Chapter 8: Play with Constructions
Answer Key with Explanations
Section A — MCQs (Answers + Brief Explanations)
b) Compass. Draws circles/arcs with fixed radius.
d) 180°. Straight angle equals 180°.
c) Ruler and compass. Standard method for bisecting segments.
b) 2 equal parts at 90°. Definition of perpendicular bisector.
d) Parallel. Remain equidistant, never meet.
b) 2,3,6. 2+3=5 < 6 violates triangle inequality.
b) Bisect at right angles. Rhombus diagonals are ⟂ and bisect.
c) 60°. 120° is doubled 60° at vertex (or 180°−60°).
b) Hexagon. Step radius around circle → 6 equal chords.
b) Divider. Transfers/compares distances.
b) 2. Two arcs from endpoints locate the third vertex.
b) Compass and ruler. Classical construction.
a) 4 equal sides and 4 right angles. Definition.
d) 77°. 30°, 45°, 75° are constructible by bisection/addition; 77° isn’t by basic steps.
a) From both arms. Intersecting arcs from each arm locate the bisector.
b) 60°. Property of equilateral triangles.
c) Rhombus. Perpendicular, bisecting diagonals define a rhombus (square needs equal diagonals too).
d) All of these. Multiple polygons can be inscribed; hexagon by stepping radius, square via diameters, etc.
b) 2.5 cm. Half of 5 cm.
b) Triangle inequality. Must hold before construction.
Section B — Assertion & Reasoning (Key + Justification)
(b) Both true; R describes a property but not the construction method.
(b) Both true; equal sides don’t explain perpendicular-bisecting diagonals.
(d) A false; R true (inequality rule shows 2,3,6 impossible).
(d) A false (protractor not essential); R true (60° then bisect).
(c) A true; R false (rectangle does not have all sides equal).
(a) Both true; R explains use of the perpendicular bisector.
(a) Both true; R explains the method (six equal chords).
(b) Both are true; R states a fact, not “why” A is true.
(b) Both true; R is an incorrect statement (protractor not necessary) → so actually (a/b?)
Correct choice: (b) Both true but R not explanation and R is false—If your pattern requires one truth value for R, mark (c) A true, R false.
(Use (c): A true, R false.)(d) A false (square diagonals are equal); R true (but irrelevant).
(d) A false (rhombus need not have right angles); R true (its diagonals are ⟂).
(d) A false (without compass not standard); R true (compass is standard).
(c) A true (one pair parallel); R false (parallelogram has two pairs).
(a) Both true; R explains A (90+60+30=180).
(a) Both true; R explains parallelogram property.
(a) Both true; R is the construction method.
(d) A false (bisector makes two equal parts); R true (definition of bisector).
(a) Both true; R explains how set-square gives a right angle.
(c) A true? No—A says “divider measures angles” → false; R true. So (d).
(d) A false (infinitely many diameters), R true.
Note on Q9 above: To avoid any ambiguity for students, evaluate as (c) A true, R false (A is correct; R wrongly claims protractor is necessary).
Section C — True/False (with Reason)
False. Compass doesn’t measure angles; protractor does.
True. Definition of perpendicular bisector.
True. Regular hexagon via stepping the radius.
True. Equality case fails triangle construction.
False. Rhombus diagonals are ⟂ but not always equal (equal in square).
True. Square ⊂ rectangle.
False. 45° is constructible by bisecting 90°.
True. Parallel via equal corresponding angles.
True. In an acute triangle, the circumcentre lies inside.
True. Divider transfers lengths.
Section D — Short Answer I (2 marks) — Suggested Steps
Perpendicular bisector (7 cm): Draw AB=7 cm. With radius > 3.5 cm, arcs from A and B intersect at P,Q. Join PQ. PQ ⟂ AB at midpoint.
60°: On line β with point O, draw an arc cutting β at A; with same radius from A cut arc at B; join OB → 60°.
30°: Make 60° first, then bisect the 60° angle.
Equilateral (5 cm): Draw AB=5 cm. Arcs radius 5 cm from A and B meet at C. Join AC, BC.
Perpendicular at a point on line: From point P on line, mark equal arcs left/right on the line; from those points draw equal arcs above; join their intersection to P.
Perpendicular from external point: From point P, draw arcs to cut line at X and Y; from X and Y draw equal arcs; join their intersection to P.
90°: Construct 60°, then construct its external 120°, bisect 60° and 120° to get 90°. (Or use the semicircle theorem.)
Bisect ∠XYZ: With centre at Y, draw an arc to cut YX, YZ at A,B; with equal radius from A and B draw arcs meet at C; join YC.
Parallel through a point: Copy angle method: recreate an equal corresponding angle at the given point → line parallel to the original.
Triangle 4–5–6 (SSS): Base 6 cm, arcs radius 4 cm and 5 cm from ends meet at third vertex; join.
120°: Make 60°, then construct an external 60° on the other side at the same vertex → 120°.
Square (4 cm): AB=4 cm; erect ⟂ at A and B; mark AD=4 cm, BC=4 cm; join CD.
Rectangle 6×4: AB=6 cm; at A construct 90° to mark AD=4 cm; through D draw line ∥ AB; through B draw line ∥ AD; intersect at C.
Divide 9 cm into 3 equal parts: Mark equal step lengths on a ray from an endpoint using compass (3 steps); join last step to other endpoint; draw parallels through intermediate step points.
Equilateral in a circle: Draw circle with centre O; pick A on circle; step off radius around circle to locate B and C; join AB, BC, CA.
Section E — Short Answer II (3 marks) — Steps + Verification/Reason
Perp. bisector (6 cm) + verify: As in D1 with AB=6 cm; check halves are 3 cm and angle is 90° (protractor or paper-folding).
