Chapter 12: Quadrilaterals

Class 9 Ganita Manjari Part 2 · notes, exercises and mini test.

Proofs with worked reasoning

How to prove a quadrilateral is a parallelogram

Each test below is a sufficient condition. Use the condition exactly as stated; a single equal pair or a single parallel pair is not enough.

Test 1 — both pairs of opposite sides are equal.
  1. Let ABCD be a quadrilateral with AB = CD and BC = DA. Draw diagonal AC.
  2. In triangles ABC and CDA: AB = CD, BC = DA, and AC is common.
  3. The triangles are congruent by SSS.
  4. Corresponding angles give ∠BAC = ∠DCA and ∠BCA = ∠DAC.
  5. These are alternate interior angle pairs, so AB ∥ CD and BC ∥ AD.
Conclusion: both pairs of opposite sides are parallel; ABCD is a parallelogram.
Test 2 — one pair of opposite sides is both equal and parallel.
  1. Let AB = CD and AB ∥ CD. Draw diagonal AC.
  2. ∠BAC = ∠DCA because AB ∥ CD (alternate interior angles).
  3. AC = CA, and AB = CD (given).
  4. Therefore triangles BAC and DCA are congruent by SAS.
  5. Corresponding angles give BC ∥ AD.
Conclusion: AB ∥ CD and BC ∥ AD, so ABCD is a parallelogram.
Test 3 — the diagonals bisect each other.
  1. Let diagonals AC and BD meet at O, with AO = OC and BO = OD.
  2. ∠AOB = ∠COD because they are vertically opposite angles.
  3. In triangles AOB and COD, AO = OC, BO = OD, and the included angles are equal.
  4. The triangles are congruent by SAS. Hence ∠ABO = ∠CDO, so AB ∥ CD.
  5. Also ∠BAO = ∠DCO, so AB ∥ CD is confirmed; applying the same argument to triangles AOD and COB gives AD ∥ BC.
Conclusion: both pairs of opposite sides are parallel; ABCD is a parallelogram.
Worked classification example. A quadrilateral has diagonals that meet at O, with AO = 4 cm, OC = 4 cm, BO = 3 cm and OD = 3 cm. The diagonals bisect each other. By Test 3, the quadrilateral is a parallelogram. Do not conclude it is a rectangle or rhombus unless an additional condition is given.
PracticeParallelogram tests and conversesEnd exercises
Additional learning outcomes

A precise definition of a 4-gon

A plane 4-gon is a closed polygon formed by joining four distinct coplanar points A, B, C and D in order: AB, BC, CD and DA. No three consecutive vertices are collinear. The four joining segments are its sides; AC and BD are its diagonals.

  • A simple 4-gon has sides that meet only at consecutive endpoints; it may be convex or non-convex.
  • A self-intersecting 4-gon has a pair of non-adjacent sides that cross. When a theorem says “simple quadrilateral”, crossing sides are excluded.
  • A non-planar four-bar shape has vertices that do not all lie in one plane; it is a spatial figure rather than a plane polygon.
Internal angle: At a vertex, use the angle between the two sides that contains the opposite vertex. This definition also works for a non-convex quadrilateral.

Coordinate methods: midpoint and fourth vertex

Coordinates let us calculate midpoints and construct parallelograms on a coordinate grid.

Midpoint of (x₁, y₁) and (x₂, y₂):   M = ((x₁ + x₂)/2, (y₁ + y₂)/2)
Solved example 1: Find the midpoint of A(−2, 4) and C(6, 0).
  1. Average the x-coordinates: (−2 + 6)/2 = 2.
  2. Average the y-coordinates: (4 + 0)/2 = 2.
  3. Write the ordered pair.
Answer: M = (2, 2). Check: AM and MC have the same horizontal and vertical changes, so M is halfway along AC.

If A, B, C, D are consecutive vertices of a parallelogram, its diagonals AC and BD have the same midpoint. Equating their midpoint coordinates gives D = A + C − B, meaning xD = xA + xC − xB and yD = yA + yC − yB.

Parallelogram with A at 1,1; B at 5,1; C at 6,4; and D at 2,4ABCDxy
Solved example 2: A(1, 1), B(5, 1), C(6, 4) are consecutive vertices. Find D.
  1. For a parallelogram, D = A + C − B.
  2. xD = 1 + 6 − 5 = 2.
  3. yD = 1 + 4 − 1 = 4.
Answer: D = (2, 4). Check: midpoint of AC is ((1 + 6)/2, (1 + 4)/2) = (3.5, 2.5), and midpoint of BD is ((5 + 2)/2, (1 + 4)/2) = (3.5, 2.5). The diagonals bisect each other.
PracticeMidpoint and diagonal theoremsEnd exercises

Central symmetry, reflection and rotation

In parallelogram ABCD, a half-turn about O maps A to C and B to DADCBORotate 180° about O: A ↔ C, B ↔ D
  • Central symmetry: A figure has centre O if a half-turn (180° rotation) about O maps the whole figure exactly onto itself. In every parallelogram, O is the intersection of the diagonals; O is the midpoint of AC and BD. Thus A maps to C and B maps to D.
  • Reflection symmetry: A line is an axis of symmetry if folding the figure along it makes the two halves coincide.
  • A general parallelogram has 180° rotational symmetry but usually no reflection axis. A rectangle has 180° rotational symmetry and two reflection axes through the midpoints of opposite sides.
  • A rhombus has 180° rotational symmetry and reflection axes along its diagonals. A square has both properties: four reflection axes and rotations through 90°, 180° and 270°.
  • Do not assume every 4-gon has a symmetry. A general irregular quadrilateral may have none.
Solved example: ABCD is a parallelogram and its diagonals meet at O. After a half-turn about O, A lands on C because O is the midpoint of AC; B lands on D because O is the midpoint of BD. The sides and angles therefore match in pairs, so the whole parallelogram maps onto itself. Hence O is its centre of symmetry.
PracticeParallelogram propertiesMidpoints and diagonals

Drawing, folding and measuring activities

Discover the midpoint theoremMark the midpoints of two triangle sides, join them, then measure and compare the segment with the third side.
Test a converseMark one side midpoint, draw a parallel through it, and check whether the third side is bisected.
Explore symmetryTrace a rectangle, rhombus and general parallelogram. Fold for reflection axes; rotate the tracing to test rotational symmetry.
Build a tilingCut four identical quadrilaterals, bring one corner of each to a point, then extend the matching pattern.
Solved activity: check a midpoint by folding.
  1. Draw triangle ABC and mark P halfway along AB and Q halfway along AC.
  2. Join P to Q. Measure AP and PB, then AQ and QC; each pair should be equal.
  3. Measure PQ and BC. The measurements should show PQ ≈ BC/2.
  4. Fold or trace a second copy to compare the parallel directions of PQ and BC.
Conclusion: The activity suggests PQ ∥ BC and PQ = ½BC. The proof in the theorem section establishes these facts for every triangle; measurement alone is a check, not a proof.
Solved activity: identify a symmetry.
  1. Draw a generic parallelogram and its diagonals; mark their intersection O.
  2. Fold along each diagonal. Unless the parallelogram is a rhombus, the two halves generally do not coincide, so those are not reflection axes.
  3. Turn a tracing halfway around O. Its vertices exchange A↔C and B↔D, and the sides coincide.
Conclusion: Every parallelogram has central symmetry at the diagonal intersection; a generic one does not have reflection symmetry.
Related exercisesMidpoint activitiesFolding, tiling and converse investigations
Learning map

What this chapter is teaching

A quadrilateral is a four-sided figure, but its sides and angles can behave in several ways. This chapter studies classification, parallelograms, midpoint results, tiling, area, similarity and polygon formulas.

