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Class 8 Ganita Prakash · Chapter 4
Class 8 Ganita Prakash · Chapter 4
Quadrilaterals
Page-wise textbook solutions (pages 82–111) · Every step · Figures · Common mistakes · Tips
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๐️ Key ideas of this chapter
- Angle sum of every quadrilateral = 360° (it splits into 2 triangles).
- Rectangle: all angles 90° ⇔ diagonals equal and bisect each other. Square: also diagonals at 90°, and they bisect the corner angles (45°).
- Parallelogram: opposite sides parallel and equal; opposite angles equal; neighbouring angles add to 180°; diagonals bisect each other.
- Rhombus: all sides equal; diagonals bisect each other at 90° and bisect the angles.
- Kite: two pairs of equal neighbouring sides. Trapezium: at least one pair of parallel sides; isosceles trapezium has equal base angles.
๐ Textbook Page 82
๐ Page 82Observe
Of five figures, (i), (ii) and (iii) are quadrilaterals but (iv) and (v) are not. Why?
1A quadrilateral is a closed figure made of exactly 4 straight line segments (sides).
2(i), (ii), (iii) have 4 straight sides. (iii) has one corner pointing inwards, but it is still a quadrilateral.
3(iv) and (v) have a curved side, so they are not quadrilaterals.
✅ Answer: Quadrilaterals need 4 straight sides; (iv) and (v) have curved sides.
๐ Textbook Page 83
๐ Page 83Carpenter's Problem
A carpenter has an 8 cm strip and wants two strips joined so that a thread round their ends makes a rectangle. How long must the other strip be, and where should they be joined?
1The strips are the diagonals of the rectangle.
2Deduction 1: ฮADC ≅ ฮDAB (SAS: AD common, ∠A = ∠D = 90°, AB = DC) → AC = BD. So diagonals are equal.
3Deduction 2: ฮAOB ≅ ฮCOD (AAS) → OA = OC, OB = OD. So diagonals bisect each other.
4Deduction 3: for ANY angle between them, equal diagonals that bisect each other give a rectangle.
✅ Answer: The other strip must also be 8 cm, and the two strips must be joined at their midpoints (4 cm from each end). Any angle between them works.
๐ Textbook Page 85
๐ Page 85Math Talk
Can AO = CO, ∠AOB = ∠COD and AD = CB be used to show ฮAOD ≅ ฮCOB?
1∠AOB and ∠COD are not angles of ฮAOD or ฮCOB at all.
2The angles we would need are ∠AOD and ∠COB — and even then the angle must be between the two known sides.
3With AO = CO and AD = CB, the angle at O is not between these sides (that would be SSA, which is not a congruence rule).
✅ Answer: No. These facts do not give SSS, SAS, ASA, AAS or RHS for ฮAOD and ฮCOB.
⚠️ Common mistake: Using any three equal parts. The equal parts must match one of the 5 congruence rules.
๐ Page 85Think
Diagonals are equal, bisect each other and meet at 60°. Find a in ฮAOB (OA = OB) and all the other angles.
1OA = OB, so the base angles of ฮAOB are equal: \(a + a + 60 = 180\)
2\(2a = 120 \Rightarrow a = 60^\circ\)
3Angles between the diagonals: 60°, 120°, 60°, 120°.
4In ฮAOD (apex 120°): base angles \(= (180 - 120) \div 2 = 30^\circ\)
5Each corner angle = 60° + 30° = 90°.
✅ Answer: a = 60°; every corner of ABCD is 90°, so ABCD is a rectangle.
๐ Textbook Page 87
๐ Page 87Generalise
If the angle between equal, bisecting diagonals is any x, is ABCD still a rectangle?
1Angles at O: \(x, x, 180 - x, 180 - x\).
2ฮAOB is isosceles: \(2a + x = 180 \Rightarrow a = 90 - \tfrac{x}{2}\)
3ฮAOD is isosceles: \(2b + (180 - x) = 180 \Rightarrow b = \tfrac{x}{2}\)
4Each corner \(= a + b = 90 - \tfrac{x}{2} + \tfrac{x}{2} = 90^\circ\)
✅ Answer: Yes — for every x, all corners are 90°.