Parallel through external point P: Copy angle of the given line at P to make equal corresponding angles → lines are parallel by the Corresponding Angles Postulate.
Equilateral in circle (r=3 cm): Draw circle; choose A; step radius to B and C; join.
Reason: Equal chords subtend equal arcs; sides equal ⇒ equilateral.
Isosceles (base 5, equal 6): AB=5; arcs radius 6 from A and B meet at C; join. Angles: ∠A = ∠B (base angles equal).
Bisect 90°: Construct 90° at O; bisect to get 45°. Check: Two 45° angles form 90°.
Rhombus (7 and 9 diagonals): Draw perpendicular diagonals intersecting at O with halves 3.5 and 4.5; join endpoints. Reason: Each side is hypotenuse of the right Ξ with legs 3.5 and 4.5 → all equal.
Square in circle (r=4): Draw circle; draw a diameter AB; construct ⟂ diameter CD; join A–B–C–D cyclically. Why 90°: Central angles subtended by quarter circles are 90°.
Trapezium (∥ sides 7 and 5): Draw base 7 cm; through a point draw line ∥ base; mark 5 cm segment on the parallel; complete non-parallel sides.
Equilateral 4.5 cm: As D4 with side 4.5 cm. Angles: Equal sides ⇒ all interior angles 60°.
Hexagon in circle (r=2.5): Step the radius around circle 6 times; join adjacent points. Reason: Each side equals radius ⇒ regular.
Section F — Long Answer (5 marks) — Full Steps + Explanation
Rhombus (diagonals 6 & 8):
Draw lines AO=3 cm and BO=4 cm perpendicular at O; extend to full diagonals AC=6, BD=8 with A–C and B–D opposite ends.
Join A–B–C–D cyclically.
Proof of equal sides: Each side (e.g., AB) is hypotenuse of right triangle with legs 3 and 4 ⇒ AB=√(3²+4²)=5 cm; similarly for all sides ⇒ rhombus.
Triangle 5–6–7 (SSS):
Draw base BC=7 cm.
With centres B and C draw arcs radii 5 cm and 6 cm; intersect at A.
Join AB, AC.
Justify: Triangle inequality holds; SSS construction fixes a unique triangle up to congruence.
Square side 6 cm; verify diagonals:
AB=6; erect ⟂ at A and B; mark AD=6, BC=6; join DC.
Verify: Measure AC and BD; both equal and bisect each other at right angles (properties of a square).
Rectangle 7×4:
AB=7; at A erect 90° to mark AD=4; draw through D line ∥ AB; through B line ∥ AD; intersect at C.
Show: Opposite sides equal by construction; parallels by equal corresponding angles.
Equilateral 6 cm:
AB=6; arcs radius 6 from A and B meet at C; join.
Explain: Rotational symmetry and SSS ⇒ all angles 60°.
Angle 75° (compass only):
Construct 60° at O; construct 90° at O; bisect the difference (or add 30° and 45°).
Explain: 75° = (30° + 45°) or (90° − 15°) via successive bisections.
Angle 135°:
Method 1: 90° + 45° (bisect 90° and add).
Method 2: 180° − 45° (construct a straight angle then subtract 45°).
Explain: Angle addition/subtraction with bisection.
Parallelogram (6, 5, included 60°):
AB=6; at A make 60° and mark AD=5; through B draw line ∥ AD; through D draw line ∥ AB; meet at C.
Explain: Opposite sides are constructed parallel and equal.
Square inscribed in circle (r=3):
Draw circle centre O; draw diameter AB; draw ⟂ diameter CD; join A–B–C–D.
Justify: AC and BD are diameters; central angles between adjacent radii are 90° → square.
Isosceles (base 5, equal 7); prove base angles equal:
AB=5; arcs radius 7 from A and B intersect at C; join.
Proof: In ΞABC, AC=BC ⇒ base angles at A and B equal (Isosceles Triangle Theorem).
Section G — Case-Based (Answers + Brief Reason)
Case 1 (Equilateral park):
Q1 c) Equilateral – all sides equal.
Q2 b) Compass – to construct equal sides via arcs.
Q3 c) 180° – triangle angle sum.
Q4 c) 12 cm – 3 × 4.
Case 2 (Rhombus kite):
Q1 b) Perpendicular and bisecting.
Q2 b) Rhombus – defined by those diagonals.
Q3 b) Compass + Ruler.
Q4 a) Equal – property of rhombus.
Case 3 (Playground rectangle):
Q1 c) 90° – rectangles have right angles.
Q2 c) 70 m – 2(20+15).
Q3 c) 300 m² – 20×15.
Q4 b) Equal and parallel – rectangle property.
Case 4 (Hexagonal tiles):
Q1 c) 6.
Q2 c) 120° – interior angle of regular hexagon.
Q3 c) 30 cm – 6×5.
Q4 b) Compass – step equal chords.
Case 5 (Clock at 3 o’clock):
Q1 c) 90° – right angle.
Q2 a) Compass method – standard perpendicular construction.
Q3 a) Protractor – for measurement.
Q4 c) Compass + Ruler – any angle can be bisected.
Image NCERT Textbook Figure
Hexagon in Circle(Page 188–189)]
NCERT Textbook Figure — Perpendicular from a Point (Page 195–196) ]
NCERT Textbook Figure — Square/Rectangle Construction (Page 195–197)]
NCERT Textbook Figure — Angle Bisector (Page 205–207)]
NCERT Textbook Figure — 60°, 30°, 90°, 120° Constructions (Page 205–207)]
NCERT Textbook Figure — SSS Triangle(Page 211–214) ] NCERT Textbook Figure — Perpendicular Bisector (Page 211–213)] ]
NCERT Textbook Figure — Rhombus by Diagonals (Page 215–216)]
