Class 8 · English Version · Mathematics
2026 Class 8 Mathematics · English Version · Chapter 8 · Chapter companion part
Construct squares, rectangles, rhombi, parallelograms and trapezia from different data, with instrument checks and explicit restrictions.
Class 8 English Version: part E 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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For adjacent sides a, b and included angle θ with0°<θ<180°, draw BC=a and BA=b on the angle ray. Through A copy a line parallel to BC and through C one parallel to BA; they meet at D. Alternatively use circles about A, C of radii a, b and select the intended intersection. With a=b the result is a rhombus; with θ=90° it is a rectangle; with both it is a square.
For side s>0 draw BC=s, construct BA⊥BC and BA=s, then complete equal-side right-angle ABCD. For a given diagonal d, bisect AC at O, draw its perpendicular, and mark OB=OD=d/2 on opposite rays. Joining A, B, C, D gives equal perpendicular bisecting diagonals, so a square. Its area is d²/2; d is not the side length.
Given adjacent lengths l, b, erect perpendiculars and complete the parallel opposite sides. Given side AB=s and diagonal AC=d>s, draw the perpendicular ray at B and intersect it with the circle about A of radius d. Complete the parallels through A, C. If d≤s the height is zero or impossible. For s=6, d=10, the other side is8 because6²+8²=10².
For positive diagonals p, q, draw AC=p, bisect at O, draw its perpendicular and mark OB=OD=q/2. Join the boundary. For side s and diagonal p, require0<p<2s; radius s arcs from A, C locate B, D on opposite sides. The other diagonal follows. For side5 and one diagonal6, the half-diagonal3 and side5 give other half4, so the other diagonal is8.
For AB=s, AC=p, BD=q, construct triangle AOB with sides AO=p/2, BO=q/2, AB=s. Require |p−q|/2<s<(p+q)/2. Extend AO and BO through O to C, D with OC=OA and OD=OB. Mutual bisection proves the parallelogram. If p=q it is a rectangle; if AO⊥BO it is a rhombus. Check the resulting vertex order.
Draw two crossing lines at the supplied angle AOB; place A, C on opposite rays from O with their given distances, and B, D on opposite rays of the second line. Join cyclic ABCD. This gives diagonals intersecting at O, not necessarily bisecting. Equality of OA, OC and OB, OD must be supplied before claiming a parallelogram.
Construct60° using an equilateral-triangle arc construction and90° by a perpendicular at the same vertex. Bisect the angle between their rays to get75°. For an 85° source angle, use a supplied angle to copy, or a protractor approximation stated as such; do not promise an exact straightedge-and-compass construction from lengths alone. Two90° rays also make a right-angle template for special shapes.
For longer base BC=a, shorter AD=b<a and two acute base angles, mark E on BC with BE=a−b. Construct angle CBA at B and angle BEA equal to the desired interior angle at C; their rays meet at A. Through A draw AD parallel to BC, mark AD=b toward C, and join DC. Triangle BAE supplies the left and right setbacks; translated EA is parallel to CD. Check ray intersection and base lengths, not just a plausible outline.
A rhombus with diagonals6 and8 has half-diagonal right triangles3,4,5. All four sides are5; area is6×8/2=24 cm². This validates both diagonal and side-plus-diagonal constructions.
A square needs only one positive side because equality and right angles determine its other measurements. This does not mean one side fixes a general quadrilateral.
75° is constructed by a bisected60°–90° gap. An85° diagram may be copied as a given angle or measured approximately; the methods are not interchangeable.
Definitions supply some conditions that a general quadrilateral would need separately.
| Target and inputs | Construction idea | Restriction |
|---|---|---|
| Square, side s | Equal sides and a perpendicular | s>0 |
| Square, diagonal d | Equal perpendicular bisecting diagonals | d>0 |
| Rectangle, adjacent l, b | Perpendicular adjacent sides then parallels | l, b>0 |
| Rectangle, side s, diagonal d | Circle meets perpendicular ray | d>s>0 |
| Rhombus, side s, diagonal p | Two radius-s circles about diagonal ends | 0<p<2s |
| Rhombus, diagonals p, q | Perpendicular diagonals bisected | p, q>0 |
| Parallelogram, side s, diagonals p, q | Half-diagonal triangle AOB | |p−q|/2<s<(p+q)/2 |
An equality at a triangle bound collapses the figure. Equal diagonal totals without equal halves do not imply a parallelogram.
For longer base a and shorter base b with acute base angles, construct triangle BAE on BE=a−b.
BC = a = 10 cm AD = b = 6 cm BE = a−b = 4 cm
The45°/45° illustration has height2 cm. It is a separate data set from the numeric area task with height4 cm.
