Class 8 · English Version · Mathematics
2026 Class 8 Mathematics · English Version · Chapter 9 · Chapter companion part
Use right-triangle equations in composite figures and prove shared-height, median and midpoint identities while detecting false claims.
Class 8 English Version:part B of the chapter9 companion. Eight teaching sections and30 original written tasks. Selected companion coverage,not complete textbook coverage; use both parts and review full proofs,drawings and exact surds with a guardian or teacher.
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For a ladder of length13 m with foot5 m from a vertical wall, the height is√(13²−5²)=12 m if wall and ground are perpendicular. A rectangle with sides8,15 has diagonal17. A square of side s has diagonal²=2s², so its area is half the area of the square on its diagonal. These are geometric idealisations; a sloping wall needs a different model.
In an original composite model, ABCF is a 10-by6 rectangle and E, D lie3 units below FC with ED=7. Then a right triangle atQ has legsCQ=3, DQ=3, soCD²=18 andCD=3√2. A pointP anywhere on AB makes triangleFPC area10×6/2=30, independent of its horizontal position. The source uses millimetres but labels one requested area square metres; keep area units consistent with lengths.
In right ABC with A=90°, let D be midpoint of AB, AD=DB=m and AC=n. ThenCD²=m²+n² whileBC²=4m²+n², soBC²=CD²+3AD². Median means it joins a vertex to the opposite-side midpoint; it need not be perpendicular. A median to the hypotenuse is a different segment from this median to a leg.
With A=90°, writeAB=u, AC=v, P midpoint of AC, Q midpoint of AB. ThenBP²=u²+v²/4 andCQ²=v²+u²/4. Their sum is5(u²+v²)/4=5BC²/4. Multiply by4 to obtain5BC²=4(BP²+CQ²). State which side each midpoint belongs to before using half lengths.
If cyclic ABCD has AC⊥BD at O, setOA=p, OC=q, OB=r, OD=s. Four right triangles giveAB²=p²+r², BC²=q²+r², CD²=q²+s², DA²=p²+s². Adding opposite pairs givesAB²+CD²=BC²+DA². Bisection and equal opposite sides are not required. This identity alone is not a claim that every such figure is a rectangle.
IfAD⊥BC at D, thenAB²=AD²+BD² andAC²=AD²+CD², soAB²−AC²=BD²−CD². AnyP on lineAD similarly givesPB²−PC²=BD²−CD², unchanged when its height varies. For rightA and anyD onAC, BC²=AB²+AC² andBD²=AB²+AD², soBC²+AD²=BD²+AC². Equal right-triangle terms cancel; do not subtract unsquared lengths.
For rightA and D, E midpoints ofAB, AC, triangleADE givesDE²=AD²+AE²=BD²+CE². In the separate crossed model, OB=9, OA=3, BC=9 withAO∥BC and A, D, C collinear acrossBO. Similar right trianglesAOD, CBD giveOD:BD=3:9 andOD+BD=9, soOD=2.25, BD=6.75. The wholeAC has horizontal9 and vertical12, soAC=15.
SourceExercise9(7) says a right triangle with altitudeAD toBC satisfiesAB²+BC²+CA²=4AD². For A-right3–4–5, the left is9+25+16=50, butAD=12/5 gives right side576/25=23.04. The statement is false as printed. A valid altitude identity isAB²+AC²=BD²+CD²+2AD². Also source short(a) listsAB=15, BC=12, AC=9 but asksBAC:the right angle isBCA, notBAC. Do not silently answer90° for the wrong vertex.
ForAB=8, AC=6 andA=90°, BC²=100. WithP midpoint ofAC andQ midpoint ofAB, BP²=73, CQ²=52, and4(73+52)=500=5BC². The theorem applies to the two smaller right triangles separately.
Exercise9(7) is false as printed; the companion shows numerical evidence and a valid related identity instead of manufacturing a proof.
For sides9,12,15, the right angle is opposite15. BAC andBCA name different vertices and cannot be exchanged silently.
Use rightABC, A=90°, AB=8, AC=6; P midpoint ofAC, Q midpoint ofAB.
| Quantity | Right-triangle calculation | Squared length in cm² |
|---|---|---|
| BC² | 8²+6² | 100 |
| BP² | 8²+3² | 73 |
| CQ² | 6²+4² | 52 |
| 4(BP²+CQ²) | 4(73+52) | 500 |
| 5BC² | 5×100 | 500 |
These entries are squared lengths. BP andCQ themselves are√73 and√52 cm; they are not73 cm and52 cm.
AD is perpendicular toBC atD.
AB²=AD²+BD² AC²=AD²+CD² AB²−AC²=BD²−CD²
This is a squared-length identity. It does not justifyAB−AC=BD−CD.
Use anA-right3–4–5 triangle, with altitudeAD to its5 cm hypotenuse.
| Expression | Calculation | Value in cm² |
|---|---|---|
| AB²+BC²+CA² | 9+25+16 | 50 |
| AltitudeAD | 2×area/BC=12/5 | 2.4 cm is a length |
| 4AD² | 4×(12/5)² | 23.04 |
The first and last values differ, so printedExercise9(7) is false. The correct summed altitude identity isAB²+AC²=BD²+CD²+2AD². A valid counterexample refutes a universal claim.
Match each letter to its explanation. Read the quantities, units and relationships, then check the counts or equations. These are original teaching models.
