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Unit IV: Geometry — Test 2

Reasoning, proof and modelling · 16 questions · 54 practice marks · 75 minutes.
MCQs are scored automatically; written answers use step-wise self/teacher review.

Section A
Question 1 (1 Mark)
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Full question paper and step-wise model answers

Practice marks are not official examination weightage. Written answers require self- or teacher assessment.

Question 1 · 1 mark

If G is the centroid on median AM, then AG:GM is:

  1. 1:1
  2. 1:2
  3. 2:1
  4. 3:1
Model answer and reasoning

(c) 2:1

  1. The centroid lies two thirds of the way from vertex A to midpoint M.
  2. Thus AG = 2AM/3 and GM = AM/3.

Question 2 · 1 mark

A simple quadrilateral with opposite angles 70° and 110° is:

  1. Necessarily a rectangle
  2. Cyclic
  3. Necessarily a rhombus
  4. Impossible
Model answer and reasoning

(b) Cyclic

  1. 70° + 110° = 180°.
  2. The converse of the cyclic-quadrilateral angle theorem gives cyclicity.

Question 3 · 1 mark

A rectangle that is not a square has:

  1. Four reflection axes
  2. Two reflection axes and half-turn symmetry
  3. No reflection axes
  4. Quarter-turn symmetry
Model answer and reasoning

(b) Two reflection axes and half-turn symmetry

  1. Its axes pass through the midpoints of opposite sides.
  2. A 180° turn exchanges opposite vertices, but a 90° turn interchanges unequal side lengths.

Question 4 · 1 mark

Among two unequal chords of the same circle, the longer chord is:

  1. Farther from the centre
  2. Closer to the centre
  3. Always a diameter
  4. At the same distance
Model answer and reasoning

(b) Closer to the centre

  1. For radius r and perpendicular distance d, half-chord length is √(r² − d²).
  2. A larger chord requires a smaller d.

Question 5 · 4 marks

State the converse of “If a quadrilateral is a square, all its angles are equal”. Decide whether it is true. Explain SSA ambiguity and give one valid special situation.

Model answer and reasoning

The converse is false; a non-square rectangle is a counterexample. RHS is a valid special situation.

  1. The converse is: if all angles of a quadrilateral are equal, it is a square.
  2. A rectangle with sides 2 cm and 3 cm has four equal angles but is not a square.
  3. For SSA, a circle fixing one side can meet the ray fixing another angle at two points, producing two possible triangles. An approximate drawing cannot settle congruence.
  4. For right triangles with equal hypotenuse and one equal leg, RHS fixes the remaining leg and gives congruence.

Question 6 · 3 marks

Using the angle sum of a triangle, prove it has at least two acute angles. Explain why it cannot have two obtuse angles or all three angles less than 60°.

Model answer and reasoning

Every non-degenerate triangle has at least two acute angles.

  1. The three positive interior angles sum to 180°.
  2. Two angles each at least 90° would leave at most 0° for the third. In particular two obtuse angles are impossible.
  3. If all three were less than 60°, their sum would be less than 180°, another contradiction.

Question 7 · 4 marks

Prove the diagonals of a parallelogram bisect each other. Then prove the converse: if the diagonals of a simple quadrilateral bisect each other, it is a parallelogram.

Model answer and reasoning

Each diagonal bisects the other; the bisection condition characterises parallelograms.

  1. For parallelogram ABCD with diagonals meeting at O, AB ∥ CD and AB = CD. Alternate angles make triangles AOB and COD congruent by ASA.
  2. Thus AO = OC and BO = OD.
  3. Conversely assume these two equalities. Vertical angles give triangles AOB and COD congruent by SAS.
  4. Corresponding alternate angles imply AB ∥ CD. Applying SAS to AOD and COB gives AD ∥ BC, so ABCD is a parallelogram.

Question 8 · 4 marks

Explain two other sufficient conditions for a simple quadrilateral to be a parallelogram: (i) one pair of opposite sides equal and parallel, (ii) opposite angles equal.

Model answer and reasoning

Both conditions imply a parallelogram.

  1. For (i), if AB = CD and AB ∥ CD, draw AC. The equal included alternate angles and common AC give SAS congruence of ABC and CDA.
  2. The other alternate angles imply BC ∥ AD.
  3. For (ii), let ∠A = ∠C and ∠B = ∠D. The 360° sum gives ∠A + ∠B = 180°.
  4. Supplementary same-side interior angles imply AD ∥ BC; using ∠B + ∠C = 180° gives AB ∥ CD.

Question 9 · 4 marks

State and prove the converse of the midpoint theorem: a line through midpoint P of AB parallel to BC meets AC at Q. Show Q is the midpoint of AC.

Model answer and reasoning

AQ = QC.

  1. Let R be the actual midpoint of AC.
  2. The midpoint theorem gives PR ∥ BC.
  3. Through P there is only one line parallel to BC, so PQ and PR are the same line.
  4. Its intersection with AC is unique; Q = R and therefore AQ = QC.

Question 10 · 5 marks

For a triangle with position vectors a, b and c, use midpoint coordinates to prove that its medians meet at a common point G and are divided 2:1 from each vertex.

Model answer and reasoning

G has position vector (a + b + c)/3 and divides every median 2:1.

