Class 8 · English Version · Mathematics
2026 Class 8 Mathematics · English Version · Chapter 8 · Chapter companion part
Classify quadrilaterals from their conditions and explain angle, side and diagonal properties without trusting appearance.
Class 8 English Version: part A of the chapter 8 companion. Eight teaching sections, 30 written tasks and a notebook challenge. Selected companion coverage, not complete textbook coverage; use all five parts. Independent proofs, compass work, drawings and nets require guardian or teacher review.
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A simple quadrilateral has four straight sides forming a closed boundary without crossings. Name ABCD around its boundary: sides AB, BC, CD, DA; diagonals AC, BD join opposite vertices. AB and CD are opposite sides; AB and BC are adjacent. Perimeter adds only the four boundary lengths. Our construction models are convex: each interior angle is less than 180°. A crossed bow-tie drawing is not one of these models.
A parallelogram has two pairs of opposite parallel sides. A rectangle is a parallelogram with a right angle; all four are then right. A rhombus has four equal sides and opposite sides parallel. A square is both a rectangle and a rhombus. A tilted square remains a square. Equal adjacent sides alone do not make an arbitrary quadrilateral a rhombus unless all four are equal.
Here trapezium means exactly one pair of opposite parallel sides. Some texts use at least one pair, which also includes parallelograms: name that convention before judging the source activity. Here a kite has two adjacent equal-side pairs of different lengths, so a rhombus is excluded. Its symmetry diagonal bisects the other at a right angle, but the other diagonal need not bisect it. Mutual bisection is not a property of every kite. For unequal side lengths pk, qk, the perimeter is2(p+q)k; a ratio question first finds k from the stated perimeter.
A diagonal inside a convex quadrilateral divides it into two triangles. Each triangle has angle sum 180°, giving 360° altogether. The split angles at the diagonal endpoints recombine to the full quadrilateral angles. Four right angles means 4×90°, not four degrees. For an ordinary concave simple quadrilateral use an internal diagonal and include its reflex interior angle.
Opposite angles are equal because parallel sides give equal alternate angles. Adjacent interior angles sum to 180° along a transversal. If angle A=68°, then C=68° and B=D=112°. Opposite sides have equal lengths. A general parallelogram need not have equal diagonals, right angles or four equal sides; a picture that looks nearly rectangular proves none of these.
In a parallelogram, AC and BD meet at O with AO=OC and BO=OD. This does not imply AO=BO: the two diagonals may differ. In a rectangle the diagonals are also equal, so all four half-diagonal lengths are equal. In a rhombus the diagonals are perpendicular but generally unequal. A square has both additional properties.
Opposite interior angles of a parallelogram are equal. Their bisector lines are parallel, or can coincide in a rhombus; do not insist they are distinct in that case. Adjacent angles sum to 180°, so their halves sum to 90° and the adjacent bisectors are perpendicular. This concerns interior bisectors, not arbitrary lines passing through the vertices.
Use markings and stated givens as the conditions. A rectangle can be verified by three right angles in a simple quadrilateral, or by first proving a parallelogram and then equal diagonals. Equal diagonals alone do not suffice: an isosceles trapezium can have them. A square needs both right-angle and equal-side information. A physical slab has measurement tolerances, so distinguish a mathematical proof from an approximate construction check.
In parallelogram ABCD with A=68°, C=68° and B=D=112°. If AC=14 and BD=10, their halves are 7 and 5; bisection does not make the diagonals equal.
The source trapezium activity depends on whether a pair means exactly one or at least one. This lesson states exactly one and explains the alternative.
Perpendicular diagonals do not by themselves imply mutual bisection. A general kite supplies a counterexample.
All entries concern nondegenerate figures; a square belongs to both special families.
| Family | Opposite sides parallel | Four equal sides | Four right angles | Diagonals |
|---|---|---|---|---|
| Parallelogram | Both pairs | Not required | Not required | Bisect |
| Rectangle | Both pairs | Not required | Yes | Equal and bisect |
| Rhombus | Both pairs | Yes | Not required | Perpendicular and bisect |
| Square | Both pairs | Yes | Yes | Equal, perpendicular, bisect |
| Kite here | Not required | No: pair lengths differ | Not required | Perpendicular; one bisects the other |
| Trapezium here | Exactly one pair | Not required | Not four | No general bisection rule |
A condition marked not required can hold in a special case only if the definitions allow that special case. Our kite/trapezium conventions exclude rhombi/parallelograms respectively.
For parallelogram ABCD with A=68°.
C = A = 68° B = D = 180°−68° = 112° 68+112+68+112 = 360°
These angle facts alone do not assert that the diagonals are equal.
Match each letter to its explanation. Read the quantities, units and relationships, then check the counts or equations. These are original teaching models.
Original labelled mathematical model, not copied source artwork. Verify stated conditions rather than judging shape from appearance.
Match each letter to its explanation. Read the quantities, units and relationships, then check the counts or equations. These are original teaching models.