๐ Textbook Page 88
๐ Page 88Think
Can we define a rectangle just as ‘a quadrilateral whose angles are all 90°’? Try to draw one with all angles 90° but opposite sides not equal.
1Join diagonal BD. ฮBAD and ฮDCB: BD is common, ∠A = ∠C = 90°.
2∠1 + ∠3 = 90° (at B) and ∠3 + ∠2 = 90° (in ฮBCD) → ∠1 = ∠2.
3So ฮBAD ≅ ฮDCB (AAS) → AD = CB and AB = DC.
✅ Answer: Impossible — all angles 90° forces opposite sides to be equal. So ‘all angles 90°’ is enough to define a rectangle.
๐ Textbook Page 90
๐ Page 90Think
Is it wrong to write ฮBAD ≅ ฮCDB? Why?
1In congruence, the ORDER of letters shows which parts match.
2B matches D, A matches C, D matches B (the right angles at A and C match).
3Correct: ฮBAD ≅ ฮDCB. Writing ฮCDB would match A with D — wrong.
✅ Answer: Yes, it is wrong. The correct statement is ฮBAD ≅ ฮDCB.
๐ Page 90Practice
Show that AB ∥ DC in rectangle ABCD.
1Take AD as a transversal of lines AB and DC.
2∠A + ∠D = 90° + 90° = 180° (interior angles on the same side).
3So AB ∥ DC.
✅ Answer: AB ∥ DC because the co-interior angles add to 180°.
๐ Page 90Observe
Four quadrilaterals with all angles 90° are shown (sides 5 × 2, 6 × 3.6, 5 × 1, 4 × 4). Are any of them NOT rectangles?
1All of them have four right angles, so all are rectangles.
2The 4 cm × 4 cm one has all sides equal too — a special rectangle called a square.
✅ Answer: None. All are rectangles; the last is also a square.
๐ Textbook Page 93
๐ Page 93Construct
Construct a square with a diagonal of 8 cm.
1Draw AC = 8 cm. Mark its midpoint O (4 cm).
2Draw a line through O perpendicular to AC (use a set-square or compass).
3Mark B and D on it with OB = OD = 4 cm.
4Join A, B, C, D.
✅ Answer: ABCD is the square (diagonals equal, bisect each other at 90°).
๐ Page 93Think
In square ABCD with diagonal AC, find ∠1, ∠2, ∠3 and ∠4 (the angles the diagonals make with the sides).
1In ฮADC: AD = DC, so ∠1 = ∠3.
2\(\angle 1 + \angle 3 + 90 = 180 \Rightarrow \angle 1 = \angle 3 = 45^\circ\)
3Same reasoning in ฮABD gives ∠2 = ∠4 = 45°.
✅ Answer: All are 45° — the diagonals of a square bisect its angles.
๐ Textbook Page 94
๐ Page 94Figure it Out · Q1
Find all the other angles inside the rectangles: (i) ABCD with ∠CAB = 30° (ii) PQRS with ∠QOR = 110° at the meeting point O of the diagonals.
1(i) ∠CAD = 90° − 30° = 60°. In ฮABC, ∠ACB = 180° − 90° − 30° = 60°.
2Diagonals are equal and bisect each other, so OA = OB → ∠OBA = 30°, ∠DBC = 60°.
3∠ACD = ∠CAB = 30° (alternate), ∠BDC = 30°, ∠ADB = 60°.
4At O: ∠AOB = 180° − 30° − 30° = 120°, ∠BOC = 60°, ∠COD = 120°, ∠AOD = 60°.
5(ii) At O: ∠POS = 110°, ∠QOP = ∠ROS = 70°.
6ฮQOR is isosceles: ∠OQR = ∠ORQ = (180° − 110°) ÷ 2 = 35°.
7ฮQOP is isosceles: ∠OQP = ∠OPQ = (180° − 70°) ÷ 2 = 55°.
