixes locus of points equidistant; intersection of two well-defined loci is unique (typically one point), so triangle is unique up to orientation.
Q9. If in Construction I BD accidentally equals less than AB+AC, what happens?
No solution — intersection won't satisfy distances; geometrically, perpendicular bisector and ray won't meet at a point satisfying required relations (or they meet but not giving required sum).
Q10. In practice set 4.2 problem 4: BC=5.2, \(\angle C=45^\circ\) and perimeter = 10. How to proceed?
PQ = perimeter = 10; put \(P,Q\) such that PQ=10. Put angles at P and Q as half of given adjacent angles to get A; then perp bisectors give B and C on PQ. (Follow Construction III).
Q11. How to bisect an angle accurately with compass and straightedge?
From vertex draw equal arcs to meet both sides; with same radius draw arcs from those two points; connect intersection with vertex — it bisects the angle.
Q12. In Construction II, why do we choose D on opposite ray for AC−AB = 3?
Because sign of difference matters: to make AD = AC we need AB < AC or vice versa; placing D on opposite ray changes relation AD=AC appropriately.
Q13. If perimeter ratio is given (Problem set 4.3 Q3: ratio 2:3:4 and perimeter 14.4), how to compute side lengths?
Let common \(=k\). Then \(2k+3k+4k=9k=14.4\Rightarrow k=1.6\). Sides: \(3.2,4.8,6.4\) cm respectively. Then construct via standard SSS.
Q14. In sum-construction, if \(BD\) equals exactly AB+AC, what can you say about A, D and C?
Point \(A\) lies between \(B\) and \(D\) on the ray and equidistant condition yields perpendicular bisector passing through A for segment DC.
Q15. For the student's accuracy: name three tips to make constructions neat.
Use sharp pencil, steady compass without changing span, draw faint construction lines and then darken final segments.
Q16. Construct concept: Given base and two adjacent angles and perimeter, why "PQ" is chosen equal to perimeter?
Because moving B and C to points on line PQ with PB=BA and CQ=AC converts perimeter into a single straight segment PQ (PB+BC+CQ) equal to required perimeter, simplifying construction.
Q17. If you have to construct triangle with PQ=11.3 and angles at P and Q known, what triangle type is PAQ (ASA)?
Yes, \(\triangle PAQ\) is ASA (two angles and included side PQ known), so A is uniquely determined.
Q18. For Construction I example numbers, verify triangle existence: BC=6.3, BD=9. Does triangle exist?
Yes — as long as perpendicular bisector of DC meets ray BD at a point that gives positive distances; typical numeric values given are valid for geometry exercise.
Q19. When using protractor to draw 40° at B, what small check assures correctness?
Measure angle using protractor both sides (clockwise and anticlockwise) or verify with constructed 60° minus 20° or by bisecting known angles to cross-check.
Q20. Short reason: Why is Construction III sometimes done with Geogebra differently?
Digital tools can perform locus intersections and algebraic placement directly, offering alternate but equivalent methods (e.g., solve for B and C coordinates rather than perpendicular bisectors).
20 Most Important 3-Mark Questions & Solutions
Q1. Fully explain Construction I (base, adjacent angle, sum of other two sides) with reason why each step works.
Steps: Draw base \(BC\). At \(B\) draw ray making given angle. Mark \(D\) on ray with \(BD=AB+AC\). Draw \(DC\). Construct perpendicular bisector of \(DC\). Intersection \(A\) of bisector and ray satisfies \(AD=AC\) (bisector) and \(BD=AB+AC=AB+AD\Rightarrow AB=BD-AD\) which matches original requirement. Join to form triangle. Uniqueness: intersection of two loci (ray and perp bisector) gives unique point.
Q2. Prove correctness of Construction II (difference type) for \(AB-AC=d\).
If \(BD=d\) and A is on ray from \(B\) so that \(BD=AB-AC\), then \(AB=AC+BD\Rightarrow AD=AC\) (since \(AD=AB-BD\)). So \(A\) must be on the perp bisector of \(DC\). Intersection of ray and that bisector gives \(A\) satisfying required relation.
Q3. Work full steps for Construction III (perimeter + adjacent angles) and justify why \(PB=BA\) and \(CQ=CA\).
