Class 8 · English Version · Mathematics
2026 Class 8 Mathematics · English Version · Chapter 10 · Chapter companion part
Use chord theorems to justify parallelism, midpoint loci, circle centres and equal chord portions.
Class 8 English Version:part B 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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Let two distinct intersecting chords share midpoint P. If P≠O, the centre-midpoint theorem makes OP perpendicular to both chord lines. Two distinct lines through P cannot both be perpendicular to one nonzero line, so P must be O. Thus both chords are diameters and their four halves equal the common radius. Merely intersecting, without mutual bisection, does not imply the centre.
For distinct parallel chords with midpoints M, N, the perpendicular bisector of each passes through O. Since the chord lines are parallel, their perpendicular bisectors have the same direction and pass through O, so they are the same line. Hence O, M, N are collinear and MN is perpendicular to both chords. If one chord is a diameter, its midpoint is O and the other bisector still determines the line.
Chords AB, AC from the same boundary point A on opposite sides of AO have equal lengths exactly when their angles with AO are equal. Equal chords and equal radii give triangles OAB, OAC congruent by SSS, hence equal angles at A. Conversely each radius triangle is isosceles; equal angles at A imply equal central angles and SAS with two radii yields AB=AC. State that B, C are distinct and the intended rays are on opposite sides.
For right ABC with C=90°, complete rectangle ACBD by drawing parallels through A and B. Its diagonals AB, CD are equal and bisect at O, so OA=OB=OC. A circle centred O with radius AB/2 passes through A, B, C. The circumcentre is the centre of this circle. Two independent chord bisectors make it unique; the right-angle vertex itself is not the centre.
Concentric circles have the same centre. On an outer-circle chord AB crossing the inner circle at C, D in order A, C, D, B, drop perpendicular OM. It bisects both chords, so AM=BM and CM=DM. Subtract:AC=AM−CM=BM−DM=BD. For outer radius 15, inner 13 and distance 12 cm, half chords are 9 and 5, so each outside portion is 4 cm. A line missing the inner circle has noC, D portions.
For equal chords AB, CD meeting inside at P, let M, N be perpendicular feet from O. Equal chords give OM=ON and equal half-chord length k. If P≠O, right triangles OMP, ONP give PM=PN=t. Each chord’s two portions therefore have lengths k−t, k+t in some order. State which short portion matches which short portion in your labelled drawing. IfP=O, all four portions are radii. Do not assert an unspecified endpoint matching.
All chords of one fixed length L<2r have midpoint distance √(r²−(L/2)²) from O; their midpoint locus is a smaller concentric circle. IfL=2r, all midpoints collapse to O, not an ordinary positive-radius circle. Equal chords AC, BD drawn from opposite diameter endpoints A, B to opposite sides give OAC≅OBD by SSS, so matching alternate angles make AC∥BD. Conversely such parallel chords give equal base angles in the two isosceles radius triangles, hence equal chords.
For fixed r, chord length L satisfiesL²=4(r²−h²). If h increases from 0 to r, L decreases. Comparing two unequal lengths in the same circle therefore reverses their distances:the larger is nearer O. The comparison does not work across different radii without more information. The diameter is the distance 0 maximum, and a tangent is the limiting one-point case.
For concentric radii 15 cm,13 cm and a crossing line12 cm from O, the half lengths are 9 cm and 5 cm. The two outer portions each measure4 cm by subtraction about the shared midpoint.
Equal diameters have only one midpoint O. Calling that a positive-radius circle would overlook a degenerate case.
For equal intersecting chords, the shorter portions match and the longer portions match. Unspecified endpoint correspondence can assert the wrong equality.
List full givens before choosing a theorem.
| Given | Bridge | Required result |
|---|---|---|
| Mutual bisection | A nonzero OP cannot be perpendicular to two distinct crossing lines | P=O |
| Parallel chords | Their perpendicular bisectors coincide through O | Midpoints and O collinear |
| Concentric crossing | Subtract outer and inner half chords at one foot | AC=BD |
| Equal intersecting chords | Equal half lengths k and equal foot distances t | Portion pair {k−t, k+t} |
| Fixed length L<2r | h²=r²−(L/2)² is constant | Midpoint locus is a concentric circle |
For diameters, the midpoint locus is one point O. The unordered portion pair does not justify arbitrary endpoint matches.
Outer radius 15 cm, inner 13 cm, perpendicular distance 12 cm.