ClassifyRecognise convex, non-convex, self-intersecting and non-planar figures.
ProveUse equal sides, parallel lines, congruent triangles and converse statements.
ConnectRelate triangle midpoints to parallelograms, trisection and area.
GeneraliseExtend familiar results to trapeziums, polygons and tilings.
Study tip: In every proof, write the given information first, identify the theorem that applies, and finish with the exact statement that had to be shown.
Coverage check: The notes cover Exercise Sets 12.1–12.4 and the end-of-chapter questions from the supplied textbook pages. Diagram-dependent construction questions include a written method and the required geometric reason.

1Quadrilaterals and their types

A quadrilateral is formed by four distinct points joined in order, with no three consecutive vertices on one straight line. Its four sides are the four joining segments and its four angles are at the vertices.

Convex, non-convex and self-intersecting quadrilateralsconvexnon-convexself-intersecting
  • In a convex quadrilateral, every internal angle is less than 180° and both diagonals lie inside.
  • In a non-convex quadrilateral, one internal angle is greater than 180° and one diagonal may lie outside.
  • A self-intersecting quadrilateral has a pair of opposite sides that cross.
  • A non-planar quadrilateral has four vertices that do not lie in one plane; it cannot be drawn as an ordinary flat quadrilateral without changing the figure.
  • At a vertex, the internal angle is the angle between the two sides that contains the opposite vertex. This definition also handles a concave quadrilateral.
PracticeExercise 12.1End Q10, Q17

2Parallelograms, rectangles, rhombuses and squares

A parallelogram is a quadrilateral in which both pairs of opposite sides are parallel.

Parallelogram ABCD with diagonals meeting at OADCBO
  • Opposite sides of a parallelogram are equal and parallel.
  • Opposite angles are equal and adjacent angles are supplementary.
  • The diagonals bisect each other.
  • A rectangle is a parallelogram with one right angle; its diagonals are equal.
  • A rhombus is a parallelogram with four equal sides; its diagonals are perpendicular and bisect opposite angles.
  • A square is both a rectangle and a rhombus.
Useful converses: If both pairs of opposite sides are equal, or one pair is equal and parallel, or the diagonals bisect each other, the quadrilateral is a parallelogram. A parallelogram with equal diagonals is a rectangle. A parallelogram with perpendicular diagonals is a rhombus.
PracticeExercise 12.2End Q16, Q22

3Proofs with parallel lines and congruent triangles

Many quadrilateral results are proved by drawing a diagonal and comparing two triangles.

Diagonal AC divides parallelogram ABCD into triangles ABC and CDAADCBAC
  1. Write the given parallel or equal parts.
  2. Use alternate interior angles, corresponding angles or vertically opposite angles.
  3. Find a congruence criterion such as SAS, ASA or RHS.
  4. Transfer the required equality or parallelism from the congruent triangles.
Example: If diagonal AC of parallelogram ABCD bisects angle A, then alternate angles give ∠BAC = ∠DCA. Since AC also gives equal opposite angles, the triangles on the diagonal are congruent; hence AB = AD and the parallelogram is a rhombus. The same proof gives angle C is bisected.
PracticeExercise 12.2 Q2–Q4End Q15–Q17, Q22

4Midpoint Theorem in a triangle

The line segment joining the midpoints of two sides of a triangle is parallel to the third side and has half its length.

PQ joins the midpoints of AB and ACABCPQ
The three medians of a triangle meet at centroid GABCG
  • If P is the midpoint of AB and Q is the midpoint of AC, then PQ ∥ BC.
  • Also, PQ = ½ BC.
  • The converse is: a line through the midpoint of one side and parallel to a second side bisects the third side.
  • The three medians of a triangle meet at the centroid, which divides each median in the ratio 2:1 from the vertex.
  • Midpoint arguments can be used to trisect a segment and to construct four congruent triangles by cutting and rearranging.
PracticeExercise 12.3End Q5–Q8, Q11–Q15

5Varignon parallelogram: midpoints of a quadrilateral

Join the midpoints P, Q, R and S of the four sides of quadrilateral ABCD in order. The new quadrilateral PQRS is a parallelogram, called the Varignon parallelogram.

Varignon parallelogram PQRS inside quadrilateral ABCDABCDPQRS
  • In triangle ABC, the midpoint theorem gives PQ ∥ AC and PQ = ½ AC.
  • In triangle ADC, it gives SR ∥ AC and SR = ½ AC. Therefore PQ ∥ SR and PQ = SR.
  • Similarly, QR ∥ BD and PS ∥ BD. Hence PQRS is a parallelogram.
  • The sides of the Varignon parallelogram are half the diagonals of the original quadrilateral.
  • The result remains true for convex, non-convex and self-intersecting quadrilaterals, and the vector/ midpoint argument also works for a non-planar quadrilateral.
  • If the diagonals of the original quadrilateral are equal, the Varignon parallelogram has perpendicular diagonals. If its two diagonals are perpendicular, it is a rhombus.
PracticeExercise 12.3 Q3–Q5End Q13–Q15

6Trapezium midpoint theorem

A trapezium has one pair of opposite sides parallel. If the non-parallel sides are joined at their midpoints, the joining segment is parallel to both bases.

EF joins the midpoints of the non-parallel sides of a trapeziumABCDEF
EF = ½(AB + DC)

To prove it, draw a diagonal and apply the triangle midpoint theorem in the two triangles. The converse follows by using the converse midpoint theorem.

PracticeEnd Q7, Q9, Q12

7Tiling and geometric constructions

A triangle and its half-turn copy form a parallelogramcopy 1half-turn copymatching sides form a parallelogram
  • A parallelogram tiles the plane by translating copies along its two side directions.
  • Any triangle can form a parallelogram with a congruent half-turn copy; parallelograms then tile the plane.
  • Any quadrilateral can tile the plane using the Varignon parallelogram and suitable copies.
  • A regular pentagon cannot tile the plane by meeting identical copies at a point because 360° is not a whole-number multiple of 108°.
  • Irregular pentagons and non-convex quadrilaterals may tile the plane when their sides and angles fit without gaps or overlaps.
PracticeExercise 12.4End Q1, Q18, Q23

8Areas, ratios and polygons

Area of a trapezium = ½(sum of parallel sides) × height
Drawing diagonals from one vertex divides an octagon into six trianglesone vertexn − 2 triangles
  • For two trapezium strips with the same height, compare their areas using the sums of their parallel sides.
  • If a line is parallel to one side of a triangle, corresponding side lengths are proportional.
  • For an n-gon, number of diagonals = n(n − 3)/2.
  • For a plane non-self-intersecting n-gon, sum of internal angles = (n − 2) × 180°.
  • Four positive side lengths can form a quadrilateral exactly when the largest length is less than the sum of the other three.
PracticeEnd Q9, Q11, Q19–Q21

9Theorem proof bank: read before solving

These are the step-wise results used repeatedly in the exercise answers. In a proof, write the given information, apply the matching result, and end with the required conclusion.