8Similarly ∠OPS = ∠OSP = 35° and ∠ORS = ∠OSR = 55°.
✅ Answer: (i) 30°s and 60°s at the corners; 120°, 60°, 120°, 60° at O. (ii) 110°, 70°, 110°, 70° at O; 35° and 55° at the corners.
๐ Page 94Figure it Out · Q2
Draw a quadrilateral whose diagonals are both 8 cm, bisect each other and meet at (i) 30° (ii) 40° (iii) 90° (iv) 140°.
1Draw AC = 8 cm and mark its midpoint O.
2At O draw a line making the given angle with AC.
3On this line mark B and D, each 4 cm from O (on opposite sides).
4Join A–B–C–D.
✅ Answer: Each figure is a rectangle; the 90° one is a square.
๐ Page 94Figure it Out · Q3
PL and AM are two perpendicular diameters of a circle with centre O. What is APML?
1PL = AM (both are diameters).
2They bisect each other at O (OP = OL = OA = OM = radius).
3They are perpendicular.
✅ Answer: APML is a square.
๐ Page 94Figure it Out · Q4
Without paper, using two sticks of equal length and a thread, how can you make an exact 90°?
1Mark the middle of each stick.
2Cross the sticks so their middles meet, and tie them there.
3Pass the thread around all four ends and pull it tight.
4The ends form a rectangle (equal diagonals that bisect each other), so each corner of the thread is exactly 90°.
✅ Answer: Use the sticks as equal diagonals joined at their midpoints; the thread corners give 90°.
๐ Page 94Figure it Out · Q5
Can ‘opposite sides parallel (and equal)’ be the definition of a rectangle?
1A parallelogram with angles 60° and 120° has opposite sides parallel and equal.
2But its angles are not 90°, so it is not a rectangle.
✅ Answer: No. Such a quadrilateral is a parallelogram — it need not be a rectangle.
๐ Page 94Think
Can a quadrilateral have three angles of 90° and the fourth angle not 90°? Why?
1The angle sum of every quadrilateral is 360° (see the next question).
2Fourth angle \(= 360 - 3 \times 90 = 90^\circ\).
✅ Answer: No — the fourth angle is forced to be 90°.
๐ Textbook Page 95
๐ Page 95Reason
Why is the sum of the angles of a quadrilateral 360°?
1Draw diagonal SM in quadrilateral SOME.
2ฮSEM: \(\angle1 + \angle2 + \angle3 = 180^\circ\); ฮSOM: \(\angle4 + \angle5 + \angle6 = 180^\circ\)
3The six angles together make the four angles of SOME.
4So the sum \(= 180 + 180 = 360^\circ\).
✅ Answer: Any quadrilateral splits into 2 triangles, so its angles add to 360°.
๐ Textbook Page 96
๐ Page 96Construct
Draw a parallelogram with sides 4 cm and 5 cm and an angle of 30° between them. Find the other angles and sides.
1Draw AB = 4 cm, then AD = 5 cm at 30° to AB.
2Through D draw a line ∥ AB; through B draw a line ∥ AD. They meet at C.
3Co-interior angles: ∠D = 180° − 30° = 150°, ∠B = 150°, ∠C = 30°.
4Opposite sides are equal: DC = 4 cm, BC = 5 cm.
✅ Answer: Angles 30°, 150°, 30°, 150°; sides 4 cm, 5 cm, 4 cm, 5 cm.
๐ Textbook Page 97
๐ Page 97Reason
Are the opposite angles of every parallelogram equal? How can we be sure?
1Let ∠P = x in parallelogram PEAR.
2∠P + ∠R = 180° (co-interior) → ∠R = 180° − x.
3∠R + ∠A = 180° → ∠A = x.
4Similarly ∠E = 180° − x = ∠R.
✅ Answer: Yes — opposite angles are always equal.
๐ Textbook Page 98
๐ Page 98Think
Is it wrong to write ฮABD ≅ ฮCBD? Why?
