CBSE Class 9 Maths Ganita Manjari Part 2 Ch 12 Quadrilaterals Long Answer and Case Study Questions with Answers

Long Answer Questions

Question 1. (i) State the chapter's definition of a quadrilateral.
(ii) The definition rules out three kinds of four-sided figure. Name each kind and explain which part of the definition excludes it.
(iii) Explain why a four-sided figure with three collinear vertices looks like a triangle or has overlapping sides.
(iv) Describe how four-sided figures are classified into skew, self-intersecting, non-convex and convex, and say which of these are the 'usual' quadrilaterals.
Answer:
(i) For four distinct points A, B, C, D in a plane, the points of AB, BC, CD, DA form quadrilateral ABCD if every such point other than A, B, C, D lies on exactly one of the four segments.
(ii) 1. Three collinear vertices: two sides lie on one line, so some points are on two segments ('exactly one' fails).
2. Self-intersecting: two sides cross; the crossing point is on two segments ('exactly one' fails).
3. Non-planar (skew): the points are not in one plane ('in a plane' fails).
(iii) If A, B, C are collinear there is no corner at B. If B is between A and C, the figure is just triangle ACD with an extra mark on a side. If B is outside segment AC, then AB and BC overlap.
(iv) First: in one plane or not (not → skew). Then: do sides cross (yes → self-intersecting)? Then: all angles below 180° (convex) or one reflex (non-convex).
The usual quadrilaterals are the convex and non-convex ones.
In simple words: Planar, no crossing sides, and no three corners in a line: that is a quadrilateral, convex or dented.

Teacher's Note:
1. Present (iv) as a step-by-step sorting.
2. Link each exclusion to a phrase in the definition.

Question 2. Prove all three parts of Theorem 1 for a parallelogram ABCD: (i) the opposite sides are equal; (ii) the opposite angles are equal; (iii) the diagonals bisect each other. Name the congruence test used in each proof, and state for (ii) the property of parallel lines that the proof depends on.
Answer:
(i) Draw AC. ∠BAC = ∠DCA (alternate, AB ∥ DC) and ∠DAC = ∠BCA (alternate, AD ∥ BC). AC is common. So △ACD ≅ △CAB by ASA, giving AB = DC and AD = BC.
(ii) No congruence is needed. AD ∥ BC with transversal AB gives ∠A + ∠B = 180° (co-interior). Similarly ∠B + ∠C = 180° and ∠C + ∠D = 180°.
From the first two, ∠A = ∠C; from the second and third, ∠B = ∠D.
Property used: co-interior angles on the same side of a transversal add to 180°.
(iii) Let diagonals meet at E. In △AED and △CEB: AD = CB (part i), ∠DAE = ∠BCE and ∠ADE = ∠CBE (alternate). So △AED ≅ △CEB by ASA.
Hence EA = EC and ED = EB: the diagonals bisect each other.
In simple words: Alternate angles give congruent triangles for sides and diagonals; co-interior angles give equal opposite angles.

Teacher's Note:
1. Part (ii) uses no congruence at all.
2. Part (iii) builds on part (i).

Question 3. Each of the three properties in Theorem 1 has a converse, and all three converses are true.
(i) Write out the three converse statements.
(ii) Prove the converse of the equal-opposite-sides property.
(iii) Prove the converse of the equal-opposite-angles property.
(iv) Prove the converse of the bisecting-diagonals property.
Answer:
(i) 1. If the opposite sides of a quadrilateral are equal, it is a parallelogram.
2. If the opposite angles of a quadrilateral are equal, it is a parallelogram.
3. If the diagonals of a quadrilateral bisect each other, it is a parallelogram.
(ii) Let AB = DC, AD = BC; draw AC. △ACD ≅ △CAB by SSS. So ∠DCA = ∠BAC and ∠DAC = ∠BCA; as alternate angles these give AB ∥ DC and AD ∥ BC.
(iii) Let ∠A = ∠C and ∠B = ∠D. Then 360° = 2(∠A + ∠B), so ∠A + ∠B = 180°; these are co-interior, so BC ∥ AD. Likewise ∠B + ∠C = 180° gives AB ∥ DC.
(iv) Let EA = EC, EB = ED. △AED ≅ △CEB by SAS (vertically opposite angles at E), so ∠DAE = ∠BCE and AD ∥ BC. Similarly △AEB ≅ △CED gives AB ∥ DC.
In each case both pairs of opposite sides are parallel, so ABCD is a parallelogram.
In simple words: Each converse ends by showing both pairs of opposite sides are parallel.

Teacher's Note:
1. SSS, angle sum, SAS: one method each.
2. Always finish with 'both pairs parallel'.