  1. The midpoint M of BC has vector (b + c)/2.
  2. The point two thirds of the way from A to M has vector a + (2/3)((b + c)/2 − a) = (a + b + c)/3.
  3. The same calculation starting from B and C produces exactly the same vector.
  4. Thus this one point lies on all three medians, proving concurrence.
  5. Its vertex-to-point and point-to-midpoint fractions are 2/3 and 1/3, so each ratio is 2:1. Coordinate calculations may be written component-wise instead of using vectors.

Question 11 · 4 marks

P, Q, R, S are midpoints of successive sides AB, BC, CD, DA of a quadrilateral. Prove PQRS is a parallelogram and describe the degenerate exception.

Model answer and reasoning

PQRS is a parallelogram when AC and BD are not parallel; parallel diagonals make it collapse to a segment.

  1. In triangles ABC and ADC, the midpoint theorem gives PQ ∥ AC, SR ∥ AC and PQ = SR = AC/2.
  2. Similarly QR and PS are parallel to BD and have length BD/2.
  3. Both pairs of opposite sides are parallel, giving the midpoint parallelogram; the reasoning does not require the original quadrilateral to be convex.
  4. If AC ∥ BD, all four midpoint sides are collinear, so the figure is a degenerate segment rather than a two-dimensional parallelogram.
Quadrilateral with side midpoints forming the Varignon parallelogramABCDPQRS

Question 12 · 4 marks

A(1,2), B(5,3), C(7,7) are consecutive vertices of parallelogram ABCD. Find D and its centre of symmetry. Compare reflection and rotation symmetries of a general parallelogram, rectangle, rhombus and square.

Model answer and reasoning

D = (3,6); centre (4,4). General parallelogram: half-turn; rectangle and rhombus: two reflection axes; square: four and quarter-turns.

  1. Diagonals have the same midpoint, so A + C = B + D coordinate-wise. D = A + C − B = (3,6).
  2. The centre is the midpoint of AC: (4,4). A 180° turn about it exchanges opposite vertices.
  3. A general parallelogram need not have a reflection axis. A non-square rectangle has two midpoint axes; a non-square rhombus has two diagonal axes.
  4. A square has four reflection axes and rotations of 90°, 180°, 270° and 360°. All parallelograms have the half-turn symmetry.

Question 13 · 4 marks

Describe a paper-cutout procedure for tiling the plane with a simple quadrilateral. Explain the role of half-turns and the angle sum; state the scope of the claim.

Model answer and reasoning

Every non-degenerate simple quadrilateral, convex or concave, admits a tiling by congruent copies.

  1. Make congruent cutouts and place neighbours by 180° turns about side midpoints so shared edges match.
  2. At the resulting vertex configurations, one copy of each of the four interior angles meets; the sum is 360° (including a reflex interior angle when concave).
  3. Continue the matching placements in repeated rows; the paired arrangements give a repeating pattern without gaps or overlaps.
  4. The tile is the filled region of a simple quadrilateral. A self-crossing four-edge figure does not automatically define the same kind of filled tile. Use paper manipulation to check the matching.

Question 14 · 4 marks

Define centre, radius, diameter, chord, arc, sector and segment of a circle. Explain why a diameter is the longest chord and give one technological and one cultural use of circles.

Model answer and reasoning

A diameter has maximum chord length 2r; circular wheels and an Ashoka Chakra motif are examples.

  1. A circle consists of points at fixed positive distance r from its centre. A radius joins centre to circle; a chord joins two points on it; a diameter is a chord through the centre.
  2. An arc is part of the circumference. A sector is bounded by two radii and an arc; a segment by a chord and an arc.
  3. For chord AB, AB ≤ AO + OB = 2r by the triangle inequality. Equality occurs for a diameter with O between the endpoints.
  4. A circular wheel maintains a constant axle height on level ground. Cultural examples include Ashoka Chakra, Dharmachakra and Sudarshan Chakra motifs.

Question 15 · 5 marks

Prove that equal chords of the same circle subtend equal central angles and are equidistant from the centre. State both converses. Then relate a 100° central angle to an inscribed angle subtending the same minor arc.

Model answer and reasoning

Equal chords give equal central angles and distances, with converses true. The specified inscribed angle is 50°.

  1. For equal chords AB and CD, triangles AOB and COD have equal radii and equal third sides; SSS gives equal central angles.
  2. Conversely equal central angles and the two pairs of radii give equal chords by SAS.
  3. Perpendiculars from O bisect the chords. Equal half-chords and equal radii give equal distances using right triangles.
  4. Conversely equal perpendicular distances and equal radii give equal half-chords, hence equal full chords.
  5. An inscribed angle with vertex on the remaining major arc is half the 100° central angle, so it is 50°. Angles in that same segment are equal.

Question 16 · 5 marks

For a cyclic quadrilateral ABCD, prove opposite angles are supplementary using the arc-angle theorem. State its converse, and find ∠C when ∠A = 78°. Explain how this tests concyclicity.

Model answer and reasoning

∠A + ∠C = 180°; ∠C = 102°. A simple quadrilateral with supplementary opposite angles is cyclic.

  1. The arcs intercepted by the opposite angles together make a complete 360° circle.
  2. Each inscribed angle is half the measure of its intercepted arc.
  3. Adding the two angles therefore gives 360°/2 = 180°.
  4. The converse says a simple quadrilateral with supplementary opposite angles has its four vertices on one circle.
  5. With ∠A = 78°, ∠C = 180° − 78° = 102°. The supplementary-angle criterion is an exact concyclicity test, not an inference from the drawing.