1In parallelogram ABCD with diagonal BD: A matches C, B matches D, D matches B.
2Correct statement: ฮABD ≅ ฮCDB.
3ฮCBD would match B with B and D with D — wrong pairing.
✅ Answer: Yes, it is wrong. Write ฮABD ≅ ฮCDB.
๐ Page 98Check
Are the diagonals of a parallelogram always equal? Do they bisect each other?
1In the 30°–150° parallelogram, the diagonal across the 150° corners is much shorter than the other one, so they are not equal.
2In parallelogram EASY: AE = YS, and two pairs of alternate angles are equal → ฮAOE ≅ ฮYOS (ASA).
3So OA = OY and OE = OS — O is the midpoint of both.
✅ Answer: Diagonals need NOT be equal, but they always bisect each other.
๐ Textbook Page 99
๐ Page 99Think
Is it wrong to write ฮAOE ≅ ฮSOY? Do the diagonals of a parallelogram meet at a special angle?
1A matches Y and E matches S, so the correct form is ฮAOE ≅ ฮYOS. ฮSOY is wrong.
2The angle between the diagonals changes from one parallelogram to another (it is 90° only for a rhombus).
✅ Answer: Yes, it is wrong; and no, there is no fixed angle in general.
๐ Textbook Page 100
๐ Page 100Think
In rhombus ABCD with ∠A = 50°, find the other angles.
1AD = AB, so in ฮADB the base angles are equal: a + a + 50 = 180 → a = 65°.
2The diagonal BD splits ∠B and ∠D into equal parts of 65°.
3∠B = ∠D = 65° + 65° = 130°; ∠C = 50°.
✅ Answer: 50°, 130°, 50°, 130°.
๐ Page 100Reason
Why is ฮGAE ≅ ฮMAE in rhombus GAME?
1GE = ME (sides of a rhombus)
2GA = MA (sides of a rhombus)
3AE is common
✅ Answer: By SSS.
๐ Textbook Page 101
๐ Page 101Venn
A rhombus and a rectangle are both parallelograms. Draw a Venn diagram. Where do squares go?
1Rectangles and rhombuses are two sets inside ‘parallelograms’.
2A square has all angles 90° (rectangle) and all sides equal (rhombus).
3So squares sit where the two sets overlap.
✅ Answer: Square = rectangle ∩ rhombus.
๐ Page 101Check
Are the diagonals of a rhombus equal?
1Draw a rhombus with angle 50°: one diagonal is clearly longer.
2They are equal only when the rhombus is a square.
✅ Answer: No, not in general.
๐ Textbook Page 102
๐ Page 102Reason
Why is ฮGEO ≅ ฮMEO in rhombus GAME? What angle do the diagonals make?
1GE = ME (sides), GO = MO (diagonals bisect each other), EO common → SSS.
2So ∠GOE = ∠MOE, and together they make 180°.
3Each is 90°.
✅ Answer: SSS; the diagonals of a rhombus meet at 90°.
๐ Page 102Figure it Out · Q1
Find the remaining angles: (i) parallelogram PEAR with ∠P = 40° (ii) parallelogram PQRS with ∠P = 110° (iii) rhombus UVWX with ∠XVU = 30° (iv) rhombus OIEA with ∠OEA = 20°.
1(i) ∠E = 180° − 40° = 140°, ∠A = 40°, ∠R = 140°.
2(ii) ∠Q = 70°, ∠R = 110°, ∠S = 70°.
3(iii) The diagonal XV bisects ∠V: ∠XVW = 30°, so ∠V = 60°; ∠X = 60° (∠UXV = ∠WXV = 30°); ∠U = ∠W = 120°.
4(iv) Diagonal OE bisects ∠E: ∠OEI = 20°, ∠E = 40°; ∠AOE = ∠EOI = 20°, ∠O = 40°; ∠A = ∠I = 140°.