Question 4. (i) Prove that a parallelogram with one right angle is a rectangle.
(ii) Prove that a parallelogram whose diagonals are equal is a rectangle.
(iii) Prove that a parallelogram whose diagonals are perpendicular is a rhombus.
(iv) Explain why 'the diagonals are equal' on its own is not enough to make a quadrilateral a rectangle, giving a counterexample.
Answer:
(i) ∠A = 90°. ∠A + ∠B = 180° (co-interior), so ∠B = 90°. Opposite angles are equal, so ∠C = ∠D = 90°. It is a rectangle.
(ii) AC = BD. In △ABC and △BAD: AB common, BC = AD, AC = BD, so they are congruent (SSS) and ∠ABC = ∠BAD. These add to 180°, so each is 90°. By (i), a rectangle.
(iii) Diagonals meet at E at right angles, and EA = EC. In △AEB and △CEB: EA = EC, ∠AEB = ∠CEB = 90°, EB common. So they are congruent (SAS) and AB = CB. Adjacent sides equal, so all four are equal: a rhombus.
(iv) In (ii) the parallelogram property supplied BC = AD. Without it, the proof fails.
Counterexample: isosceles trapezium A(0, 0), B(10, 0), C(7, 4), D(3, 4). AC = BD = √65, but its angles are about 53°, 53°, 127°, 127°, so it is not a rectangle.
In simple words: Equal diagonals make a parallelogram a rectangle, but not every quadrilateral.

Teacher's Note:
1. Point out where 'parallelogram' was used.
2. Dropped hypotheses cause false converses.

Question 5. In a parallelogram ABCD the diagonal BD bisects ∠B.
(i) Prove that BD also bisects ∠D.
(ii) Prove that ABCD is a rhombus.
(iii) Given that ∠B = 70°, find all three angles of △ABD.
(iv) State one extra condition that would make ABCD a square.
Answer:
(i) AB ∥ DC: ∠ABD = ∠BDC (alternate). AD ∥ BC: ∠ADB = ∠DBC (alternate). Given ∠ABD = ∠DBC.
So ∠BDC = ∠ABD = ∠DBC = ∠ADB, hence ∠ADB = ∠BDC: BD bisects ∠D.
(ii) In △ABD, ∠ABD = ∠ADB, so AD = AB (sides opposite equal angles). Adjacent sides equal, so all four sides are equal: a rhombus.
(iii) ∠ABD = 70° ÷ 2 = 35°. ∠A = 180° − 70° = 110°. ∠ADB = 180° − 110° − 35° = 35°.
(iv) Any one of: one angle is 90°; or the diagonals AC and BD are equal.
In simple words: A diagonal that bisects an angle of a parallelogram makes it a rhombus.

Teacher's Note:
1. Chain the equal angles carefully.
2. The two 35° angles confirm part (ii).

Question 6. (i) State the Midpoint Theorem.
(ii) Prove it, using the line through C parallel to BA, and name the congruence test you use.
(iii) Use the theorem to prove that joining the midpoints of all three sides of △ABC divides it into four congruent triangles.
(iv) In such a triangle, AB = 12 cm, BC = 16 cm and CA = 18 cm. Find the perimeter of the middle triangle and its ratio to the perimeter of △ABC.
Answer:
(i) The segment joining the midpoints of two sides of a triangle is parallel to the third side and half its length.
(ii) P, Q are midpoints of AB, AC. The line through C parallel to BA meets PQ produced at R.
In △APQ and △CRQ: AQ = CQ, ∠AQP = ∠CQR (vertically opposite), ∠APQ = ∠CRQ (alternate). So they are congruent by AAS: PQ = QR, CR = AP.
AP = BP, so CR = BP and CR ∥ BP: BCRP is a parallelogram, so PR ∥ BC and PR = BC. Hence PQ ∥ BC and PQ = ½PR = ½BC.
(iii) With P, Q, R the midpoints of AB, AC, BC: QR = ½AB, RP = ½CA, PQ = ½BC.
△APQ, △PBR, △QRC and △RQP each have sides ½AB, ½BC, ½CA. So all four are congruent (SSS).
(iv) Sides of middle triangle: 6, 8, 9 cm. Perimeter = 23 cm. △ABC perimeter = 46 cm. Ratio = 1 : 2.
In simple words: Midpoint segments are half the sides, so the four small triangles are identical half-size copies.

Teacher's Note:
1. Include a labelled sketch in the exam.
2. Perimeter ratio is always 1:2.