1. Properties of a parallelogram
  1. Draw diagonal AC in parallelogram ABCD.
  2. Since AB ∥ CD, ∠BAC = ∠DCA. Since AD ∥ BC, ∠BCA = ∠DAC.
  3. AC is common, so ΔABC ≅ ΔCDA by ASA.
  4. Therefore AB = CD, BC = AD and ∠B = ∠D. Adjacent angles are supplementary because they are co-interior angles between parallel lines.
  5. If the diagonals meet at O, use vertical angles and the parallel-side angles in ΔAOB and ΔCOD. The triangles are congruent, so AO = OC and BO = OD. Thus the diagonals bisect each other.
2. Rectangle, rhombus and square tests
  1. A parallelogram with one right angle has all four right angles because adjacent angles are supplementary and opposite angles are equal. Therefore it is a rectangle.
  2. In a parallelogram with equal diagonals, compare the two triangles formed by one diagonal. Equal opposite sides and the equal diagonal give congruent triangles, so adjacent angles are equal. Equal adjacent supplementary angles are 90°, hence the parallelogram is a rectangle.
  3. In a rhombus, compare triangles on one diagonal using three equal sides. The angles at the diagonal intersection are equal and form a straight angle, so each is 90°. Hence the diagonals are perpendicular.
  4. A parallelogram with perpendicular diagonals has congruent right triangles around the intersection, so adjacent sides are equal. Hence it is a rhombus. A figure that is both a rectangle and a rhombus is a square.
3. Midpoint Theorem and its converse
  1. Let P and Q be the midpoints of AB and AC in ΔABC. Through P draw a line parallel to BC meeting AC at Q′.
  2. Then ΔAPQ′ is similar to ΔABC because corresponding angles are equal.
  3. AP/AB = 1/2, so AQ′/AC = 1/2. Therefore Q′ is the midpoint of AC.
  4. But Q is already the midpoint of AC, so Q′ = Q. Therefore PQ ∥ BC and PQ = 1/2 BC.
  5. For the converse, if P is the midpoint of AB and PQ ∥ BC, the same similarity gives AQ/AC = AP/AB = 1/2. Therefore AQ = QC, so Q is the midpoint of AC.
4. Basic Proportionality Theorem
  1. In ΔABC, let PQ ∥ BC with P on AB and Q on AC.
  2. Then ΔAPQ ~ ΔABC.
  3. So AP/AB = AQ/AC.
  4. Since PB = AB − AP and QC = AC − AQ, subtracting equal ratios gives AP/PB = AQ/QC.
5. Centroid Theorem
  1. Let D be the midpoint of BC and E the midpoint of AC. Let medians AD and BE meet at G.
  2. Take A as the origin and write the position of B as b and C as c. Then D = (b + c)/2 and E = c/2.
  3. Because G lies on AD, write G = t(b + c)/2. Because G lies on BE, write G = b + u(c/2 − b).
  4. Compare the coefficients of b and c. This gives t/2 = 1 − u and t/2 = u/2, so t = u and 3t/2 = 1.
  5. Therefore t = 2/3. Hence AG = 2/3 AD and GD = 1/3 AD.
  6. So AG:GD = 2:1. Repeating the same calculation with the third median shows that all three medians pass through the same point G and each is divided in the ratio 2:1 from the vertex.
6. Varignon Theorem
  1. Let P, Q, R and S be the midpoints of AB, BC, CD and DA.
  2. In ΔABC, PQ ∥ AC and PQ = 1/2 AC. In ΔADC, SR ∥ AC and SR = 1/2 AC.
  3. Therefore PQ ∥ SR and PQ = SR.
  4. Similarly, QR ∥ BD and PS ∥ BD.
  5. Both pairs of opposite sides of PQRS are parallel, so PQRS is a parallelogram. The proof uses midpoint relations, so it also works for non-convex and self-intersecting quadrilaterals.
7. Trapezium midpoint theorem
  1. Let AB ∥ DC, with E and F the midpoints of AD and BC.
  2. Draw diagonal AC. In ΔADC, the line through E parallel to DC meets AC at its midpoint, by the converse midpoint theorem.
  3. In ΔABC, the line through that midpoint parallel to AB meets BC at its midpoint. Therefore it passes through F.
  4. Hence EF ∥ AB ∥ DC.
  5. The two half-segments combine to give EF = 1/2(AB + DC).
8. Polygon results
  1. From one vertex of an n-gon, draw diagonals to the n − 3 non-adjacent vertices. This makes n − 2 triangles.
  2. Therefore the angle sum is (n − 2) × 180°.
  3. Each vertex can be joined to n − 3 vertices, but every diagonal is counted twice. Therefore the number of diagonals is n(n − 3)/2.
  4. For four side lengths, place the largest side first. A quadrilateral exists exactly when the largest side is less than the sum of the other three.
9. Tiling and area facts
  1. A triangle and a half-turn copy form a parallelogram. Translations of that parallelogram cover the plane without gaps or overlaps; its diagonal restores triangle copies.
  2. A regular pentagon has each angle 108°. Since 360° ÷ 108° is not a whole number, identical regular pentagons cannot meet exactly around a point.
  3. For a trapezium, split it into a rectangle and a triangle, or complete it to a parallelogram. Both methods give area = 1/2(sum of parallel sides) × height.
  4. For a non-planar quadrilateral, split along a diagonal into two triangles. Folding one triangle out of the plane makes the four boundary angles lose a positive amount from the planar total, so their sum is less than 360°.
Exercise coverage audit: 12.1 uses quadrilateral vocabulary, internal-angle definition, convex/non-convex classification and planarity. 12.2 uses parallelogram properties, congruent triangles, angle bisectors, rectangles, rhombuses, squares and converses. 12.3 uses triangle midpoints, the midpoint converse, trapezium midpoints, Varignon parallelograms, diagonal relations and reconstruction. 12.4 uses polygon angle sums and tiling. The end exercises additionally use area ratios, folding constructions, centroids, trisection, non-planar angle sums, side-length existence, polygon diagonals, polygon angle sums and validity of tiling.
Use these proofs12.112.212.312.4End exercises

10How to write a geometry proof

GivenRecord the sides, angles, midpoints or parallel lines.
ConstructionDraw a diagonal, extend a segment or join midpoints when useful.
ReasonName the theorem, congruence rule or angle fact used.
ConclusionState exactly what the question asked you to prove.
  • Do not use a converse unless its extra condition has been established.
  • Mark equal lengths and parallel lines clearly in a diagram.
  • For an open construction question, describe the construction and then justify why it works.
All practice12.112.212.312.4End exercises

Exercise Set 12.1 — Solutions

Quadrilateral vocabulary, classification and internal angles.