Draw \(PQ= \) perimeter. At \(P\) draw ray inside with angle \(=\tfrac12\angle B\) (i.e., \(\angle APB=\angle PAB\)). At \(Q\) draw ray with angle half of \(\angle C\). Their intersection is \(A\). Because \(PAQ\) constructed with given angles, reflecting that \(PB\) is congruent to \(BA\) (PB constructed on PQ to represent AB), B is point on perp bisector of \(AP\) (equidistance). So B found as intersection of perp bisector and PQ, similarly C.
Q4. Demonstrate with algebra how to split a given perimeter among three sides in problem set 4 Q3 (ratio 2:3:4 and perimeter 14.4) and then how you'd construct.
Compute \(k\) from \(2k+3k+4k=9k=14.4\Rightarrow k=1.6\). Sides: \(3.2,4.8,6.4\). Construction: use SSS — draw one side \(6.4\), use compass radii 3.2 and 4.8 from endpoints to get third vertex intersection.
Q5. For Construction I, show algebraically that BD=AB+AC implies AD=AC if A lies on perp-bisector of CD.
If A lies on perp bisector of CD, then \(AD=AC\). Given \(BD=AB+AC\). But \(AB=AD\) only when A chosen such that \(AD=AB\) — in the scheme we choose A to satisfy both relations; with perpendicular bisector giving \(AD=AC\), substituting: \(BD=AB+AD\Rightarrow AB=BD-AD\) consistent with geometrical placement.
Q6. Construct and justify triangle when base=7 cm, \(\angle B=40^\circ\), \(AC-AB=3\) (ex. in book). Summarize differences vs previous case.
Draw \(BC=7\). Draw ray at \(B\). On opposite ray mark \(D\) with \(BD=3\). Construct perp bisector of \(DC\); intersection with ray gives \(A\) (so \(AD=AC\)). Differences: sign of difference changed position of D (opposite ray).
Q7. For practice set 4.3 Q1 (angles 70° and 80° and perimeter 9.5), outline the full construction and give one check to confirm correctness.
Draw \(PQ=9.5\). At \(P\) put half of remaining angle? Actually compute \(\angle APQ=\tfrac12(180^\circ-\angle Q-\angle R)\) etc — follow Construction III: build PAQ with \(\angle APQ=\tfrac12\angle Q\), \(\angle AQP=\tfrac12\angle R\). Construct perp bisectors to find Q,P positions of B and C. Check: compute distances \(AB+BC+CA\) with compass; should equal 9.5.
Q8. Why perpendicular bisector is central in all three constructions?
Because the bisector encodes "equidistance" relations (AD=AC or PB=BA etc.), which transform sum/difference/perimeter constraints into simple locus intersections (line vs bisector).
Q9. Show one worked numeric check: Construction I example (BC=6.3, BD=9). After constructing A via method, compute AB+AC numerically (conceptually) to confirm 9.
Measure AB and AC with compass; the sum equals BD by construction. (Numerically: if AB=3.2 and AC=5.8 then sum = 9.0 matching BD.)
Q10. Problem set 4 Q4 (PQ−PR = 2.4, QR=6.4, \(\angle PQR=55^\circ\)): sketch method.
Treat as difference-type with base \(QR\) and difference \(PQ-PR\) given. Draw \(QR\), ray at \(Q\) at \(55^\circ\); mark point \(D\) on appropriate ray with BD=2.4; perp bisector of DC intersection yields P.
Q11. Explain how to handle ambiguous orientation when given angles adjacent to base (which side to put A — above or below BC)?
Choose orientation per diagram instruction; both give congruent triangles mirrored across BC; usually book fixes A on one side; using ray (not full line) fixes side explicitly.
Q12. For Construction III explain why \(\angle APB = \angle PAB\) (they both equal half of given angle at B).
Because PB is chosen as mirror representation of AB on PQ: triangle \(APB\) is isosceles with \(PB=BA\) so base angles equal; using exterior-angle reasoning gives each = half of given \(\angle B\).
Q13. In difference construction if AB
Switch which side is considered minuend/subtrahend; place D on opposite ray so BD equals magnitude of difference and proceed accordingly (choose orientation so BD positive).
Q14. Using triangle inequality show why some combinations of base and sum/difference may be impossible.
Triangle inequality: each side < sum of other two. If given sum/difference leads to violation (e.g., BD too small or too large), construction impossible — check numerically before constructing.
Q15. For construction accuracy: how to mark D on ray precisely with compass?
Set compass to required length BD on ruler, then with center at B draw an arc on the ray to locate D without changing span.
Q16. Prove uniqueness for Construction III (perimeter + base angles): why points B and C are unique?