Question 7. (i) State the converse of the Midpoint Theorem.
(ii) Prove it by retracing the proof of the Midpoint Theorem backwards.
(iii) Prove it again, this time using the fact that there is exactly one line through a given point parallel to a given line.
(iv) The second proof uses the Midpoint Theorem itself. Explain why this is not circular reasoning.
Answer:
(i) The line through the midpoint of one side of a triangle, parallel to another side, bisects the third side (and the segment is half the parallel side).
(ii) P is the midpoint of AB; the line through P parallel to BC meets AC at Q. Draw CR ∥ BA meeting PQ produced at R.
PR ∥ BC and CR ∥ BP, so BCRP is a parallelogram and CR = BP = PA.
In △APQ and △CRQ: ∠AQP = ∠CQR, ∠APQ = ∠CRQ, PA = RC, so they are congruent (AAS). Hence AQ = QC, and PQ = ½PR = ½BC.
(iii) Let M be the midpoint of AC. By the Midpoint Theorem PM ∥ BC. Lines PM and PQ both pass through P parallel to BC, so they coincide. Thus Q = M.
(iv) Circular reasoning would use the converse to prove itself. Here the Midpoint Theorem was already proved separately, and a new fact (uniqueness of the parallel) is added. Using a theorem to prove its converse is allowed.
In simple words: A theorem can help prove its converse as long as the converse itself is never assumed.

Teacher's Note:
1. Explain 'circular' in one line.
2. The uniqueness of the parallel is the new ingredient.

Question 8. (i) State the Centroid Theorem.
(ii) Taking the medians CP and BQ of △ABC meeting at M, and letting X and Y be the midpoints of BM and CM, prove that CM:MP = 2:1.
(iii) Explain how the argument extends to show that the third median also passes through M.
(iv) The three medians of a triangle measure 15 cm, 18 cm and 21 cm. Find the two parts of each median.
Answer:
(i) The three medians of a triangle meet at one point (the centroid), which divides each median 2:1, the longer part next to the vertex.
(ii) In △ABC: PQ ∥ BC and PQ = ½BC. In △MBC: XY ∥ BC and XY = ½BC (Midpoint Theorem twice).
So PQ ∥ XY and PQ = XY: PQYX is a parallelogram.
Its diagonals PY and QX bisect each other at M, so MP = MY.
Y is the midpoint of CM, so CY = YM. Then CM = 2YM = 2MP, i.e. CM:MP = 2:1 (similarly BM:MQ = 2:1).
(iii) Repeat with medians AR and BQ meeting at N: BN:NQ = 2:1. Only one point divides BQ in the ratio 2:1 from B, so N = M. Hence AR also passes through M.
(iv) 15 cm → 10 cm and 5 cm; 18 cm → 12 cm and 6 cm; 21 cm → 14 cm and 7 cm.
In simple words: A hidden parallelogram inside the triangle shows the centroid cuts each median 2:1.

Teacher's Note:
1. Draw the figure first; the parallelogram is the key.
2. Longer part always at the vertex.

Question 9. (i) State the Midpoint Theorem for Quadrilaterals.
(ii) Prove it for a quadrilateral ABCD with midpoints P, Q, R, S of AB, BC, CD, DA.
(iii) Prove that PR and QS bisect each other.
(iv) State the condition on the diagonals of ABCD for PQRS to be a rhombus, the condition for it to be a rectangle, and the condition for it to be a square.
Answer:
(i) The midpoints of the four sides of any quadrilateral are the vertices of a parallelogram (the Varignon parallelogram).
(ii) Draw AC. In △ABC, PQ ∥ AC and PQ = ½AC. In △ADC, SR ∥ AC and SR = ½AC.
So PQ ∥ SR and PQ = SR: PQRS is a parallelogram (equal parallel sides test).
(iii) PR and QS are the diagonals of the parallelogram PQRS, and the diagonals of a parallelogram bisect each other.
(iv) Rhombus: AC = BD. Rectangle: AC ⊥ BD. Square: AC = BD and AC ⊥ BD.
In simple words: Midpoints always give a parallelogram; its shape depends on the lengths and angle of the diagonals.

Teacher's Note:
1. Sides depend on diagonal lengths, angles on diagonal directions.
2. Part (iii) follows in one line.