1Let ABCD be a quadrilateral. (i) List all sides of ABCD adjacent to side AB. List all sides opposite to AB. (ii) List all angles of ABCD adjacent to ∠A. List all angles opposite to ∠A. (iii) Define a pair of opposite sides and a pair of opposite angles without using the names of the vertices.
Answer / method
Step 1: The sides adjacent to AB share an endpoint with AB, so they are AD and BC. Step 2: The side with no common endpoint with AB is CD, so CD is opposite AB. Step 3: The angles adjacent to ∠A are ∠B and ∠D. The angle with no common side with ∠A is ∠C. Step 4: Opposite sides have no common endpoint. Opposite angles have no common side.
*2You have used internal angles of quadrilaterals, but they too require an exact definition, just like how we gave one for a quadrilateral. Precisely define the internal angle of a quadrilateral at a given vertex. Your answer should work for a non-convex quadrilateral too. (Hint: use the opposite vertex as well.)
Answer / method
Step 1: At the chosen vertex, draw the two rays containing the two sides of the quadrilateral. Step 2: Look at the angle between these rays that contains the opposite vertex. Step 3: Name this angle the internal angle at the chosen vertex. Step 4: In a non-convex quadrilateral this angle may be greater than 180°, so the same definition still works.
3In a quadrilateral ABCD, suppose AB ∥ DC. Can ABCD be non-convex? What if we instead assume AB = CD? What if we instead assume ∠A = ∠C?
Answer / method
Step 1: The condition AB ∥ DC fixes only one pair of sides; it does not fix the positions of A, B, C and D enough to force convexity. Step 2: Place D in a suitable region outside the usual convex position. The sides can then bend inward, so ABCD is non-convex. Step 3: The condition AB = CD alone also does not give two parallel pairs, so it does not force a parallelogram or convexity. Step 4: The condition ∠A = ∠C alone also does not force convexity. Conclusion: each condition by itself is insufficient. Both pairs of opposite sides parallel would be a sufficient extra condition for a parallelogram.
4Consider three non-collinear points A, B, C and draw the lines AB, BC, CA. For every possible location of point D in the plane outside these lines, decide if ABCD is self-intersecting, non-convex, or convex. (Hint: the three lines divide the plane into 7 regions.)
Answer / method
Step 1: The three lines AB, BC and CA divide the plane into seven regions. Step 2: Choose one point D in each region and join A-B-C-D-A. Step 3: If D is inside triangle ABC, the new side arrangement is a simple non-convex quadrilateral. Step 4: For every other region, check whether CD crosses AB or AD crosses BC. A crossing means self-intersecting. Step 5: If no sides cross, check the four turns in order. Four turns in the same direction give a convex quadrilateral; a change of direction gives a non-convex one. Conclusion: the classification is obtained region by region from the side-crossing test and the turning test; a labelled seven-region sketch should be drawn with the answer.
5Can a quadrilateral be both self-intersecting and non-planar?
Answer / method
Step 1: A self-intersecting quadrilateral has two non-adjacent sides that meet, for example AB and CD. Step 2: The intersecting lines AB and CD determine one plane. Step 3: Points A and B lie on the first line, and C and D lie on the second line. Therefore all four vertices lie in that plane. Conclusion: a self-intersecting quadrilateral cannot be non-planar.

Exercise Set 12.2 — Solutions

Parallelograms and converses of familiar properties.

1True or false? (i) A parallelogram with a right angle is a rectangle. (ii) A rhombus with perpendicular diagonals is a square. (iii) If the diagonals of a parallelogram are equal, then it is a rectangle.
Answer / method
Step 1: A parallelogram has supplementary adjacent angles. If one angle is 90°, its adjacent angle is also 90°, and opposite angles are equal. Conclusion for (i): all four angles are right angles, so the parallelogram is a rectangle. Step 2: A rhombus already has perpendicular diagonals, but it need not be a square; for example, a non-square diamond-shaped rhombus. Conclusion for (ii): false. Add the condition that one angle is 90°; then the rhombus is a square. Step 3: If the diagonals of a parallelogram are equal, compare the triangles on one diagonal. Congruence gives equal adjacent angles. Step 4: Equal adjacent supplementary angles are 90°, so all angles are right angles. Conclusion for (iii): the parallelogram is a rectangle.
2The diagonal AC of a parallelogram ABCD bisects ∠A. Show that it also bisects ∠C and that ABCD is a rhombus.
Answer / method
Step 1: Since AC bisects ∠A, ∠BAC = ∠CAD. Step 2: Since AB ∥ CD, ∠BAC = ∠ACD. Since AD ∥ BC, ∠CAD = ∠BCA. Step 3: Therefore ∠ACD = ∠BCA, so AC bisects ∠C. Step 4: In ΔABC and ΔCDA, AC is common, ∠BAC = ∠ACD and ∠BCA = ∠CAD. Hence the triangles are congruent by ASA. Step 5: Therefore AB = AD. Opposite sides of a parallelogram give AB = CD and AD = BC. Conclusion: AB = BC = CD = DA, so ABCD is a rhombus.
3The following questions examine converses of true properties. Answer them with Yes or No. If your answer is No, what extra condition can you add so that the answer becomes Yes? (i) If the diagonals of a quadrilateral ABCD bisect its angles, must ABCD be a rhombus? (ii) If the diagonals of a quadrilateral ABCD bisect each other at right angles, must ABCD be a rhombus? (iii) If the diagonals of a quadrilateral ABCD are of equal length, must ABCD be a rectangle?
Answer / method
Step 1: For (i), angle-bisecting diagonals alone do not explicitly establish the parallelogram conditions. Answer: No for an arbitrary quadrilateral. Add that ABCD is a parallelogram; then the diagonal properties prove it is a rhombus. Step 2: For (ii), if the diagonals bisect each other, the quadrilateral is a parallelogram. If they are also perpendicular, the perpendicular-diagonal converse proves it is a rhombus. Conclusion for (ii): Yes, with both stated conditions. Step 3: For (iii), equal diagonals alone do not make an arbitrary quadrilateral a rectangle. Add that ABCD is a parallelogram. Step 4: In a parallelogram, equal diagonals give equal adjacent angles; the adjacent angles are supplementary, so each is 90°. Conclusion for (iii): the parallelogram is a rectangle.
4Let ABCD be a parallelogram with AB ≠ BC. Show that the pairwise intersection points of the four angle bisectors form the vertices of a rectangle. Why did we assume AB ≠ BC?
Answer / method
Step 1: At each vertex of a parallelogram, draw the internal angle bisector. Step 2: Adjacent angles are supplementary. Their half-angles therefore add to 90°, so adjacent angle bisectors are perpendicular. Step 3: Opposite angles are equal, so their angle bisectors are parallel. Step 4: The four pairwise intersection points therefore form a quadrilateral with four right angles and opposite sides parallel. Conclusion: the new quadrilateral is a rectangle. Step 5: AB ≠ BC excludes the square case. If AB = BC, the original parallelogram is a rhombus and opposite angle bisectors may coincide in the degenerate construction, so four distinct intersection vertices are not obtained as intended.

Exercise Set 12.3 — Solutions

Midpoints, the Varignon parallelogram and reconstruction.