✅ Answer: (i) 140°, 40°, 140° (ii) 70°, 110°, 70° (iii) ∠V = ∠X = 60°, ∠U = ∠W = 120° (iv) ∠E = ∠O = 40°, ∠A = ∠I = 140°
๐ Page 102Figure it Out · Q2
Using diagonal properties, construct a parallelogram with diagonals 7 cm and 5 cm meeting at 140°.
1Draw AC = 7 cm; mark its midpoint O (3.5 cm).
2At O draw a line making 140° with OC.
3On it mark B and D with OB = OD = 2.5 cm.
4Join A–B–C–D. (Diagonals bisect each other → parallelogram.)
✅ Answer: ABCD is the required parallelogram.
๐ Page 102Figure it Out · Q3
Construct a rhombus with diagonals 4 cm and 5 cm.
1Draw AC = 5 cm and mark its midpoint O (2.5 cm).
2Draw the perpendicular to AC through O.
3Mark B and D on it with OB = OD = 2 cm.
4Join A–B–C–D. (Diagonals bisect each other at 90° → rhombus.)
✅ Answer: ABCD is the required rhombus (each side ≈ 3.2 cm).
๐ Textbook Page 103
๐ Page 103Geoboard
Two rubber bands, equal in length and perpendicular, cross at their middles. Join the ends — what do you get? Then extend one diagonal by 2 cm on both sides — what now?
1Equal diagonals, bisecting each other at 90° → a square.
2After extending one diagonal on both sides, the diagonals still bisect each other at 90° but are no longer equal.
3Diagonals bisecting at 90° → a rhombus.
✅ Answer: First a square, then a rhombus.
๐ Textbook Page 104
๐ Page 104Joining Triangles
Join two equilateral triangles of side 8 cm. What quadrilateral do you get?
1All four outer sides are 8 cm.
2Angles: 60°, 120° (60° + 60°), 60°, 120°.
✅ Answer: A rhombus with angles 60° and 120°.
๐ Page 104Joining Triangles
Join two isosceles triangles (8 cm, 8 cm, 6 cm) in different ways. What quadrilaterals do you get?
1Join along the 6 cm sides: four sides of 8 cm → a rhombus.
2Join along an 8 cm side, one triangle turned around: sides 8, 6, 8, 6 opposite each other → a parallelogram.
3Join along an 8 cm side as a mirror image: sides 6, 6 together and 8, 8 together → a kite.
✅ Answer: Rhombus, parallelogram and kite.
๐ Textbook Page 105
๐ Page 105Joining Triangles
Join two scalene triangles (6 cm, 9 cm, 12 cm). What quadrilaterals can you get?
1You can join along the 6, 9 or 12 cm side, and each time either turn one triangle (rotation) or flip it (mirror).
2Turning gives a parallelogram (3 different ones).
3Flipping gives a kite (3 different ones: sides 9-9-12-12, 6-6-12-12, 6-6-9-9). Some may point inwards (a ‘dart’).
✅ Answer: Up to 6 shapes: 3 parallelograms and 3 kites.
๐ Page 105Kite
In kite ABCD (AB = BC, CD = DA), show that diagonal BD bisects ∠ABC and ∠ADC, bisects AC and is perpendicular to it.
1ฮABD ≅ ฮCBD (SSS: AB = CB, AD = CD, BD common) → ∠ABD = ∠CBD and ∠ADB = ∠CDB.
2ฮAOB ≅ ฮCOB (SAS: AB = CB, ∠ABO = ∠CBO, BO common) → AO = OC and ∠AOB = ∠COB.
3∠AOB + ∠COB = 180°, so each is 90°.
✅ Answer: BD bisects the angles at B and D, and is the perpendicular bisector of AC.
๐ Textbook Page 106
๐ Page 106Trapezium
In trapezium PQRS (PQ ∥ SR), can you find the remaining angles without measuring?
1PS is a transversal of PQ and SR → ∠S + ∠P = 180°.
2QR is a transversal → ∠R + ∠Q = 180°.
✅ Answer: Yes: each angle = 180° − the angle next to it on the same slanting side.