Question 10. ABCD is a trapezium with AB ∥ DC and AB ≠ DC. The points E and F are the midpoints of the non-parallel sides AD and BC.
(i) Prove that EF ∥ AB.
(ii) Prove that EF = \( \frac{AB + DC}{2} \).
(iii) Given AB = 11 cm and DC = 19 cm, find EF.
(iv) State the converse of the result in (i) and say whether it is true.
Answer:
(i) Draw AC and let the line through E parallel to DC meet AC at G. In △ADC, E is the midpoint of AD and EG ∥ DC, so G is the midpoint of AC (converse of the Midpoint Theorem) and EG = ½DC.
In △ABC, G and F are midpoints of AC and BC, so GF ∥ AB and GF = ½AB. Both EG and GF are parallel to AB, so E, G, F lie on one line and EF ∥ AB.
(ii) EF = EG + GF = ½DC + ½AB = \( \frac{AB + DC}{2} \).
(iii) EF = \( \frac{11 + 19}{2} \) = 15 cm.
(iv) Converse: if E is the midpoint of AD and the line through E parallel to AB meets BC at F, then F is the midpoint of BC.
It is true: apply the converse of the Midpoint Theorem in △ADC and then in △ABC.
In simple words: The diagonal splits EF into two midpoint segments, so EF is the average of the parallel sides.

Teacher's Note:
1. G is the link between the two triangles.
2. EF lies between the two parallel sides in length.

Question 11. (i) Explain why the angles meeting at any vertex of a tiling must total 360°, and why this makes a quadrilateral a promising shape for tiling.
(ii) Describe the chapter's first method of tiling with a quadrilateral, using a rotation through 180°.
(iii) Explain why the rotated copy fits exactly along the shared edge.
(iv) Explain why repeating the procedure at every edge covers the plane with no gaps and no overlaps.
Answer:
(i) The tiles around a vertex fill one full turn with no gap and no overlap, so their angles total 360°. A quadrilateral's four angles already total 360°, so four copies (one of each angle) can fit round a point.
(ii) Place one copy. Mark the midpoint of an edge. Lay a second copy on top and rotate it 180° about that midpoint. Repeat at every edge of every copy.
(iii) A half-turn about the midpoint maps the edge onto itself, swapping its ends. So the new copy shares the whole edge with the old one.
(iv) At each vertex the four angles appear once each and total 360°, so they close up. Along each edge two copies meet exactly. So there is no gap or overlap, and continuing outwards covers the plane.
The chapter calls this an outline rather than a full proof.
In simple words: Half-turns make edges match; the 360° angle sum makes corners close.

Teacher's Note:
1. Use a cardboard quadrilateral to show it.
2. Works for any quadrilateral shape.

Question 12. The chapter's second method of tiling uses the Varignon parallelogram.
(i) Explain what is observed when the Varignon parallelogram is drawn inside every copy of the quadrilateral in a completed tiling.
(ii) Describe how the tiling can instead be built by starting from a grid of Varignon parallelograms.
(iii) Explain why the gaps left between the placed copies are themselves congruent copies of the original quadrilateral.
(iv) Compare the two methods, saying what each one makes easy to see.
Answer:
(i) The Varignon parallelograms fit together into a regular parallelogram grid, like the pinned-stick grid. The irregular tiling and the grid are the same picture seen two ways.
(ii) Draw the grid of Varignon parallelograms first. Place a copy of the quadrilateral on grid cells so that its four side-midpoints sit on the four vertices of a cell. A 3 × 3 block of such copies touch only at vertices.
(iii) Each gap is bounded by midpoints of the surrounding copies, and these form a parallelogram congruent to the Varignon parallelogram. The original quadrilateral can be rebuilt from its Varignon parallelogram, so each gap is a congruent copy of it.
(iv) Method 1 shows clearly why edges match and angles close up, but builds the tiling piece by piece.
Method 2 shows the overall structure: a parallelogram grid in disguise, repeating regularly. It needs the reconstruction argument to explain the fit.
In simple words: Every quadrilateral tiling hides a parallelogram grid made of its Varignon parallelograms.

Teacher's Note:
1. Drawing helps more than words here.
2. Compare 'local' (Method 1) and 'global' (Method 2) views.

Question 13. (i) State the condition a regular polygon must satisfy in order to tile the plane on its own.
(ii) Find the internal angle of a regular triangle, square, pentagon and hexagon.
(iii) Decide which of those four can tile the plane, and give the number of tiles meeting at each vertex.
(iv) Describe what is meant by an aperiodic tiling, and name the shape found in 2023 together with the team who found it.
Answer:
(i) Its interior angle must divide 360° exactly.
(ii) Using \( \frac{(n-2) \times 180^\circ}{n} \): triangle 60°, square 90°, pentagon 108°, hexagon 120°.
(iii) 360 ÷ 60 = 6, 360 ÷ 90 = 4, 360 ÷ 108 = 3⅓, 360 ÷ 120 = 3.
Triangle tiles with 6 at a vertex, square with 4, hexagon with 3. The pentagon does not tile (36° gap with three, overlap with four).
Polygons with 7 or more sides have angles between 120° and 180°, so no whole number of them makes 360°.
(iv) An aperiodic tiling covers the plane but never repeats. The 'hat' (2023) by Smith, Myers, Kaplan and Goodman-Strauss was the first single shape to do this.
In simple words: Only triangles, squares and hexagons have angles that fit exactly into 360°.

