Class 8 · English Version · Mathematics
2026 Class 8 Mathematics · English Version · Chapter 10 · Chapter companion part
Identify circle features and prove how a chord, its midpoint and its distance from the centre are related.
Class 8 English Version:part A of the chapter10 companion. Eight teaching sections and30 original written tasks. Selected companion coverage,not complete textbook coverage; use all four parts. General proofs,exact pi expressions,compass drawings and practical measurements need guardian or teacher review.
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A circle is the set of plane points at one fixed positive distance from centre O. That distance is radius r. The filled circular region is a disk. A compass fixes its point at O and keeps its opening r; its pencil traces the boundary. A wheel rim and a clock-hand tip suggest this model, while a coin face also includes the inside. Real traced edges and physical measurements are approximate.
A chord PQ is the straight segment joining two distinct boundary points. The two boundary paths between P, Q are arcs. For a non-diameter chord, the shorter is the minor arc and the longer the major arc; a third point names the intended path. The two arcs together make one whole boundary. A segment from the centre to the circle is a radius, not a chord unless both endpoints lie on the circle.
A diameter is a chord through O, with length d=2r. Its two arcs are semicircles of equal length. The circumference C is the length of the complete boundary, not the straight diameter or the filled area. Tracing a marked rim through one complete revolution, without slipping, measures one circumference along the ground. A fixed-length clock hand traces a circle only when its pivot is fixed.
For a non-diameter chord AB with midpoint M, join OA, OB, OM. Triangles OAM, OBM have OA=OB (radii), AM=BM (midpoint), and OM common. SSS gives equal angles OMA, OMB; they form a straight angle, so each is 90°. Thus OM⊥AB. For a diameter, M=O and the zero segment OM has no direction; do not call it a perpendicular line.
If OM is perpendicular from O to chord AB, triangles OMA, OMB are right with equal hypotenuses OA=OB and shared leg OM. RHS (right angle, hypotenuse, one side) congruence gives AM=BM. Thus the perpendicular bisects the chord. A chord’s perpendicular bisector passes through the centre. Two distinct nonparallel chords can locate the centre by intersecting their perpendicular bisectors.
Let equal chords AB, CD have perpendicular feet E, F from O. By the converse, AE=AB/2 and CF=CD/2, so AE=CF. Right triangles OEA, OFC have equal radii as hypotenuses and equal half-chord legs. RHS gives OE=OF. Distance to a line means its perpendicular distance, not a sloping segment to an endpoint. A radius and a half chord play different roles.
For two chords in the same circle, OE=OF and equal radii make right triangles OEA, OFC congruent by RHS. Hence AE=CF; double the half lengths to get AB=CD. The same-circle or equal-radius condition is essential:equal distances in different-sized circles need not give equal chords. The two chord-distance theorems are converses and need separate premises.
For a non-diameter chord AB, triangle OAB gives AB<OA+OB=2r. A diameter attains 2r. With chord distance h from O and half length x, Pythagoras gives x²+h²=r², so length=2√(r²−h²). A genuine nonzero chord requires 0≤h<r. At h=r the line is tangent, meeting the circle once; h>r gives no boundary intersection. A straight line meets a circle in at most two points.
For r=5 cm and centre-to-chord distance 3 cm, the half-chord triangle is 3–4–5. The half length is 4 cm and the whole chord 8 cm; the perpendicular foot is the midpoint.
The printed radius 3/distance 6 question has no chord. Its answer list must not turn an impossible object into a length 0 answer.
Measured tracing, clock tips and wheel rims illustrate circles; their practical errors do not change the mathematical definition.
AB is not a diameter; M is its midpoint.
| Statement | Reason |
|---|---|
| OA=OB | Radii of one circle |
| AM=BM | Given midpoint |
| OM=OM | Shared side |
| OAM ≅ OBM | SSS congruence |
| ∠OMA=∠OMB=90° | Equal supplementary angles |
For the converse, perpendicular OM is given first; use RHS to prove AM=BM. Do not assume the conclusion. A diameter has M=O, so its zero join has no direction.
A line is at perpendicular distance h from a radius r centre.
x²+h²=r² L=2x=2√(r²−h²) 0≤h<r for a nonzero chord
At h=r there is one tangent point. At h>r there is no intersection, not a chord of length zero. The printed radius 3/distance 6 task has no chord.
Match each letter to its explanation. Read the quantities, units and relationships, then check the counts or equations. These are original teaching models.
Original geometry, not copied artwork. Curved sectors retain their arcs. Perspective cylinders are explicitly schematic; the separate cut layout uses one length scale.
Match each letter to its explanation. Read the quantities, units and relationships, then check the counts or equations. These are original teaching models.
Original geometry, not copied artwork. Curved sectors retain their arcs. Perspective cylinders are explicitly schematic; the separate cut layout uses one length scale.
Match each letter to its explanation. Read the quantities, units and relationships, then check the counts or equations. These are original teaching models.