๐ Textbook Page 107
๐ Page 107Reason
In isosceles trapezium UVWX (UX = VW), why is ฮUXY ≅ ฮVWZ?
1XY and WZ are perpendicular to UV, and XWZY is a rectangle, so XY = WZ.
2UX = VW (given, the hypotenuses).
3∠XYU = ∠WZV = 90°.
✅ Answer: By RHS, so ∠U = ∠V.
๐ Page 107Figure it Out · Q1
Find all the sides and angles of the quadrilateral made by joining two equilateral triangles of side 4 cm.
1All outer sides are 4 cm.
2Two corners are 60° (one angle of each triangle).
3The other two corners are 60° + 60° = 120°.
✅ Answer: Sides 4 cm each; angles 60°, 120°, 60°, 120° (a rhombus).
๐ Page 107Figure it Out · Q2
Construct a kite with diagonals 6 cm and 8 cm.
1Draw AC = 6 cm and mark its midpoint O.
2Draw the perpendicular to AC through O.
3Mark B on it, say 2 cm from O, and D on the other side so that BD = 8 cm (OD = 6 cm).
4Join A–B–C–D.
✅ Answer: ABCD is a kite (AB = CB, AD = CD). Choosing OB = OD = 4 cm would give a rhombus instead.
๐ Page 107Figure it Out · Q3
Find the remaining angles: (i) trapezium with bottom angles 135° and 105° (top ∥ bottom) (ii) isosceles trapezium with one angle 100°.
1(i) Each top angle + the bottom angle on the same slanting side = 180°.
2Top-left = 180° − 135° = 45°; top-right = 180° − 105° = 75°.
3(ii) The angle next to 100° along a slanting side = 180° − 100° = 80°.
4In an isosceles trapezium the two angles on the same parallel side are equal: so 100°, 100°, 80°, 80°.
✅ Answer: (i) 45° and 75° (ii) 100°, 80°, 80°
๐ Page 107Figure it Out · Q4
Draw a Venn diagram of parallelograms, kites, rhombuses, rectangles and squares. (i) Which quadrilateral is both a kite and a parallelogram? (ii) Can a quadrilateral be both a kite and a rectangle? (iii) Is every kite a rhombus?
1(i) A kite that is also a parallelogram has all four sides equal → a rhombus (squares included).
2(ii) Yes — a square is both a kite and a rectangle.
3(iii) No. Every rhombus is a kite, but a kite need not have all sides equal.
✅ Answer: (i) Rhombus (ii) Yes, a square (iii) No — rhombus ⊂ kite.
⚠️ Common mistake: Saying ‘no’ to (ii). A square has AB = BC and CD = DA, so it IS a kite.
๐ Page 107Figure it Out · Q5
PAIR and RODS are rectangles with ∠IRO = 30°. Find ∠IOD.
1In ฮRIO, ∠RIO = 90° (corner of PAIR).
2\(\angle ROI = 180 - 90 - 30 = 60^\circ\)
3∠ROD = 90° (corner of RODS).
4\(\angle IOD = 90 - 60 = 30^\circ\)
✅ Answer: ∠IOD = 30°
๐ Textbook Page 108
๐ Page 108Figure it Out · Q6
Construct a square with diagonal 6 cm without a protractor.
1Draw AC = 6 cm.
2With compass (radius more than 3 cm) draw arcs from A and C above and below; join the crossing points — this is the perpendicular bisector of AC through O.
3Mark B and D on it with OB = OD = 3 cm.
4Join A–B–C–D.
✅ Answer: ABCD is a square (equal diagonals bisecting at 90°).
๐ Page 108Figure it Out · Q7
CASE is a square; U, V, W, X are the midpoints of its sides. What is UVWX? Find other ways to draw a square inside a square.
1Let the side be s. Each corner triangle is right-angled with legs \(\tfrac{s}{2}, \tfrac{s}{2}\) — all four are congruent (SAS).
2So UV = VW = WX = XU.
3Each corner triangle is isosceles right → its acute angles are 45°.
4At U: 45° + ∠VUX + 45° = 180° → ∠VUX = 90°. Same at every vertex.