Teacher's Note:
1. Make a small table of n, angle and 360 ÷ angle.
2. Aperiodic is a recent discovery.

Question 14. (i) Explain what is meant by a diagonal of an n-gon.
(ii) Make a table of the number of diagonals for n = 3, 4, 5, 6, 7 and 8.
(iii) State the formula for the number of diagonals of an n-gon.
(iv) Justify the formula by counting the diagonals from each vertex.
Answer:
(i) A diagonal is a segment joining two vertices that are not adjacent.
(ii) n = 3: 0; n = 4: 2; n = 5: 5; n = 6: 9; n = 7: 14; n = 8: 20.
(iii) Number of diagonals = \( \frac{n(n-3)}{2} \).
(iv) From one vertex, diagonals go to all vertices except itself and its two neighbours: n − 3. For n vertices that is n(n − 3), but each diagonal is counted from both ends, so divide by 2.
Check: n = 6 gives 9 and n = 3 gives 0, matching the table.
In simple words: Each vertex sends n − 3 diagonals; halve the total to avoid double counting.

Teacher's Note:
1. Draw the hexagon to verify 9.
2. A triangle has no diagonals.

Question 15. (i) State the sum of the internal angles of a triangle, a quadrilateral, a pentagon and a hexagon.
(ii) State the formula for the sum of the internal angles of an n-gon.
(iii) Prove the formula by cutting the n-gon into triangles from one vertex.
(iv) The chapter notes that a self-intersecting quadrilateral has an angle sum of less than 360°. Explain why the proof in (iii) does not apply to such a figure.
Answer:
(i) Triangle 180°, quadrilateral 360°, pentagon 540°, hexagon 720°.
(ii) Sum = (n − 2) × 180°.
(iii) From one vertex draw all n − 3 diagonals. They cut the polygon into n − 2 triangles.
The angles of these triangles together make up all the interior angles of the polygon exactly once.
So the sum is (n − 2) × 180°.
(iv) The proof needs the diagonals to lie inside and the triangles to cover the polygon exactly once. A self-intersecting figure has no clear inside; the triangles overlap or leave parts uncovered, so the argument fails.
In simple words: n − 2 triangles give (n − 2) × 180°, but only when the triangles fit inside without overlapping.

Teacher's Note:
1. Show a hexagon cut into 4 triangles.
2. Know the limits of a proof.

Question 16. (i) State the condition on three positive numbers for a triangle with those side lengths to exist.
(ii) State the corresponding condition on four positive numbers for a quadrilateral with those side lengths to exist.
(iii) Three sides of a quadrilateral measure 2 cm, 5 cm and 11 cm. Find the range of possible lengths of the fourth side.
(iv) Hence decide whether the fourth side could measure 1 cm, 10 cm or 100 cm, giving a reason in each case.
Answer:
(i) Each number must be less than the sum of the other two. If a ≤ b ≤ c, it is enough that a + b > c.
(ii) Each number must be less than the sum of the other three. If a ≤ b ≤ c ≤ d, it is enough that a + b + c > d.
(iii) d < 2 + 5 + 11 = 18, and 11 < 2 + 5 + d gives d > 4. So 4 cm < d < 18 cm.
(iv) 1 cm: No; 1 + 2 + 5 = 8 cannot reach across 11.
10 cm: Yes; it lies between 4 and 18.
100 cm: No; 2 + 5 + 11 = 18 cannot reach across 100.
In simple words: The longest side must be shorter than the other three put together.

Teacher's Note:
1. Only the largest side needs checking.
2. Think of three sticks trying to span the fourth.

Question 17. ABCD is a parallelogram. Two points P and Q are taken on the diagonal BD so that DP = BQ, with P nearer to D and Q nearer to B.
(i) Prove that △APD ≅ △CQB.
(ii) Prove that APCQ is a parallelogram.
(iii) State which of the four tests for a parallelogram you used in (ii).
(iv) Explain what happens to the figure APCQ when P and Q both move to the midpoint of BD.
Answer:
(i) AD = CB and AD ∥ CB. With BD as transversal, ∠ADP = ∠CBQ (alternate). DP = BQ (given).
So △APD ≅ △CQB (SAS).
(ii) By CPCT, AP = CQ and ∠APD = ∠CQB.
∠APQ = 180° − ∠APD and ∠CQP = 180° − ∠CQB (linear pairs), so ∠APQ = ∠CQP.
These are alternate angles for AP and CQ with transversal PQ, so AP ∥ CQ. With AP = CQ, APCQ is a parallelogram.
(iii) The equal parallel sides test.
(iv) P and Q meet at the midpoint of BD, which is also the midpoint of AC. APCQ collapses into the segment AC, a figure with zero area.
In simple words: Equal pieces cut from the diagonal give congruent triangles, and so a parallelogram APCQ.