Original geometry, not copied artwork. Curved sectors retain their arcs. Perspective cylinders are explicitly schematic; the separate cut layout uses one length scale.
Match each letter to its explanation. Read the quantities, units and relationships, then check the counts or equations. These are original teaching models.
Original geometry, not copied artwork. Curved sectors retain their arcs. Perspective cylinders are explicitly schematic; the separate cut layout uses one length scale.
Match each letter to its explanation. Read the quantities, units and relationships, then check the counts or equations. These are original teaching models.
Original geometry, not copied artwork. Curved sectors retain their arcs. Perspective cylinders are explicitly schematic; the separate cut layout uses one length scale.
Show your working on paper. Check your answer, then choose Next.
6
The fixed centre-to-boundary distance is the radius.
12
d=2r=12.
7.5
r=d/2=7.5.
2
There are two paths along the circle between the endpoints.
2
Each is half of the complete boundary.
90
Centre-to-midpoint OM is perpendicular to the chord.
16
The perpendicular bisects, so AB=2AM.
4
Equal chords in one circle are equidistant from its centre.
10
The converse chord-distance theorem gives equality.
18
A diameter has length 2r=18.
0
The line is outside the disk; no chord exists. This does not assign length 0 to a chord.
1
It is a tangent, not a nonzero chord.
8
Half length=√(25−9)=4; double it.
12
Half chord 5 gives √(169−25)=12.
12
The half-chord right triangle gives √(169−25)=12.
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The pencil traces the circle boundary. The whole filled face is a disk. A radius joins the chosen centre to the boundary. A physical tracing is only an approximate model of the ideal circle.
Name the boundary, inside and distance separately.
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The same endpoints have two boundary paths. A non-diameter chord has a shorter minor arc and a longer major arc; name a third point on a path to distinguish them. The chord is the straight join, not either curved path.
Draw and label both paths without confusing an arc with an area.
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The straight chord passes through O and contains two radii end-to-end. A half-turn about O exchanges its two boundary arcs, so their lengths are equal and each is half the circumference.
Use a centre point and two opposite endpoints.
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OA=OB, AM=BM and OM common give OAM≅OBM by SSS. The matching angles at M are equal and supplementary, so each is 90°. Therefore OM⊥AB.
State why the chord is not a diameter and why equal angles are supplementary.
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With foot M, the triangles OMA, OMB have right angles, equal radius hypotenuses and shared leg OM. RHS gives AM=BM.
Do not assume the midpoint condition while proving it.
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The perpendiculars bisect both chords, so AE=CF. Radii OA=OC are equal hypotenuses of right triangles OEA, OFC. RHS implies OE=OF.
Mark perpendicular feet and match corresponding sides.
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Right triangles OEA, OFC have equal radius hypotenuses and equal distance legs. RHS gives AE=CF. Since each is a half chord, AB=2AE=2CF=CD.
State the common-radius assumption.
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For noncollinear O, A, B, triangle inequality gives AB<OA+OB=2r. A diameter is exactly2r. The strict inequality cannot be applied to the degenerate triangle when AB itself is a diameter.
Separate the strict case from equality.
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The entire line is farther from the centre than every circle point can be. It has no circle intersection and no chord. Reporting a zero-length chord would confuse this outside line with a tangent at distance 3 cm.
Identify the impossible premise; do not select an answer option merely because it says0.
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The perpendicular bisects the chord. Half lengthx satisfiesx²+h²=r², so AB=2√(r²−h²). Require r>0 and 0≤h<r for a nonzero chord.
Keep the half-length and full-length steps separate.
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Draw two distinct nonparallel chords and construct each perpendicular bisector. Their intersection is equidistant from the relevant endpoints and is the circle centre. Two parallel chord bisectors can coincide, so they do not give two independent lines.
Retain compass arcs and choose suitable chords.
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Keep a fixed positive compass opening, draw the circle, and choose four points on its boundary. Measure their joins to O:each should approximate the same radius. Small physical differences reflect drawing/measurement precision, not four different ideal radii.
Record instrument units and distinguish practical checks from proof.
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At distance 0, the chords through centres are diameters. Radius3 and radius 5 circles give lengths6 and 10 although both centre distances are 0. Equal-radius or same-circle context is required.
State both radii and both lengths.
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Its midpoint is O, so OM has length 0 and no direction. A perpendicular line through O can bisect the diameter, but the zero segment itself is not a directed perpendicular.
Explain the degenerate join rather than applying the theorem mechanically.
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Take the perpendicular foot from O to the line. A boundary point must lie at distance √(r²−h²) on one side or the other of the foot. There are at most two such positions; one if tangent and none if outside.
Use the fixed radius and both possible directions along the line.
Make a labelled circle and prove one theorem with its converse.
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This original Kidemy practice follows the topic order of 2026 Class 8 Mathematics · English Version, chapter 10, printed pages 159–170. It does not replace the complete chapter or its practical teaching.
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