5Other ways: mark points the SAME distance (say 2 cm) from each corner, going around in the same direction. Join them — you always get a square.
✅ Answer: UVWX is a square.
๐ Page 108Figure it Out · Q8
A quadrilateral has four equal sides and one angle 90°. Is it a square?
1Four equal sides → it is a rhombus, so it is a parallelogram.
2Neighbouring angles add to 180° → the angles next to 90° are 90°; the opposite one is also 90°.
✅ Answer: Yes, it is a square.
๐ Page 108Figure it Out · Q9
What type of quadrilateral has both pairs of opposite sides equal? Justify.
1Draw diagonal AC in ABCD with AB = CD, BC = AD.
2ฮABC ≅ ฮCDA (SSS).
3So ∠BAC = ∠DCA → AB ∥ CD (alternate angles), and ∠BCA = ∠DAC → BC ∥ AD.
✅ Answer: A parallelogram.
๐ Page 108Figure it Out · Q10
Is the angle sum 360° even for a quadrilateral with one corner pointing inwards (like an arrowhead)?
1Join the inward corner D to the opposite corner B. This diagonal lies inside the shape.
2It splits the shape into two triangles: 180° + 180°.
3The angles of the triangles make up exactly the 4 angles (the inward angle at D is more than 180°).
✅ Answer: Yes, still 360°.
๐ก Tip: Choose the diagonal that lies INSIDE the shape.
๐ Page 108Figure it Out · Q11
True or false?
(i) Equal diagonals that bisect each other → must be a square.
(ii) Three right angles → must be a rectangle.
(iii) Diagonals bisect each other → must be a parallelogram.
(iv) Perpendicular diagonals → must be a rhombus.
(v) Opposite angles equal → must be a parallelogram.
(vi) All angles equal → a rectangle.
(vii) Isosceles trapeziums are parallelograms.
(i) Equal diagonals that bisect each other → must be a square.
(ii) Three right angles → must be a rectangle.
(iii) Diagonals bisect each other → must be a parallelogram.
(iv) Perpendicular diagonals → must be a rhombus.
(v) Opposite angles equal → must be a parallelogram.
(vi) All angles equal → a rectangle.
(vii) Isosceles trapeziums are parallelograms.
1(i) False — it must be a rectangle; a square also needs the diagonals at 90°.
2(ii) True — the fourth angle = 360° − 270° = 90°.
3(iii) True — proved by congruent triangles and alternate angles.
4(iv) False — a kite also has perpendicular diagonals.
5(v) True — if ∠A = ∠C = x and ∠B = ∠D = y, then 2x + 2y = 360°, so x + y = 180° → opposite sides are parallel.
6(vi) True — each angle = 360° ÷ 4 = 90°.
7(vii) False — only one pair of sides is parallel.
✅ Answer: (i) F (ii) T (iii) T (iv) F (v) T (vi) T (vii) F
๐ Textbook Page 111
๐ Page 111Puzzle: Which Quad?
Fold a sheet into a quarter and make a triangular crease at the folded corner. Open it: what shape do the creases make? How do you get nested shapes? How do you get a square?
1The two folds are perpendicular lines through the centre, and the crease is copied in all four quarters.
2The crease ends become the corners; the fold lines are the diagonals, which bisect each other at 90° → a rhombus.
3Several creases at different distances from the corner → several rhombuses one inside another.
4For a square, the diagonals must also be equal: make the crease cut off EQUAL lengths on both folded edges (a 45° crease).
✅ Answer: A rhombus; a 45° crease (equal lengths on both edges) gives a square.
๐ Where do we see quadrilaterals?
- Carpenters and builders check a frame is rectangular by measuring that the diagonals are equal.
- Kites, tiles, window grills, bridges (parallelogram linkages), folding stands.
๐ Link to higher classes
- Class 9: proofs with congruence, mid-point theorem, areas of parallelograms.
- Class 10: coordinate geometry — proving a shape is a parallelogram or rhombus using distances and slopes.
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