Once A constructed (ASA) with PQ fixed, perpendicular bisectors of AP and AQ intersect PQ at unique points B and C respectively (perp bisector crosses a line at most once), so B and C unique.
Q17. For Construction III give a short alternate method using similarity or algebra (if you know coordinates).
Place PQ on x-axis with P at 0 and Q at perimeter \(s\); compute coordinates of A solving two ASA angle constraints; then B and C are intersections of perpendicular bisectors — algebraic intersection gives coordinates directly (Geogebra does this).
Q18. When constructing triangles from SSS, SAS etc., which instrument is must and why?
Compass and straightedge: compass for copying lengths and arcs; straightedge for drawing straight lines and rays precisely.
Q19. Give one short proof that perpendicular bisector passes through circumcenter for a triangle.
Perp bisector of two sides meet at point equidistant from endpoints of both sides — that's circumcenter; equidistant from all three vertices so it's center of circumscribed circle.
Q20. Student question: When constructing with given perimeter and angles, how to check final triangle perimeter quickly?
Use compass to transfer AB and AC to measure along PQ or sum AB+BC+CA by stepping with compass and compare with original perimeter PQ; they must match.
All Textbook Exercise Questions — Perfect Solutions
Construction I — Examples & Practice (solved)
Ex: Construct \(\triangle ABC\) where \(BC=6.3\) cm, \(\angle B=75^\circ\), \(AB+AC=9\) cm — solution (concise and exact):
- Draw \(BC\) of length \(6.3\) cm.
- At \(B\), draw ray \(BP\) making \(75^\circ\) with \(BC\) (use protractor).
- On ray \(BP\) mark point \(D\) such that \(BD=9\) cm (use compass set to 9 cm and mark).
- Join \(D\) to \(C\) to get segment \(DC\).
- Construct perpendicular bisector of \(DC\): draw arcs from \(D\) and \(C\) with same radius \(> \tfrac12 DC\), join intersection points of arcs to get bisector.
- Name intersection of bisector and ray \(BP\) as \(A\).
- Join \(A\) to \(B\) and \(C\). Then \(AB+AC=BD=9\) by construction and \(BC=6.3\), \(\angle B=75^\circ\) satisfied.
Practice set 4.2 — Solutions
Q1. Construct \(\triangle PQR\) with \(QR=4.2\) cm, \(\angle Q=40^\circ\), \(PQ+PR=8.5\) cm — steps:
Same pattern as Construction I: draw \(QR=4.2\). At \(Q\) draw ray making \(40^\circ\). Mark \(D\) on ray with \(QD=8.5\). Join \(D\) to \(R\). Construct perpendicular bisector of \(DR\). Intersection of ray and that bisector is \(P\). Join to get triangle.
Q2. Construct \(\triangle XYZ\) with \(YZ=6\), \(XY+XZ=9\), \(\angle XYZ=50^\circ\) — steps:
Draw \(YZ=6\). From \(Y\) draw ray at \(50^\circ\). Mark \(D\) on ray with \(YD=9\). Join \(DR\) where \(R\) is \(Z\) (rename). Construct perp bisector of \(DZ\). Intersection gives \(X\). Join to obtain \(\triangle XYZ\).
Q3. Construct \(\triangle ABC\) where \(BC=6.2\), \(\angle ACB=50^\circ\), \(AB+AC=9.8\) — steps:
Same as Ex: draw \(BC=6.2\); at \(C\) draw ray making \(50^\circ\); mark \(D\) with \(CD=9.8\); construct perp bisector of \(DB\); intersection with ray is \(A\); join. (Note: ensure correct vertex used for given adjacent angle.)
Q4. Construct \(\triangle ABC\) with \(BC=5.2,\ \angle ACB=45^\circ\) and perimeter \(10\) — approach:
Use Construction III: draw \(PQ=10\). At the point corresponding to C put half-angles accordingly (compute half of adjacent angles to set rays). Construct \(PAQ\) as ASA then perp bisectors yield \(B\) and \(C\) on PQ. (Numeric specifics depend on which adjacent angles are given in actual problem statement.)
Construction II — Examples & Practice (solved)
Ex: Construct \(\triangle ABC\) with \(BC=7.5\), \(\angle ABC=40^\circ\), \(AB-AC=3\).
Steps:
- Draw \(BC=7.5\) cm.
- At \(B\) draw ray \(BL\) making \(40^\circ\) with \(BC\).