AM=√(15²−12²)=9 cm CM=√(13²−12²)=5 cm AC=AM−CM=4 cm=BD
The point order is A, C, M, D, B. If the line misses the inner circle, C and D do not exist.
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.
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.
7
Their crossing is the centre and each half is a radius.
7
Opposite-side perpendicular distances add:3+4.
1
Same-side distances subtract:4−3.
35
SSS congruence gives the matching angle.
9
The centre is the hypotenuse midpoint, at half its length from each vertex.
9
√(15²−12²)=9.
5
√(13²−12²)=5.
4
Subtract the same-side half lengths:9−5.
5
The unordered portion pair is the same:{5,11}.
11
Match the longer members of the equal unordered pairs.
3
√(25−4²)=3.
1
Every diameter midpoint is the same point O; the locus is not a positive-radius circle.
9
The given equal-chord condition fixes BD.
12
Half length 5 gives √(169−25)=12.
5
Half length 12 gives √(169−144)=5; the longer chord is nearer O.
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Let their common midpoint P differ from O. Then OP is perpendicular to both distinct intersecting chord lines, which is impossible. Therefore P=O, and both chords are diameters.
State distinct chord lines and avoid assigning a direction to OP once P=O.
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Both perpendicular bisectors pass through O and have the same direction. They coincide as one line through both distinct midpoints. If one midpoint is O, use the other bisector to determine this line.
Explain why parallel chords do not give two independent lines for locating O.
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Triangles OAB, OAC have common OA, equal radii OB=OC and given AB=AC. SSS makes corresponding angles at A equal.
Write the correspondence before taking the angles.
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Each of OAB, OAC is isosceles by its equal radii. Equal angles at A give equal corresponding base angles and hence equal included angles at O. With OA common, OB=OC, SAS gives equal chord bases AB=AC.
Do not use an invalid ambiguous side-side-angle congruence rule.
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Complete a rectangle with the triangle as one half. Its equal, bisecting diagonals show the midpoint O of the hypotenuse is equidistant from all three triangle vertices. The circle of that radius goes through them, and chord-bisector uniqueness fixes its centre.
Do not assume the centre is the right-angle vertex.
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The perpendicular from their common centre bisects both chords at the same M. Then AM=BM, CM=DM, and AC=AM−CM=BM−DM=BD.
Use the stated point order and verify that the line intersects both circles.
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Let the half length be k. Equal perpendicular distances and shared OP give PM=PN=t by RHS. The portions on each chord are k−t, k+t. Pair short with short, long with long. AtP=O all four equal r.
If a foot equals P, then both feet do andt=0; handle that case without a degenerate triangle proof.
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The perpendicular-foot distanceh=√(r²−(L/2)²) is the same positive value for every such chord, so their midpoints form a circle centred O. For L=2r, h=0 and the locus is only O.
Define locus as the complete set of positions, including that every direction can occur.
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OA=OB, OC=OD and AC=BD make OAC≅OBD by SSS. Matching angles between the chords and diameter are equal alternate angles, so AC∥BD. The opposite-side placement selects the correct alternate-angle configuration.
Draw endpoints cyclically and do not omit the side-of-diameter condition.
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Parallelism gives equal alternate angles OAC, OBD. Radius triangles OAC, OBD are isosceles, so their remaining base angles also match; their central angles match. OA=OB, OC=OD and that central angle give SAS, so AC=BD.
State both equal radii and the matching included angle.
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For each chord L²=4(r²−h²). The larger L has largerL², so its h² is smaller; nonnegative distances preserve the comparison. Thus its perpendicular distance is smaller.
Use a fixed commonr and explain why positive square roots preserve order.
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Their lengths are 2√(25−9)=8 cm and 2√(25−16)=6 cm. The nearer chord is longer. If the chords are parallel, the line separation is 1 cm on the same side of O or7 cm on opposite sides.
Do not add distances until the placement is specified.
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Mutual bisection forces the crossing to O, so each chord is a diameter 14 cm and each half is 7 cm. Equal total chord lengths alone would not force the crossing to O.
Separate the intersection, equality and bisection premises.
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Midpoint O of the 10 cm hypotenuse has distance 5 cm to both its endpoints and to the right-angle vertex by the rectangle argument. The circle has radius 5 cm, not10 cm.
Show a labelled construction and all three equal joins.
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No. A fixed chord length is needed. Different lengths have different perpendicular centre distances; all chord midpoints together fill the disk interior, while a fixed non-diameter length gives one concentric circle.
State which quantity is fixed before naming the locus.
Build a proof portfolio for parallel and concentric chords.
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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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