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.
22
2(7+4)=22.
105
Adjacent parallelogram angles sum to180°.
36
6²=36.
50
Its equal perpendicular diagonals give10×10/2=50.
8
10²−6²=64, whose positive square root is8.
28
2(6+8)=28.
8
Half-diagonal3 gives other half4 in a 3–4–5 right triangle; double4.
5
Half-diagonals3,4 make a right triangle with hypotenuse5.
6
O is the midpoint of AC.
10
(12+8)/2=10; the side must be strictly smaller.
7
A, O, C are collinear with O between:4+3=7.
7
5+2=7; equal total diagonals do not imply bisection.
75
60+(90−60)/2=75.
4
BE=a−b=10−6.
32
(10+6)×4/2=32.
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Draw BC=7 and BA=4 on a 60° ray. Through A draw a parallel to BC; through C a parallel to BA. Their intersection D completes the figure. The construction supplies both pairs; check opposite lengths and angle.
Show how parallels or compass copies are made and retain their marks.
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Draw BC=6, erect a perpendicular at B, mark BA=6. Circles about A, C each radius6 give the intersection D away from B; join AD, DC. All sides equal and angle B right, so the rhombus is a square.
Do not choose B as the second circle intersection.
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Draw AC=10, bisect at O, erect the perpendicular at O and mark B, D at 5 cm on its opposite rays. Both diagonals are equal, perpendicular and bisected, so cyclic ABCD is a square.
Show perpendicular-bisector arcs and all four half lengths.
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The adjacent-side method erects perpendiculars6 and8. The diagonal method intersects the perpendicular at B with the circle about A radius10; its positive height is8 from6²+8²=10². Complete parallels and verify.
Use independent input conditions rather than assuming the answer.
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A nonzero perpendicular side creates a hypotenuse strictly longer than either leg. A5 cm diagonal cannot reach the line6 cm from its centre; at 6 cm the circle is tangent at B and the other side has length0.
Give both impossible and equality cases.
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Draw AC=6, construct its perpendicular bisector at O, mark OB=OD=4. Join the four endpoints cyclically. The perpendicular half-diagonal triangles have legs3,4, so all sides5.
Verify mutual bisection and equal sides.
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Draw AC=6. Radius5 arcs from A, C meet at B, D on opposite sides. Join cyclically. Require0<6<2×5, or in general0<p<2s, for distinct intersections and a nondegenerate rhombus.
Retain both arc pairs and show all four equal radius lengths.
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Build AOB with AO=6, BO=4, AB=5;2<5<10 verifies the triangle. Extend through O to C, D with OC=6, OD=4. The diagonals bisect, so ABCD is a parallelogram.
Show the half-lengths and both strict bounds.
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Half-diagonal lengths are6 and4. Their difference is2, so side2 equals the lower bound rather than exceeding it. The three points are collinear and no nondegenerate parallelogram results.
Five or three numbers can satisfy an equality but still fail construction.
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Draw crossing lines at 100°, put A, C on opposite rays with distances4,3 and B, D on the other pair with5,2. Join cyclically. The halves differ, so the diagonals do not bisect and it is not a parallelogram.
Label O and retain the given angle; do not infer from equal diagonal totals7.
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Construct60° and90° rays from one baseline at the same vertex. Bisect the30° gap with equal-radius arc intersections. Its direction is60+15=75°.
Show the angle-bisector arcs, not only a protractor reading.
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A given drawn85° angle can be copied by equal-radius chord transfer. If using a protractor, describe the result as a measured approximation. This lesson does not claim an exact85° construction from a unit segment alone.
Name the available input and instrument; do not silently substitute methods.
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Draw BC=10, mark BE=4 toward C. At B make a 45° ray above BC; at E make a ray above EB making45° to EB. They meet at A,2 cm above the baseline. Through A draw AD∥BC, AD=6 toward C; join DC. Both base angles are45° and the bases are10,6.
Verify the height2 and avoid assuming the separate4 cm-height numeric task is this construction.
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Mark BE=a−b on BC=a. Triangle BAE uses angle ABE=x and BEA=y. Its rays meet above BC. Through A draw a parallel to BC and AD=b toward C; then EA∥CD because E and C differ by b, so angle BCD=y.
Check the acute-angle conditions and the translated side direction.
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Transfer7 and4 end-to-end to make diagonal11. Bisect it, erect its perpendicular and mark the other diagonal halves5.5 on opposite rays. Join the four endpoints. The equal perpendicular bisecting diagonals define the square; its area is121/2=60.5 cm².
Do not use11 as the square’s side.
Make a construction portfolio with one general and three special figures.
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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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