Teacher's Note:
1. Linear pairs turn one angle equality into another.
2. Discuss the limiting case in (iv).

Question 18. ABCD is a parallelogram in which AB ≠ BC. The bisectors of the four internal angles are drawn, and P, Q, R and S are the points in which consecutive bisectors meet.
(i) Show that the bisectors of two adjacent angles of a parallelogram meet at right angles.
(ii) Hence prove that PQRS is a rectangle.
(iii) Explain why the condition AB ≠ BC is needed.
(iv) State what PQRS becomes when ABCD is a rectangle.
Answer:
(i) ∠A + ∠B = 180° (co-interior). In △ABP, ∠PAB + ∠PBA = \( \frac{\angle A}{2} \) + \( \frac{\angle B}{2} \) = \( \frac{180^\circ}{2} \) = 90°.
So ∠APB = 180° − 90° = 90°.
(ii) The same holds for every pair of adjacent angles, so the angles of PQRS at P, Q, R and S are all 90°. Hence PQRS is a rectangle.
(iii) If AB = BC, ABCD is a rhombus; its bisectors are its diagonals and all meet at one point. Then P, Q, R, S coincide and there is no rectangle.
(iv) If ABCD is a rectangle, each bisector makes 45° with the sides and PQRS is a square.
In simple words: Half of 180° is 90°, so neighbouring bisectors always meet at right angles.

Teacher's Note:
1. The key is halving a supplementary pair.
2. Check the special cases in (iii) and (iv).

Question 19. The Midpoint Theorem, applied to △ABC with P on AB and Q on AC, has two assumptions and two conclusions: Assumptions — (P MID) P is the midpoint of AB; (Q MID) Q is the midpoint of AC. Conclusions — (PRLL) PQ ∥ BC; (HALF) PQ = ½BC.
(i) Write out in full the statement 'if (PRLL) and (HALF) are true, then (P MID) and (Q MID) are true', and state whether it is true.
(ii) Write the converse of the Midpoint Theorem in terms of these four named conditions.
(iii) There is one further combination of two assumptions and two conclusions still to be examined. Write it out in words.
(iv) Explain what this exercise shows about the idea of 'the' converse of a theorem.
Answer:
(i) In △ABC, let P be on AB and Q on AC. If PQ ∥ BC and PQ = ½BC, then P is the midpoint of AB and Q is the midpoint of AC. This is true (proved by running the Midpoint Theorem proof backwards).
(ii) If (P MID) and (PRLL), then (Q MID), and (HALF) follows too.
(iii) Remaining pair: (P MID) and (HALF). 'If P is the midpoint of AB and PQ = ½BC, must Q be the midpoint of AC, or must PQ be parallel to BC?'
(iv) A theorem with several assumptions and conclusions has several converses, not just one. Each must be judged true or false separately. The chapter's chosen converse is simply the most useful one.
In simple words: Swapping assumptions and conclusions can be done in many ways, giving many converses.

Teacher's Note:
1. List the pairings systematically.
2. Each converse needs its own proof or counterexample.

Question 20. Write an account of why the converse of a theorem must always be proved separately, drawing all your illustrations from this chapter. Your account must include (i) the three properties of a parallelogram whose converses all turn out to be true; (ii) a property of a quadrilateral whose converse is false, with a counterexample; (iii) the strategy of running a proof backwards, and the chapter's warning about when it fails; (iv) the case of the Midpoint Theorem, whose converse is proved using the theorem itself; (v) what a student should do before assuming that a converse is true.
Answer:
'If X then Y' and 'if Y then X' are different claims; proving one says nothing about the other.
(i) Parallelogram properties with true converses: equal opposite sides (Theorem 2, by SSS), equal opposite angles (Theorem 3, by the 360° sum), bisecting diagonals (Theorem 4, by SAS). Their truth turns properties into tests.
(ii) A rectangle has equal diagonals, but 'equal diagonals ⇒ rectangle' is false. The isosceles trapezium A(0, 0), B(10, 0), C(7, 4), D(3, 4) has AC = BD = √65 and no right angle.
(iii) Running a proof backwards can work (Theorem 2 reverses Theorem 1(a), using SSS instead of ASA). But each step must be checked: 'x = y ⇒ x² = y²' cannot be reversed, since 3 and −3 have equal squares.
(iv) The converse of the Midpoint Theorem is proved using the theorem plus a new fact: only one line through a point is parallel to a given line. This is not circular, because the converse is never assumed.
(v) Write the converse in 'if … then …' form; look for a counterexample; if none appears, try to prove it, checking every reversed step; remember a theorem may have several converses.
In simple words: Some converses are true and some are false, so each one must be tested on its own.

