Class 8 · English Version · Mathematics
2026 Class 8 Mathematics · English Version · Chapter 8 · Chapter companion part
Choose a construction from explicit givens, check existence and justify the selected circle or ray intersections.
Class 8 English Version: part D 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 general quadrilateral commonly needs five suitably placed independent measurements. Four sides can flex while their lengths stay fixed. Some five-measurement lists are inconsistent or ambiguous; special families need fewer because their definitions supply conditions. Label exactly which angle and diagonal are given and check a simple noncrossing result. Do not treat a rough sketch as an exact construction.
For AB=3, BC=5, CD=4, DA=4 cm and ABC=90°, draw BC=5 and a perpendicular at B, mark A3 cm along it. Intersect circles centred A, C each of radius4 cm at D, choosing the point giving cyclic convex ABCD. Join AD, DC. The other intersection may lie outside the intended region. Verify all four lengths and the90° included angle; the angle location is essential.
Let BD=6, BA=DA=5, BC=DC=4 cm. Draw BD. Radius5 arcs about B, D locate A above it; radius4 arcs locate C below it. Join cyclic A, B, C, D. For a triangle with side radii r, s and centre distance e require |r−s|<e<r+s. Check this for both triangles. Equality gives a flat triangle; the source lists upper-sum bounds but the lower-difference bounds are also necessary.
Specify AB, AD, CD and diagonals BD, AC. First construct ABD by SSS on BD. Then circles centred A, D with radii AC, CD locate C; choose the point on the opposite side of BD from A when this exists and produces the required convex order. Both triangle stages need full inequalities and a final side-of-line check. Giving only the first triangle’s condition, as the source summary does, does not guarantee the second construction.
Given AB=3, BC=5, CD=4 cm with ABC=90°, BCD=90°, draw BC, erect same-side perpendicular rays at B, C, mark A, D at distances3,4 and join AD. General given angles are copied at the specified vertices. Check A, D on the intended side, noncrossing boundary and every supplied length. The final AD is determined, not an extra freely chosen length.
Given AB=3, BC=5 cm and ABC=90°, BCD=90°, BAD=90°, build triangle ABC using the first included angle. At C draw the ray making the supplied BCD angle; at A draw the ray making BAD. Their forward intersection is D. Parallel rays or an intersection behind a ray fail this construction. For a convex result the remaining angle360°−A−B−C must be between0° and180°.
With a proposed triangle sides2,3 and diagonal6,2+3<6 so the arcs do not meet. With sides1,5 and diagonal2, their sum exceeds2 but |5−1|=4>2, so one circle lies inside the other without meeting. Angles20°,30°,40° force a 270° fourth angle, ruling out the intended convex quadrilateral. State convexity when making that conclusion; angle sum alone does not prohibit every concave figure.
An instrument construction records the baseline, copied angles, compass centres, radii and chosen intersections. A proof then matches each given to a construction step. If two non-adjacent sides and three right angles are given, a rectangle with only one side length fixed can still have many widths; such data may be insufficient. Numerical coordinates in our models verify the illustrated construction, while your independent compass drawing needs review.
A diagonal construction with radii1 and5 and centre distance2 fails:2<1+5 is true, but2>|5−1| is false. Both strict bounds are necessary before drawing arcs.
The chapter’s introductory list describes useful data patterns. Compatibility, full inequalities and placement still need checking.
Our numerical coordinates check the SVG model. Plotting rounded points is not a substitute for the stated compass steps.
The position of each side, angle and diagonal matters.
| Method | First fixed structure | Remaining point |
|---|---|---|
| Four sides+included angle | The two adjacent sides and angle | Two circle radii from the fixed endpoints |
| Four sides+diagonal | The specified diagonal | One triangle on each side |
| Three sides+two diagonals | One SSS triangle on the given diagonal | Second circle intersection from two named centres |
| Three consecutive sides+included angles | Baseline and its two angle rays | Mark both endpoint lengths, join |
| Two adjacent sides+three angles | One SAS triangle | Forward intersection of two angle rays |
Check full triangle inequalities, ray directions, distinct vertices and the required convex/noncrossing result. A list of five values alone does not certify a construction.
Two circle radii r, s and centre distance e form a nondegenerate triangle only between strict bounds.
|r−s| < e < r+s For r=1, s=5, e=2: 4 < 2 < 6 is false
The upper-sum inequalities printed in the source are necessary but not sufficient by themselves.
BD=6 cm, AB=AD=5 cm, BC=CD=4 cm. Every coloured arc is centred on the labelled endpoint with the stated radius.