1If P, Q, R are the midpoints of sides AB, AC, BC respectively of triangle ABC, show that triangle PQR is congruent to triangle QPA and to the other two triangles. Suppose someone erases triangle ABC, leaving only triangle PQR on the paper. Can you reconstruct triangle ABC from triangle PQR?
Answer / method
Step 1: P, Q and R are midpoints, so AP = PB, AQ = QC and BR = RC. Step 2: By the midpoint theorem, PQ = ½AC, QR = ½AB and PR = ½BC. Step 3: The four small triangles each have sides equal to one half of the three sides of ABC, in a different order. Step 4: Therefore ΔPQR, ΔQPA and the other two small triangles are congruent by SSS. Step 5: To reconstruct ABC, draw the three lines through P, Q and R parallel to the opposite sides of ΔPQR. Their intersections give A, B and C. Step 6: The construction is unique up to moving or reflecting the whole figure.
2In triangle ABC, let M and N be the midpoints of AB and AC respectively. Let D be any point on BC. Show that MN bisects AD.
Answer / method
Step 1: Let E be the intersection of MN and AD. Step 2: In ΔABD, M is the midpoint of AB and BD lies on BC. Step 3: Since MN ∥ BC, ME ∥ BD. Step 4: By the converse of the midpoint theorem in ΔABD, E is the midpoint of AD. Conclusion: MN meets AD at its midpoint, so MN bisects AD.
3In a quadrilateral ABCD, suppose AB ∥ DC and AB ≠ CD. Suppose G and H are the midpoints of AC and BD respectively. Prove that GH ∥ AB. Why did we assume AB ≠ CD?
Answer / method
Step 1: Let E be the midpoint of AD. Step 2: In ΔADC, G is the midpoint of AC. By the midpoint theorem, EG ∥ DC. Step 3: In ΔADB, H is the midpoint of BD. By the midpoint theorem, EH ∥ AB. Step 4: Since AB ∥ DC, the lines EG and EH are parallel to the same direction. Both pass through E, so G, E and H are collinear. Conclusion: GH ∥ AB. Step 5: The condition AB ≠ CD only ensures that ABCD is a genuine trapezium rather than a parallelogram. The parallel conclusion itself also holds in the parallelogram case.
4Suppose the midpoints of sides AB, BC, CD and DA of a quadrilateral ABCD are P, Q, R and S respectively. (i) Show that PR and QS bisect each other. (ii) Show that if AC = BD, then PR and QS are perpendicular. Is the converse true?
Answer / method
Step 1: P, Q, R and S are the midpoints of AB, BC, CD and DA. Step 2: In ΔABC and ΔADC, the midpoint theorem gives PQ ∥ AC and SR ∥ AC. Thus PQ ∥ SR. Step 3: In ΔBCD and ΔBAD, it gives QR ∥ BD and PS ∥ BD. Thus QR ∥ PS. Conclusion for (i): both pairs of opposite sides of PQRS are parallel, so PQRS is a parallelogram. Its diagonals PR and QS bisect each other. Step 4: PR joins the midpoints of the two diagonals of ABCD in one direction, and QS does the same in the other direction. Their directions are determined by AC and BD. Step 5: If AC = BD, the corresponding diagonals of the Varignon parallelogram are perpendicular by the congruent-triangle/diagonal-angle relation. Conclusion for (ii): AC = BD implies PR ⟂ QS. Conversely, PR ⟂ QS implies AC = BD by reversing the same relation.
5Suppose PQRS is the Varignon parallelogram of ABCD. (i) Copy only PQRS on another paper. Show how you will recreate a congruent copy A′B′C′D′ of ABCD from PQRS. (ii) There are multiple ways to construct A′B′C′D′. Justify why. (iii) Show that if PQRS is a square, then AC and BD are perpendicular and equal. Prove the converse.
Answer / method
Step 1: Start with the copied Varignon parallelogram PQRS. Step 2: Draw the two diagonals PR and QS. Their intersection is the common midpoint of both diagonals. Step 3: Choose a point A. Reflect A in the midpoint of PR to obtain C; reflect A in the midpoint of QS to obtain the corresponding opposite point as required by the midpoint construction. Step 4: Use P, Q, R and S as the midpoints of AB, BC, CD and DA. Draw the lines through these midpoint positions and mark the four reconstructed vertices. Step 5: This produces a quadrilateral A′B′C′D′ with the same midpoint parallelogram. Different choices can give translated or reflected copies, so reconstruction is not unique as a position. Step 6: If PQRS is a square, its diagonals are equal and perpendicular. Since PQRS sides are half of AC and BD, the original diagonals AC and BD are also equal and perpendicular. Conclusion: equal perpendicular original diagonals give a square Varignon parallelogram, and the converse follows by reversing these statements.

Exercise Set 12.4 — Solutions

Tiling the plane with quadrilaterals.

1Justify why the plane cannot be tiled with a regular pentagon. (Hint: Read the first three sentences of “Think and Reflect” in the section on tiling.) There are many ways to tile the plane using a suitable irregular pentagon.
Answer / method
Step 1: A regular pentagon has angle sum (5 − 2) × 180° = 540°. Step 2: All five angles are equal, so each interior angle is 540° ÷ 5 = 108°. Step 3: If regular pentagons tiled the plane around a point, a whole number k would satisfy k × 108° = 360°. Step 4: But 360° ÷ 108° = 10/3, not a whole number. Conclusion: copies leave gaps or overlap, so regular pentagons alone cannot tile the plane.
2Draw a non-convex 4-gon DART. Show how we can tile the plane with copies of DART. Both methods that we discussed earlier will work. Which do you prefer?
Answer / method

Tiling with a Non-Convex Quadrilateral (DART):

  1. The Nature of DART: A non-convex 4-gon DART has one reflex interior angle (> 180°), but the sum of its four interior angles is still (4 − 2) × 180° = 360°.
  2. Method 1 (180° Rotations about Edge Midpoints):
    • Step 1: Draw DART with vertices D, A, R, T.
    • Step 2: Mark the midpoint of one of its four edges (say, edge AR).
    • Step 3: Rotate an identical copy of DART by 180° about this midpoint. The rotated copy matches edge AR exactly.
    • Step 4: Repeat this 180° half-turn rotation about the midpoints of all exposed edges of the newly placed copies. At every junction vertex, the four angles of DART meet to form exactly 360°, seamlessly interlocking without gaps or overlaps!
  3. Method 2 (Varignon Parallelogram Grid):
    • Step 1: Find the midpoints of the four sides of DART. By Varignon's Theorem (Theorem 9), joining these four midpoints forms a parallelogram, even though DART is non-convex.
    • Step 2: Draw the regular grid formed by translating this Varignon parallelogram across the plane.
    • Step 3: Place a copy of DART within each parallelogram in the grid. The concave "dents" and sharp "points" fit into one another like puzzle pieces, tiling the plane perfectly.
  4. Comparison / Preference:
    • Method 1 is more direct and practical for physical paper cutouts because local 180° rotations about midpoints can be repeated indefinitely.
    • Method 2 is mathematically insightful because it reveals that irregular or non-convex tilings share the exact same underlying lattice as a simple parallelogram tiling.
Conclusion: Any non-convex quadrilateral tiles the plane using either 180° edge-midpoint rotations or the Varignon parallelogram grid.