Original exact-coordinate model, not source artwork. Read the angle and side labels; a page orientation does not identify the hypotenuse.
Match each letter to its explanation. Read the quantities, units and relationships, then check the counts or equations. These are original teaching models.
Original exact-coordinate model, not source artwork. Read the angle and side labels; a page orientation does not identify the hypotenuse.
Match each letter to its explanation. Read the quantities, units and relationships, then check the counts or equations. These are original teaching models.
Original exact-coordinate model, not source artwork. Read the angle and side labels; a page orientation does not identify the hypotenuse.
Match each letter to its explanation. Read the quantities, units and relationships, then check the counts or equations. These are original teaching models.
Original exact-coordinate model, not source artwork. Read the angle and side labels; a page orientation does not identify the hypotenuse.
Match each letter to its explanation. Read the quantities, units and relationships, then check the counts or equations. These are original teaching models.
Original exact-coordinate model, not source artwork. Read the angle and side labels; a page orientation does not identify the hypotenuse.
Show your working on paper. Check your answer, then choose Next.
12
√(169−25)=12.
17
√(64+225)=17.
98
7²+7²=98.
30
FC=10 mm and its perpendicular distance toAB is6 mm, so area30mm².
18
3²+3²=18; CD=3√2 mm.
52
AD=4; CD²=4²+6²=52.
48
100−52=48, also3×4².
73
8²+3²=73.
52
6²+4²=52.
500
4(73+52)=500=5×100.
54
(9+25)+(16+4)=54.
55
The common25 cancels:64−9=55.
6.75
OD:BD=1:3 andOD+BD=9, soBD=27/4.
50
9+16+25=50, regardless of which leg is3 or4.
23.04
Area6 givesAD=12/5;4×(12/5)²=576/25=23.04.
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Vertical wall and horizontal ground give a right angle. The ladder is the hypotenuse, soh²=13²−5²=144 andh=12 m. This ideal model ignores ladder thickness and wall/ground unevenness.
Draw the right angle and identify the positive length.
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For sides, the diagonal d satisfiesd²=s²+s²=2s². Original square area iss²=d²/2, half the diagonal-square aread².
Compare areas, not side lengths.
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FC is fixed10 andAB∥FC at separation6. EveryP onAB has the same perpendicular height6 aboveFC, so its area is10×6/2=30 square units.
P must remain on the stated parallel line.
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CD=√18=3√2 mm. Any area computed from these millimetre lengths is in mm². The source square-metre label beside its millimetre figure is inconsistent; convert explicitly if metres are requested.
Do not silently combine millimetre lengths with square-metre values.
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LetAD=DB=m, AC=n. Right trianglesCAD, CAB giveCD²=m²+n² andBC²=(2 m)²+n². Subtraction givesBC²−CD²=3m².
Use the midpoint condition, not an unsupported perpendicular median.
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WriteAB=u, AC=v. BP²=u²+v²/4 andCQ²=v²+u²/4. Add to get5(u²+v²)/4=5BC²/4; multiply by4.
Show both small right triangles and their half-side lengths.
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LetOA=p, OC=q, OB=r, OD=s. ThenAB²+CD²=p²+r²+q²+s², whileBC²+DA²=q²+r²+p²+s². The same four terms establish equality.
No midpoint or equal-side assumption is needed.
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AB²=AD²+BD², AC²=AD²+CD². Subtract to cancel the identicalAD². Lengths are nonnegative distances, so the squared relations remain valid when a foot is on the extended line.
State each right triangle before subtraction.
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PB²=PD²+BD² andPC²=PD²+CD², so their difference isBD²−CD². The analogousA equations give exactly that difference. Hence it is independent ofP height.
Explain what remains fixed whenP moves.
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BC²=AB²+AC² andBD²=AB²+AD². AddAD² to the first orAC² to the second:both equalAB²+AC²+AD².
Identify the two common-right-angle triangles.
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TriangleADE is right atA, soDE²=AD²+AE². Midpoints giveAD=BD andAE=CE; substituting yields the claim.
Use the midpoint equalities only after the right-triangle equation.
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Right trianglesAOD, CBD are similar, soOD/BD=3/9=1/3. SinceOD+BD=9, OD=2.25, BD=6.75. A andC differ horizontally9 and vertically3+9=12, soAC=√(81+144)=15 cm.
Use the entire vertical separation forAC, not9−3.
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The squared-side sum is50. Area6 gives altitudeAD=12/5, so4AD²=23.04, which differs from50. A correct general altitude identity isAB²+AC²=BD²+CD²+2AD², obtained by adding the two right-triangle equations.
A counterexample refutes the printed universal statement; never invent a proof for it.
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Since9²+12²=15², the longest sideAB is the hypotenuse and its opposite angleBCA is90°. BAC is acute, not90°; if inverse trigonometry is available, cos(BAC)=9/15=3/5 gives about53.13°. State the naming issue.
Do not silently changeBAC toBCA when reporting the answer.
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Check givens, vertex order, units and calculations. Present the counterexample explicitly, then offer a correctly stated related identity with a proof. Record any interpretation change and ask a teacher to review the original wording.
Reasoned correction is part of learning; an answer key is not permission to assert something false.
Make a proof-and-error notebook using one median and one shared-height model.
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This original Kidemy practice follows the topic order of 2026 Class 8 Mathematics · English Version, chapter 9, printed pages 152–158. It does not replace the complete chapter or its practical teaching.
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