Teacher's Note:
1. Use headings (i) to (v) in the answer.
2. One true and one false example make the point clearly.

Case Study Based Questions

Question 21. The Pinned Stick Grid: Two sets of long sticks, one set horizontal and the other vertical, are pinned together wherever they cross so that they can turn freely about the pins. The grid formed by the sticks tiles a region of the plane with squares. One student holds the bottom stick fixed while another pushes the vertical side sideways.
(a) State what remains true of the two sets of sticks after the grid has been pushed over.
(b) State what has changed about the angle between the two sets.
(c) Name the shape that each cell of the grid has become.
(d) Explain why the region is still tiled, with no gaps and no overlaps.
Answer:
(a) Each set is still a set of parallel sticks.
(b) The angle between the sets is no longer 90°.
(c) Each cell is now a parallelogram.
(d) At every pin the four cell angles still make 360°, and cells still meet edge to edge. Nothing is stretched or torn, so no gaps or overlaps appear.
In simple words: Pushing the grid turns squares into parallelograms, but they still fit together perfectly.

Teacher's Note:
1. Only the angle changes.
2. Link to the four-bar linkage idea.

Question 22. Angles in a Parallelogram: In parallelogram ABCD, the angle at A measures 108°.
(a) Find ∠C.
(b) Find ∠B.
(c) Find ∠D.
(d) Name the property of a parallelogram used in part (a) and the property of parallel lines used in part (b).
Answer:
(a) ∠C = ∠A = 108°.
(b) ∠B = 180° − 108° = 72°.
(c) ∠D = ∠B = 72° (check: 108 + 72 + 108 + 72 = 360).
(d) (a): opposite angles of a parallelogram are equal. (b): co-interior angles add to 180° (AD ∥ BC, transversal AB).
In simple words: Opposite angles match; neighbouring angles add to 180°.

Teacher's Note:
1. Always check the 360° total.
2. Name properties in full words.

Question 23. The Midpoint Triangle: In △ABC, AB = 14 cm, BC = 18 cm and CA = 20 cm. The points P, Q and R are the midpoints of AB, AC and BC respectively, and are joined to form △PQR.
(a) Find PQ.
(b) Find QR.
(c) Find RP.
(d) Find the ratio of the perimeter of △PQR to the perimeter of △ABC.
Answer:
(a) PQ = ½BC = 9 cm.
(b) QR = ½AB = 7 cm.
(c) RP = ½CA = 10 cm.
(d) 26 : 52 = 1 : 2.
In simple words: Each side of the inner triangle is half the side it runs parallel to.

Teacher's Note:
1. Match each segment with the side it does not touch.
2. Ratio is always 1:2.

Question 24. Finding the Centroid: In △ABC the medians AR and BQ meet at the centroid M. The median AR measures 24 cm and the median BQ measures 30 cm.
(a) Find AM.
(b) Find MR.
(c) Find BM.
(d) Find MQ.
Answer:
(a) AM = ⅔ × 24 = 16 cm.
(b) MR = ⅓ × 24 = 8 cm.
(c) BM = ⅔ × 30 = 20 cm.
(d) MQ = ⅓ × 30 = 10 cm.
In simple words: Two-thirds of each median lies on the vertex side of the centroid.

Teacher's Note:
1. Check 16:8 and 20:10 are both 2:1.
2. Vertex part is the longer one.

Question 25. Midpoints of a Kite-Shaped Plot: A four-sided plot of land ABCD has diagonals measuring AC = 20 m and BD = 14 m. Posts are placed at the midpoints P, Q, R and S of the four boundaries and joined by a fence.
(a) Find the length PQ of the fence joining the midpoints of AB and BC.
(b) Find the length QR.
(c) Find the total length of the fence PQRS.
(d) State what shape the fence would be if the two diagonals of the plot were equal in length.
Answer:
(a) PQ = ½AC = 10 m.
(b) QR = ½BD = 7 m.
(c) 2(10 + 7) = 34 m (= AC + BD).
(d) A rhombus.
In simple words: The fence is a parallelogram whose sides are half the diagonals.