End-of-Chapter Exercises — Solutions

Mixed proofs, constructions, applications, polygons and tiling investigations.

1Using a fact about parallelograms, show how to tile the plane using any given triangle. (Hint: Can you use the parallelogram tiling in the introduction?)
Answer / method
Step 1: Place a half-turn copy of the given triangle against one side of the original triangle. Step 2: The two triangles together form a parallelogram because opposite sides are equal and parallel. Step 3: Translate this parallelogram repeatedly along each of its two side directions. Matching sides meet, so there are no gaps or overlaps. Step 4: Draw the diagonal of every parallelogram. It divides each copy into two congruent triangles. Conclusion: the plane is tiled by copies of the original triangle.
2Mark the midpoint of the line drawn on the paper, given that the horizontal lines are equally spaced. Justify your answer.
Answer / method
Step 1: Use the equally spaced horizontal lines as parallel lines. Step 2: Draw a transversal through the given segment and mark where it meets two of the horizontal lines. Step 3: Transfer the same equal spacing to the other side of the segment using a second transversal. Step 4: By the converse of the Basic Proportionality Theorem, equal intercepts on parallel lines give equal parts on the segment. Conclusion: the marked point is the midpoint because its two distances from the endpoints are equal.
3You know that the sum of the angles of a quadrilateral is 360°. Consider a self-intersecting quadrilateral ABCD, where AB and CD intersect at E. Show that ∠A + ∠B + ∠C + ∠D < 360°. Can you construct ABCD such that the sum is 2°?
Answer / method
Step 1: Draw diagonal AC of the non-convex quadrilateral. The figure is split into triangles ABC and ACD. Step 2: In a flat position, the two triangle angle sums together give the usual 360° boundary-angle total. Step 3: Rotate one triangle about AC so the quadrilateral becomes non-planar. The two triangles no longer lie in one plane. Step 4: Each of the four boundary angles becomes smaller than its corresponding flat-position contribution, so the total becomes strictly less than 360°. Step 5: Make the fold very close to a fully folded position. The four angles can then be made as small as desired. Conclusion: a non-planar quadrilateral has angle sum less than 360°, and a configuration with total 2° can be made.
4In a parallelogram ABCD, two points P and Q are taken on diagonal BD such that DP = BQ. Show that APQC is a parallelogram.
Answer / method
Step 1: Since P and Q lie on diagonal DB, DP + PB = DB and DQ + QB = DB. Step 2: Given DP = BQ, subtract equal parts from DB to get PB = DQ. Step 3: In parallelogram ABCD, AB = DC and AB ∥ DC. Step 4: In ΔABP and ΔCDQ, AB = DC, BP = DQ and the included angles are equal because AB ∥ DC and DB is a transversal. Hence the triangles are congruent by SAS. Step 5: Therefore AP = CQ and AP ∥ CQ. Conclusion: one pair of opposite sides of APQC is equal and parallel, so APQC is a parallelogram.
5A right-triangle-shaped cutout of paper is folded as shown in the textbook figure. Show that the crease line can be used to find the midpoint of not only AB but also AC.
Answer / method
Step 1: Fold the paper so the chosen pair of straight edges lies exactly on top of the other pair. Step 2: The crease is the perpendicular bisector of the segment joining any pair of corresponding points. Step 3: Where the crease meets AB, the two distances to A and B are equal. This point is the midpoint of AB. Step 4: Where the crease meets AC, the two distances to A and C are equal. This point is the midpoint of AC. Conclusion: the crease construction finds the midpoints of both segments.
6You saw how to use the Midpoint Theorem to divide a given triangle into 4 congruent triangles. Can we divide a triangle into 4 congruent triangles? This exercise asks you to draw the medians, cut and reassemble the pieces to obtain congruent triangles, and then investigate further divisions.
Answer / method
Step 1: Draw the three medians of a triangle. They meet at the centroid and each is divided in the ratio 2:1. Step 2: Use one median to cut the triangle into two equal-area triangles. Step 3: Use the other median and the midpoint theorem to divide the pieces again into equal triangles. Step 4: Four resulting pieces have equal corresponding sides and angles, so they are congruent. Step 5: Rearrange the four pieces into the required larger shape. Cutting and rearranging preserves area and congruence. Conclusion: the construction gives four congruent triangles without using computer code.
7A more general midpoint theorem and its converse. In a quadrilateral ABCD, suppose AB ∥ DC. Let E be the midpoint of AD. A line through E meets BC at F. (i) If EF ∥ AB, show that F is the midpoint of BC and conclude that EF = (AB + CD)/2. (ii) If F is the midpoint of BC, show that EF ∥ AB.
Answer / method
Step 1: In trapezium ABCD, let E be the midpoint of AD and draw EF ∥ AB. Step 2: In ΔADC, the converse midpoint theorem shows that the point where EF meets AC is the midpoint of AC. Step 3: In ΔABC, the midpoint theorem shows that the line through this midpoint parallel to AB meets BC at its midpoint. Hence F is the midpoint of BC. Step 4: The two parallel base lengths are combined by the two midpoint segments, so EF = 1/2(AB + DC). Step 5: Conversely, if F is the midpoint of BC, apply the same argument with the two triangles on diagonal AC. Conclusion: EF ∥ AB ∥ DC and the midpoint formula follows.
8The diagonals AC and BD of a parallelogram ABCD intersect at O. A line through O meets AB and CD at P and Q respectively. Show that O is the midpoint of PQ.
Answer / method
Step 1: Let O be the intersection of diagonals AC and BD. In a parallelogram, AO = OC and BO = OD. Step 2: Since P and Q lie on AB and CD and OP and OQ are the corresponding cut segments, compare ΔAOP and ΔCOQ. Step 3: Opposite sides are equal and parallel, and vertically opposite angles at O are equal. Therefore the triangles are congruent. Step 4: Hence OP = OQ. Conclusion: O is the midpoint of PQ. Therefore the diagonals AC and BD of the parallelogram bisect the segment joining the two points.
9ABCD is a trapezium with parallel sides AD = 3 cm and BC = 5 cm. E and F are the midpoints of the non-parallel sides. Find the ratio of the areas of quadrilaterals AEFD and EBCF.
Answer / method
Step 1: The trapezium midpoint formula is EF = 1/2(AD + BC). Step 2: Substitute AD = 3 cm and BC = 5 cm: EF = 1/2(3 + 5) = 4 cm. Step 3: The upper trapezium has parallel sides 3 cm and 4 cm. The lower trapezium has parallel sides 4 cm and 5 cm. Step 4: Their heights are equal, so their area ratio is (3 + 4):(4 + 5) = 7:9. Conclusion: EF = 4 cm and the area ratio is 7:9.
10Consider 4 points A, B, C, D in the plane with no three collinear. Answer the following questions, considering different possible positions of the points: (i) How many different quadrilaterals do they form if quadrilaterals are allowed to be self-intersecting or non-convex? (ii) How many are self-intersecting? How many are convex?
Answer / method
Step 1: If A, B, C and D are the vertices of a convex quadrilateral, fix one cyclic order. Reversing the order does not make a new quadrilateral, so there are three orders to consider. Step 2: Of these three orders, one follows the boundary and is convex. Step 3: The other two orders join opposite vertices in the wrong order, so their sides cross and they are self-intersecting. Step 4: If one point lies inside the triangle formed by the other three, each of the three orders gives a simple non-convex quadrilateral. Conclusion: the answer depends on the placement of the four points: convex position gives one convex and two self-intersecting orders; one point inside the other three gives three non-convex orders.
11Suppose P is a point on side AB of triangle ABC and the line through P parallel to BC meets AC at Q. Assume AP = 1 and AQ = √5. (i) If PB = 2, find QC. (ii) If PB = 1/3, find QC. (iii) If PB = 2/3, find QC. (iv) Show that if AP/PB is rational, then AP/PB = AQ/QC.
Answer / method
Step 1: Since PQ ∥ BC, ΔAPQ ~ ΔABC. Therefore AP/AB = AQ/AC. Step 2: (i) AP = 1 and PB = 2, so AB = 3 and AP/AB = 1/3. Thus QC/AC = 2/3. Since AQ = √5, QC = 2√5. Step 3: (ii) PB = 1/3, so AB = 1 + 1/3 = 4/3 and AP/AB = 3/4. Thus QC/AC = 1/4. Since AC = AQ ÷ (3/4) = 4√5/3, QC = √5/3. Step 4: (iii) PB = 2/3, so AB = 5/3 and AP/AB = 3/5. Thus QC/AC = 2/5. Since AC = 5√5/3, QC = 2√5/3. Step 5: (iv) From AP/AB = AQ/AC, write AB = AP + PB and AC = AQ + QC. Subtract the equal ratios to obtain AP/PB = AQ/QC.
12Suppose ABCD is a parallelogram and M, N are midpoints of AB and CD respectively. (i) Show that segments DM and BN trisect AC. (ii) Use part (i) to find a procedure to trisect any given segment PQ. Find two other ways to trisect PQ, one using Exercise 11 and a third using the Centroid Theorem.
Answer / method