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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.
5
Five is the usual general count; compatibility and placement still matter.
16
3+5+4+4=16; a successful drawing must match these lengths.
90
It is the stated included angle at B, copied in the construction.
3
Symmetry places each intersection above or below the midpoint,3 cm from either end horizontally.
4
The altitude bisects6 to3; 5²−3²=16, so altitude4.
8
A nondegenerate triangle needs e<4+4=8.
16
5+5+6=16.
16
5+5+6=16; the second triangle requires its own check.
1
Their heights are3 and4 above the same baseline.
90
360−270=90.
75
360−110−80−95=75.
4
|5−1|=4; an upper-sum check alone would miss this restriction.
270
360−90=270; this is not a convex interior angle.
16
2(5+3)=16; changing the width would change this result.
18
2(5+4)=18, illustrating that a fixed length alone does not fix the figure.
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A hinged four-side frame can change angles and diagonals while preserving side lengths. For example equal fixed sides can make rhombi with different included angles. Another compatible condition is needed.
Name what changes and what stays fixed.
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Set B=(0,0), C=(5,0), A=(0,3) as a verification model. Radius4 circles about A, C meet; choose the intersection to the right and above giving convex ABCD. Join AD, DC and check distances. The compass construction uses these lengths, not plotted decimal estimates.
Preserve arcs and verify all five givens.
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Draw BD=6. Equal radius5 arcs locate A on one side; equal radius4 arcs locate C on the other. Join the boundary and verify both triangle inequalities and four sides.
Do not place both apex points on the same chosen side without checking the shape.
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If |r−s|≥e one circle is inside the other or internally tangent, so no nondegenerate triangle exists. Require |r−s|<e<r+s with positive lengths. Sides1,5, e=2 pass the sum test but fail the difference test.
Give a concrete counterexample and both strict inequalities.
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Construct ABD on BD using AB, AD. Then intersect the circle about A of radius AC with the circle about D of radius CD. Check |AB−AD|<BD<AB+AD and |AC−CD|<AD<AC+CD. Choose a C on the opposite side of BD from A and verify a convex noncrossing boundary.
The first triangle’s success alone does not prove the whole construction works.
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Build ABD with sides5,5,6. Its apex A is4 cm above the midpoint of BD. Circles about A, D of radii8,5 locate C on the opposite side,4 cm below BD. All four sides become5, diagonals6 and8.
Show both stages and check C rather than assuming its final position.
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Draw BC=5, same-side perpendicular rays at B, C, mark BA=3 and CD=4, then join AD. The labelled coordinates B(0,0), C(5,0), A(0,3), D(5,4) verify a convex figure with the requested givens.
Do not add a fourth side length that was not given.
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Build BC=5 and BA=3 on a perpendicular. Draw the perpendicular ray at C and a ray at A perpendicular to AB directed toward that C ray. They meet at D, making a 3-by5 rectangle.
Use forward rays, correct vertices and all three supplied angles.
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The supporting lines might meet behind a specified ray endpoint. That point does not satisfy the chosen interior-angle rays. Parallel rays may never meet. Verify a forward intersection and final cyclic interior angles.
Distinguish lines from rays.
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The shortest path along the two sides has length2+3=5, less than6. Triangle inequality fails, so no circle intersection constructs a nondegenerate triangle.
Do not describe failure as a bad drawing technique.
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The fourth would be270°, which exceeds180° and is reflex. Therefore no convex quadrilateral has those angles. The angle sum alone does not rule out every concave interpretation.
State exactly the convexity assumption.
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Rectangles of length5 and widths3 or4 both satisfy those opposite-side lengths and angles, but have different perimeters16 and18. An adjacent width is still missing.
Give two noncongruent examples with the same stated data.
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List the named givens, baseline, angle vertices, compass centres/radii, intersection choice and final joins. Then verify each given individually and the simple convex boundary.
A final outline without steps is incomplete.
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After placing the included-angle triangle, compute or compare the distance between the remaining circle centres. Their radii must satisfy both strict intersection inequalities; select an intersection producing the intended boundary, not a repeated or crossed vertex.
A five-number count does not guarantee existence or the intended orientation.
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Copy60° at the specified baseline endpoint and75° at the other with the correct interior orientation. Mark the given adjacent side lengths on those rays, join the endpoints and verify the resulting convex shape.75° can be constructed by bisecting between60° and90°.
Keep the positions of both angles explicit; do not swap them.
Construct one valid figure and document one impossible data set.
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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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