Original labelled mathematical model, not copied source artwork. Verify stated conditions rather than judging shape from appearance.
Match each letter to its explanation. Read the quantities, units and relationships, then check the counts or equations. These are original teaching models.
Original labelled mathematical model, not copied source artwork. Verify stated conditions rather than judging shape from appearance.
Match each letter to its explanation. Read the quantities, units and relationships, then check the counts or equations. These are original teaching models.
Original labelled mathematical model, not copied source artwork. Verify stated conditions rather than judging shape from appearance.
Match each letter to its explanation. Read the quantities, units and relationships, then check the counts or equations. These are original teaching models.
Original labelled mathematical model, not copied source artwork. Verify stated conditions rather than judging shape from appearance.
Match each letter to its explanation. Read the quantities, units and relationships, then check the counts or equations. These are original teaching models.
Original labelled mathematical model, not copied source artwork. Verify stated conditions rather than judging shape from appearance.
Show your working on paper. Check your answer, then choose Next.
18
Add the four boundary lengths: 3+4+5+6.
2
Its two opposite-vertex joins are the diagonals.
90
360−75−85−110=90.
112
Adjacent angles sum to 180°.
68
Opposite angles are equal.
7
The diagonals bisect each other, so AO=14/2.
11
BD=2BO.
13
Rectangle diagonals are equal.
90
The diagonals meet perpendicularly.
26
Its four sides are equal: 4×6.5.
4
One right angle in a parallelogram forces all four.
6
Two sides are2k and two are3k, so10k=30, k=3 and2k=6.
34
Its interior bisector divides 68° into two equal 34° angles.
90
Half of two supplementary angles adds to 90°.
90
360−3×90=90.
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Sides AB, BC, CD, DA; diagonals AC, BD; adjacent AB, BC; opposite AB, CD. A different correct pair is acceptable.
Follow boundary order and exclude diagonals from perimeter.
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A square has four right angles and four equal sides. A 3-by-5 rectangle lacks four equal sides; a non-right rhombus lacks four right angles.
Give conditions and genuine counterexamples.
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With exactly one parallel-side pair, no. With at least one pair, yes. The source wording a pair does not resolve the convention on its own.
Name the convention; do not treat a terminology difference as a failed proof.
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A kite symmetry diagonal bisects the other at right angles. The reverse bisection need not hold. For vertices (0,3),(2,0),(0,−1),(−2,0), the vertical diagonal has unequal halves 3 and 1 at the crossing.
Identify which diagonal is bisected. This clarifies ambiguous source Exercise 8.2(1).
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Draw an internal diagonal, sum the two triangle angle sums, then recombine the two split endpoint angles. Total is 180°+180°=360°.
Show how each quadrilateral angle is counted once.
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A=C=68° by opposite-angle equality; B=D=112° by supplementary adjacent angles. They total 360°.
Include the two reasons and a check.
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Bisection compares halves within each diagonal. One diagonal could be 14 and the other 10, with half lengths 7 and 5.
Keep the two equalities separate.
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All four families have mutual bisection. Rectangle adds equal lengths; rhombus adds perpendicularity; square adds both. A general rectangle need not have perpendicular diagonals.
Use the property table and identify properties that are not universal.
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If adjacent angles are α and 180°−α, their halves sum to 90°. With the common side as transversal, the triangle between the bisectors has remaining angle 90°.
Use interior rays and show the half-angle calculation.
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Their supporting lines have the same direction; they can coincide in a rhombus, where a diagonal bisects both opposite angles. Distinct parallel lines is too strong without excluding that case.
Distinguish equal direction from two distinct lines.
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No. An isosceles trapezium can have equal diagonals without four right angles. First prove a parallelogram, then use equal diagonals, or use sufficient right-angle conditions.
State the missing premise and a counterexample.
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Three right interior angles imply the fourth is right. Alternatively establish opposite-side parallelism and equal diagonals; a parallelogram with equal diagonals is a rectangle. Real measurements have tolerance.
Do not claim equal diagonals alone are enough.
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Rotation changes its orientation, not its four equal side lengths or right angles. Diamond is a visual description, not a replacement for its mathematical conditions.
Use properties rather than page orientation.
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Let the unequal lengths be2k,3k. Each occurs twice, so2(2k+3k)=30 gives k=3; the lengths are6 cm,9 cm, each repeated. Their sum6+6+9+9 checks30.
Do not divide the perimeter by the ratio sum without accounting for both equal pairs.
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Its boundary crosses and does not enclose the same simple convex region. The intended interior angles and diagonal split must be defined differently. Our models require cyclic noncrossing vertices and nondegenerate triangles.
State the assumption that fails.
Make a property notebook with a labelled example and a counterexample.
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This original Kidemy practice follows the topic order of 2026 Class 8 Mathematics · English Version, chapter 8, printed pages 133–151. It does not replace the complete chapter or its practical teaching.
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