Step 1: In parallelogram ABCD, let M and N be the midpoints of AB and CD. Step 2: Draw diagonal AC. In ΔABC, the midpoint theorem gives MN ∥ AC and MN = 1/2 AC. Step 3: The lines DM and BN meet AC at points that divide AC into equal parts, so AC is split into three equal segments. Step 4: To trisect any segment PQ, place it as a diagonal of a parallelogram and repeat the same midpoint construction. Step 5: Another method uses the midpoint theorem directly; a third uses the centroid, which divides a median in the ratio 2:1 and therefore supplies the required one-third point. Conclusion: the segment can be divided into three equal parts by these constructions.
13Is the Midpoint Theorem for Quadrilaterals true when the quadrilateral is non-convex? How about when it is self-intersecting? (i) Explain why. (ii) In a special case the Varignon parallelogram becomes a single segment. When will this happen? (iii) Does the reasoning apply even when ABCD is non-planar?
Answer / method
Step 1: Join the midpoints of any quadrilateral. The Varignon proof uses the midpoint theorem in the four triangles formed by its diagonals. Step 2: Those midpoint segments remain parallel to the diagonals in the same way even when the quadrilateral is non-convex or self-intersecting. Step 3: Therefore the midpoint quadrilateral is still a parallelogram. In the non-planar case, the four midpoints also lie in one plane because they form the parallelogram determined by the two diagonal midpoint directions. Step 4: The Varignon parallelogram becomes a single segment when the two original diagonals are parallel, because all its sides then have one direction. Conclusion: the theorem extends beyond ordinary convex quadrilaterals, with that special degeneracy.
14Let P, Q, R, S be four points on sides AB, BC, CD and DA respectively of a quadrilateral ABCD. Suppose PQRS is a parallelogram. Must P, Q, R and S be midpoints of the respective sides?
Answer / method
Step 1: Assume P, Q, R and S divide AB, BC, CD and DA in the same fraction t, where t is not 1/2. Step 2: In a parallelogram, equal fractional movement along consecutive sides produces PQ ∥ SR and QR ∥ PS. Step 3: Thus PQRS is a parallelogram even though P, Q, R and S are not the midpoints. Conclusion: No. A parallelogram condition alone does not force the four points to be midpoints.
15(i) Complete the proof of the Midpoint Theorem by extending PQ beyond Q until S. By how much should PQ be extended? (ii) Give a similar proof of the converse of the Midpoint Theorem, starting by extending PQ to a suitable point S.
Answer / method
Step 1: Extend PQ beyond Q to S so that QS = PQ. Step 2: Since Q is the midpoint of AC, A, Q and C are collinear and AQ = QC. Since P, Q and S are collinear and PQ = QS, Q is the midpoint of PS. Step 3: The vertical angles ∠AQP and ∠CQS are equal, and AQ = QC, PQ = QS. Hence ΔAPQ ≅ ΔCSQ by SAS. Step 4: Corresponding angles give AP ∥ CS. Thus APCS is a parallelogram, so PS ∥ AC and the midpoint relation is obtained. Step 5: For the converse, start with P as the midpoint of AB and PQ ∥ BC. Similar triangles APQ and ABC give AQ/AC = AP/AB = 1/2. Conclusion: AQ = QC, so Q is the midpoint of AC.
*16Review all the properties of a rhombus, rectangle and square proved in Grade 8. Formulate a converse of each. Decide if the converse is true.
Answer / method
Step 1: Write each property in the form If the conditions hold, then the conclusion follows. Step 2: Reverse the conditions and conclusion to form a possible converse. Step 3: Equal opposite sides in both pairs imply a parallelogram; one pair of equal parallel sides also implies a parallelogram; and diagonals that bisect each other imply a parallelogram. Step 4: Within a parallelogram, equal diagonals imply a rectangle, perpendicular diagonals imply a rhombus, and being both gives a square. Step 5: Equal diagonals or perpendicular diagonals alone do not classify an arbitrary quadrilateral. Conclusion: every converse must include all necessary extra conditions, especially the parallelogram condition.
*17Show that the sum of angles of a non-planar quadrilateral is always less than 360°. Can you find a non-planar quadrilateral ABCD for which ∠A + ∠B + ∠C + ∠D = 2°? Think of a diagonal as a hinge around which two triangles can rotate.
Answer / method
Step 1: Draw diagonal AC and regard it as a hinge joining triangles ABC and ADC. Step 2: In a flat quadrilateral, the two triangles lie in one plane and the four boundary angles have total 360°. Step 3: Rotate one triangle about AC so the quadrilateral becomes non-planar. The boundary angles at the four outer vertices become smaller than their flat-position values. Step 4: Therefore ∠A + ∠B + ∠C + ∠D < 360°. Step 5: Continue the rotation towards a folded position. The four angles can be made very small, so a configuration with total 2° is possible.
*18Use two coloured copies of the 4-gon SOME sharing a vertex to investigate the third method of tiling the plane. Prove the exact match using four parallelograms, then repeat the procedure along each diagonal in both directions.
Answer / method
Step 1: Take two coloured copies of the four-gon SOME with a common vertex. Step 2: Match equal corresponding sides. The spaces between the copies have opposite sides equal and parallel, so they are parallelograms. Step 3: Slide one copy along a diagonal until its corresponding vertices coincide with the second copy. The matching sides fit exactly. Step 4: Repeat the same placement along both diagonal directions. Step 5: Adjacent copies meet on complete sides, their interiors do not overlap, and the repeated pattern covers the plane. Conclusion: the four-parallelogram construction proves the tiling.
*19For a 4-gon with three side lengths 2, 5 and 11, can the fourth side be 100? Can it be 10? Can it be 1? What are the possible lengths of the fourth side? For positive numbers a, b, c, d, how will you decide whether a 4-gon exists?
Answer / method
Step 1: Let the fourth side be x. The four lengths are 2, 5, 11 and x. Step 2: A quadrilateral exists only if the largest length is less than the sum of the other three. Step 3: If x is not the largest, the condition is 11 < 2 + 5 + x, so x > 4. Also x < 2 + 5 + 11 = 18. Step 4: If x is the largest, the condition is x < 2 + 5 + 11 = 18; the lower restriction is still x > 4. Conclusion: 4 < x < 18. Thus x = 100 is impossible, x = 10 is possible, and x = 1 is impossible. Step 5: In general, arrange a, b, c and d so the largest is L. A quadrilateral exists exactly when L is less than the sum of the other three.
*20Counting diagonals of a polygon. (i) Define a diagonal of an n-gon. How many diagonals does an n-gon have? Make a table for small values of n. (ii) Guess and justify a formula. How many diagonals are added when the number of sides is increased by 1?
Answer / method
Step 1: A diagonal joins two non-adjacent vertices of a polygon. Step 2: From each vertex of an n-gon, there are n − 3 possible diagonals. Step 3: This counts every diagonal twice, once from each endpoint. Step 4: Therefore the number of diagonals is n(n − 3)/2. Step 5: For n = 3, 4, 5, 6, 7 and 8, the numbers are 0, 2, 5, 9, 14 and 20. Step 6: Increasing from n sides to n + 1 sides adds (n + 1)(n − 2)/2 − n(n − 3)/2 = n − 1 diagonals.
*21What is the sum of the angles of a plane non-self-intersecting n-gon? Find the next few values after 180° for n = 3 and 360° for n = 4. Find a formula in terms of n and prove it.
Answer / method
Step 1: Draw diagonals from one vertex of a plane non-self-intersecting n-gon to every non-adjacent vertex. Step 2: These diagonals divide the polygon into n − 2 triangles. Step 3: Each triangle has angle sum 180°. Step 4: Therefore the polygon angle sum is (n − 2) × 180°. Step 5: For n = 3, 4, 5, 6 and 7, the sums are 180°, 360°, 540°, 720° and 900°.
*22Study the multiple converses of the Midpoint Theorem. Express the theorem in “If … then …” form, name its conditions and conclusions, and decide which possible converse statements are true.
Answer / method
Step 1: The Midpoint Theorem says: if P and Q are the midpoints of AB and AC, then PQ ∥ BC and PQ = 1/2 BC. Step 2: A true converse is: if P is the midpoint of AB and PQ ∥ BC, then Q is the midpoint of AC. The proof is the similarity proof in the theorem bank. Step 3: A length statement such as PQ = 1/2 BC alone is not enough; the position of Q or a parallel-line condition is also needed. Step 4: A statement that only one point is a midpoint must be checked with the missing parallel condition before it is accepted. Conclusion: reverse both the hypotheses and conclusion carefully, and list every extra condition used.
*23Validity of tiling methods. Show using reasoning that each of the three tiling methods studied in the chapter produces a tiling of the plane.
Answer / method
Step 1: For a parallelogram, translate copies along both side directions. Equal translated sides meet exactly, so there are no gaps or overlaps. Step 2: For a triangle, join it to a congruent half-turn copy to make a parallelogram. Tile the plane with those parallelograms, then use each diagonal to recover triangle copies. Step 3: For a quadrilateral, join the side midpoints to make a Varignon parallelogram and repeat the compatible copy construction along its diagonals. Step 4: In every method, check that copies cover the whole plane, adjacent copies share complete matching sides, and interiors do not overlap. Conclusion: these three checks establish that each construction is a genuine tiling.
*24What fraction of the square is shaded? (Fig. 12.43)
Answer / method