Teacher's Note:
1. Fence length = sum of diagonals.
2. Equal diagonals → rhombus.

Question 26. A Trapezium Window: A window frame ABCD has AB ∥ DC, with AB = 8 cm and DC = 18 cm. A strengthening bar EF is fixed joining the midpoints of the two slanting sides AD and BC.
(a) Find the length of the bar EF.
(b) State whether EF is parallel to AB, and give the reason.
(c) State what shape the frame would be if AB were also 18 cm.
(d) Name the result of this chapter that part (a) generalises.
Answer:
(a) EF = \( \frac{8 + 18}{2} \) = 13 cm.
(b) Yes. Using diagonal AC, the converse of the Midpoint Theorem and then the Midpoint Theorem show EF ∥ AB.
(c) A parallelogram (equal and parallel opposite sides).
(d) The Midpoint Theorem (the case DC = 0 gives EF = ½AB).
In simple words: The bar is the average of the top and bottom of the frame.

Teacher's Note:
1. The answer lies between 8 and 18.
2. Trapezium result generalises the triangle result.

Question 27. Tiling a Courtyard: A courtyard is to be paved with identical tiles, each an irregular quadrilateral whose four angles measure 70°, 85°, 110° and 95°.
(a) Verify that these four angles could indeed be the angles of a quadrilateral.
(b) State how many tiles must meet at each vertex of the tiling, and why.
(c) Describe briefly how the tiler should place each new tile against one already laid.
(d) Explain why this method works for a tile of any quadrilateral shape.
Answer:
(a) 70 + 85 + 110 + 95 = 360°, as required.
(b) Four tiles, one giving each angle, since the four angles total 360°.
(c) Rotate the new tile 180° about the midpoint of an edge of a laid tile, so it shares that whole edge.
(d) A half-turn always matches the edge, and every quadrilateral's angles total 360°, so corners always close up, whatever the shape.
In simple words: Turn each new tile half a turn about the middle of an edge, and it always fits.

Teacher's Note:
1. Two facts do all the work.
2. Even non-convex tiles work.

Question 28. Choosing a Regular Tile: A designer wishes to pave a floor using copies of a single regular polygon, with no gaps and no overlaps.
(a) Find the internal angle of a regular hexagon.
(b) Find how many regular hexagons meet at each vertex of a hexagonal tiling.
(c) Find the internal angle of a regular pentagon.
(d) Explain why the designer cannot use regular pentagons.
Answer:
(a) \( \frac{(6-2) \times 180^\circ}{6} \) = 120°.
(b) 360 ÷ 120 = 3.
(c) \( \frac{(5-2) \times 180^\circ}{5} \) = 108°.
(d) 360 ÷ 108 = 3⅓, not whole: three give 324° (36° gap), four need 432° (overlap).
In simple words: Hexagon angles fit into 360° exactly; pentagon angles do not.

Teacher's Note:
1. Whole-number test decides.
2. Honeycomb uses hexagons.

Question 29. Diagonals and Angles: A student is investigating polygons with six and eight sides.
(a) Find the number of diagonals of a hexagon.
(b) Find the sum of the internal angles of a hexagon.
(c) Find each internal angle of a regular hexagon.
(d) Find the number of diagonals of an octagon.
Answer:
(a) \( \frac{6 \times 3}{2} \) = 9.
(b) (6 − 2) × 180° = 720°.
(c) 720° ÷ 6 = 120°.
(d) \( \frac{8 \times 5}{2} \) = 20.
In simple words: Use n(n − 3)/2 for diagonals and (n − 2) × 180° for angle sums.

Teacher's Note:
1. Two formulas, four quick answers.
2. Divide by 6 only for a regular hexagon.

Question 30. The Pantograph: A pantograph is a linkage of four rods pinned together, used by draughtsmen to copy a drawing at a different size.
(a) State what a pantograph does to a drawing.
(b) Name the four-sided shape on which the linkage is based.
(c) State the property of that shape which is preserved as the linkage moves, and which makes the device work.
(d) Name the mathematician whose parallel motion linkage, based on the same idea, formed part of his steam engine.
Answer:
(a) It makes an exact scaled copy: same shape, different size.
(b) A parallelogram.
(c) Opposite sides stay parallel as it moves, so the copy is not distorted.
(d) James Watt.
In simple words: A moving parallelogram keeps its sides parallel, which lets it copy drawings to scale.

Teacher's Note:
1. A real-life use of parallelograms.
2. Shape stays parallel even as angles change.