- On ray \(BL\) mark \(D\) such that \(BD=3\) cm.
- Join \(DC\). Construct perpendicular bisector of \(DC\).
- Intersection of bisector and ray \(BL\) is \(A\). Join \(A\) to \(B\) and \(C\).
Practice set 4.2 additional answers (concise):
- \(\triangle XYZ\): \(YZ=7.4,\ \angle XYZ=45^\circ,\ XY-XZ=2.7\) — draw \(YZ\), ray at \(Y\), mark \(D=2.7\) on ray, perp bisector of \(DZ\) gives \(X\).
- \(\triangle PQR\): \(QR=6.5,\ \angle PQR=60^\circ,\ PQ-PR=2.5\) — similar: draw \(QR\), at \(Q\) ray, locate D with 2.5 on opposite ray if sign needed, get \(P\) as intersection.
- \(\triangle ABC\): \(BC=6,\ \angle ABC=100^\circ,\ AC-AB=2.5\) — same difference-type procedure with orientation per sign of difference.
Construction III — Examples & Practice (solved)
Ex: Construct \(\triangle ABC\) such that \(AB+BC+CA=11.3\), \(\angle B=70^\circ,\ \angle C=60^\circ\) — solution:
- Draw \(PQ=11.3\) cm.
- At \(P\) draw ray making \(35^\circ\) with \(PQ\) (since \(\angle APB=\angle PAB=35^\circ\)).
- At \(Q\) draw ray making \(30^\circ\) (since \(\angle AQC=\angle QAC=30^\circ\)).
- Intersection of those rays is \(A\).
- Construct perpendicular bisector of \(AP\); its intersection with line \(PQ\) is \(B\) (since \(PB=BA\)).
- Construct perpendicular bisector of \(AQ\); intersection with \(PQ\) is \(C\) (since \(CQ=CA\)).
- Join \(AB,AC,BC\). Completed triangle satisfies the perimeter condition because \(PQ=PB+BC+CQ = AB+BC+AC\).
Practice set 4.3 answers (concise):
- \(\triangle PQR\) with \(\angle Q=70^\circ,\ \angle R=80^\circ,\) perimeter \(9.5\): Use Construction III with PQ=9.5, draw half-angles at P and Q, intersection gives A, then perpendicular bisectors locate B and C.
- \(\triangle XYZ\) with \(\angle Y=58^\circ,\ \angle X=46^\circ\), perimeter 10.5 — same method.
- \(\triangle LMN\) with \(\angle M=60^\circ,\ \angle N=80^\circ\), perimeter 11 — same method (PQ=11 etc.).
Problem Set 4 — Solutions
Q1. Construct \(\triangle XYZ\) such that \(XY+XZ=10.3\), \(YZ=4.9\), \(\angle XYZ=45^\circ\).
Use Construction I: Draw \(YZ=4.9\), at \(Y\) draw ray \(45^\circ\), mark \(D\) with \(YD=10.3\), join \(DR\) where \(R=Z\), construct perp bisector of \(DZ\), intersection with ray gives \(X\).
Q2. Construct \(\triangle ABC\) with \(\angle B=70^\circ,\ \angle C=60^\circ,\) and perimeter \(11.2\).
Same as Construction III with \(PQ=11.2\): draw rays at P (\(35^\circ\)) and Q (\(30^\circ\)), intersection A, perp bisectors of \(AP\) and \(AQ\) intersect PQ at \(B,C\) respectively.
Q3. Perimeter 14.4 and side ratio \(2:3:4\) — construct the triangle.
Compute \(k=14.4/9=1.6\). Sides are \(3.2,4.8,6.4\). Construct triangle with SSS: draw largest side \(6.4\), from its endpoints draw arcs radius 3.2 and 4.8; intersection is third vertex.
Q4. Construct \(\triangle PQR\) with \(PQ-PR=2.4,\ QR=6.4,\ \angle PQR=55^\circ\).
Draw \(QR=6.4\). At \(Q\) draw ray \(55^\circ\). Mark D on ray with \(QD=2.4\) (direction depending on which is larger). Join \(DR\). Perp bisector of \(DR\) intersects ray \(Q\) giving \(P\). Join to finish.
Final notes & study tips: Always read orientation (which adjacent angle is given at which vertex). Draw faint construction arcs first; label every step; verify final distances with compass. For numeric checks use triangle inequality before starting. For digital practice use GeoGebra.