Geometric Solution to the Shaded Square Puzzle (Fig. 12.43):

Fig 12.43: A square partitioned by segments joining vertices to midpoints of opposite sidesABCDM₁M₂M₃M₄
  1. Figure Description: In square ABCD with side length s, midpoints of AB, BC, CD, DA are M₁, M₂, M₃, M₄. Segments are drawn from each vertex to the midpoint of the opposite side in cyclic order: AM₂, BM₃, CM₄, and DM₁. These 4 segments intersect to enclose a central tilted quadrilateral.
  2. Symmetry & Shape: By 90° rotational symmetry about the center, the central region is an equiangular, equilateral quadrilateral with right angles — a square.
  3. Method 1 — Right-Triangle Similarities:
    • Let side of square ABCD be s = 1 (Total Area = 1).
    • In right ΔABM₃, legs are AB = 1 and BM₃ = 1/2. Hypotenuse AM₃ = √(1² + (1/2)²) = √5 / 2.
    • The intersecting transversals are perpendicular because their slopes are 1/2 and −2 (product is −1).
    • In the similar right triangles formed along AM₃, the ratio of short leg to long leg to hypotenuse is 1 : 2 : √5.
    • The side length x of the central shaded square equals (1/2) / (√5 / 2) = 1 / √5.
    • Therefore, the area of the shaded square is:
    ext{Shaded Area} = x^2 = \left( rac{1}{\sqrt{5}} ight)^2 = rac{1}{5}
  4. Method 2 — Visual Dissection into 5 Equal Squares:
    • The outer region around the central shaded square consists of 4 right triangles with legs in ratio 1:2 and 4 right trapezoids.
    • Each outer right triangle can be cut and translated into the adjacent trapezoid to assemble a square identical to the central shaded square!
    • This dissects the entire original square into exactly 5 congruent squares of identical size.
    • Since the shaded square is 1 of these 5 equal squares, its fraction is exactly 1/5.
Answer: Exactly 1/5 (or 20%) of the square is shaded.
Diagram note: The textbook figures are represented by clean diagrams in the study notes. Construction questions that depend on a photographed figure include the required geometric construction and justification in their answers.

Mini Test - 10 Questions

Select one option for each question.

Q1. The diagonals of a parallelogram:

Q2. The line joining the midpoints of two sides of a triangle is:

Q3. A square is both a:

Q4. The Varignon quadrilateral formed by side midpoints is always a:

Q5. The sum of the interior angles of a quadrilateral is:

Q6. The number of diagonals of an n-gon is:

Q7. A regular pentagon cannot tile the plane because:

Q8. The area of a trapezium is:

Q9. For a quadrilateral with four given positive side lengths, the largest side must be:

Q10. The centroid divides each median in the ratio:

Score: 0 / 10