Class 8 · English Version · Mathematics
2026 Class 8 Mathematics · English Version · Chapter 10 · Chapter companion part
Choose the correct circular region or boundary and derive curved, open and closed cylinder surface areas from their parts.
Class 8 English Version:part D 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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An annulus is the region between concentric circles with outer radius R and inner radius r, where R>r≥0. Its area is π(R²−r²), notπ(R−r)². Ring width w=R−r, so R=r+w when width is given. The source Example4 text uses radii 9 and 4, but its inner picture label says8. Withπ≈3.14, the text gives 204.10 cm²; the picture readingr=8 gives 53.38 cm². Show the discrepancy before choosing an interpretation.
In the source creative ring, the inner radius is 3 m and a2 m label lies across the surrounding band. Reading2 m as band width gives outer radius 5 m. Reading it as outer radius would be inconsistent because2<3. For fresh inner radius 4 m, width3 m, the outer radius is 7 m and ring area 33πm². At50 BDT/m², cost is 1650πBDT; underπ≈3.14 that is 5181 BDT.
Attach an outward semicircle to every side of a square of sides. Each semicircle has radiuss/2 and arc length πs/2. The outer boundary is four arcs, total2πs. The square sides are internal joins and are not added to that perimeter. Area is a different question:the square plus four non-overlapping semicircular disks gives s²+πs²/2. Verify that the added regions do not overlap before adding areas.
For a radius 8 cm board with two radius 1 cm circular holes and one3 cm-by 1 cm rectangle removed, the remaining area is 64π−2π−3=62π−3 cm². This assumes all cutouts lie inside and do not overlap. The source mouth picture looks oval while its written data specify a rectangle; use the stated rectangle dimensions and record that the illustration is schematic. Do not invent an ellipse formula for rectangular data.
Rotate a rectangle about one fixed side of height h; the other dimensionr becomes the cylinder radius. The fixed side is the axis and the opposite rotating side is a generator of lengthh. A right circular cylinder has circular parallel bases and an axis perpendicular to them. The resulting radius isr, not the circumference of a rolled sheet. Rotating a square of sides givesr=h=s.
Cut along one straight generator and unroll the wall. The rectangle has height h and width equal to base circumference2πr. Curved surface area is 2πrh. A cut-surface layout shows this rectangle plus two disks of radius r for a closed cylinder. Each disk rim joins one long rectangle edge after rolling; they are not flat polygon faces hinged along straight shared edges. Perspective circles may look elliptical in the drawing, but represent circular bases.
A closed cylinder has area 2πrh+2πr²=2πr(h+r). An open-top cylinder with one base has2πrh+πr². A wall only has2πrh. These ideal areas ignore thickness, glue tabs, waste and interior/exterior double coating unless specified. Surface area is in square units; volumeπr²h is a different quantity and is not the surface-area formula.
Use one length unit before squaring or multiplying. Check R>r for a ring, d=2r for a circle, and whether a cylinder task requests wall, one cap or two. Forr=3 cm, h=7 cm with π≈3.14, the curved area is 131.88 cm², one-base total 160.14 cm² and closed total 188.40 cm². A cost multiplies area by price per square unit; do not charge a surface rate per cubic unit.
A radius 3 cm, height7 cm cylinder has wall area 42π and each cap9πcm². Wall, one-base and closed totals are 42π,51π,60π cm² respectively. Underπ≈3.14 they are 131.88,160.14,188.40 cm².
Example4 has inner 4 in its text and inner 8 in its picture. The two interpretations and their different answers remain visible.
The two disks close curved rims after the rectangular wall is rolled. Our diagram is a dimension-matched cut layout.
Example4 uses outer radius 9; compare the text and image readings using π≈3.14.
| Reading | Area expression | Area in cm² |
|---|---|---|
| Text inner radius 4 | π(81−16)=65π | 204.10 |
| Image inner radius 8 | π(81−64)=17π | 53.38 |
These are incompatible inputs, not two correct answers for the same fixed ring. State which reading is being used. In creative5, inner radius 3 plus band width2 instead gives outer radius 5.
Radius3 cm and height7 cm; ideal surfaces have no thickness or joining allowance.
| Surface | Exact area (cm²) | Using π≈3.14 (cm²) |
|---|---|---|
| Wall only | 42π | 131.88 |
| One cap | 9π | 28.26 |
| Wall + one cap | 51π | 160.14 |
| Wall + two caps | 60π | 188.40 |
The difference between adjacent totals is a physical circular cap. Volume uses cubic units and is a different calculation.
One wall rectangle plus two circular caps.
A=2πrh+2πr² =2πr(h+r) r=3, h=7 ⇒ A=60π cm²
The cut layout uses exact width6π cm≈18.85 cm. With the convention π≈3.14 the calculated width is 18.84 cm; do not confuse the approximation with exact π.
Compare the generating rectangle, the cylinder and the cut surface pieces.
Manual stages; no autoplay. The disks are separate cut pieces, not straight-edge hinged polygon faces.
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.
103.62
3.14×(49−16)=103.62.
7
R=r+w=4+3.
33
R²−r²=49−16=33.
5181
Area33×3.14=103.62 m²; cost×50=5181.
50.24
Four half-circumferences total2πs=2×3.14×8=50.24.
32
Each radius 4 semicircle has area 8π; four give32π.
191.68
62×3.14−3=191.68.
3
The other rectangle dimension becomes the radius.
7
The fixed axis side becomes its height.
131.88
2×3.14×3×7=131.88.
18.84
The width is 2πr=18.84, not the diameter 6.
188.4
131.88+2×28.26=188.4.
160.14
131.88+28.26=160.14.
56.52
2πr²=2×3.14×9=56.52.
200
0.02×10,000=200 cm².
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Subtract the inner disk area from the outer disk. Squaring the width instead would describe a different disk and omit the cross term. For R=7, r=4, the coefficient is 49−16=33, not9.
Name both complete disks and their common centre.
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The text says R=9, r=4, so area 65π≈204.10 cm² underπ≈3.14. The image says inner 8, which would give17π≈53.38 cm². State which input you use; the two readings cannot be silently combined.
Retain the discrepancy alongside the corrected calculation.
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If2 m is band width, outer radius is 3+2=5 m and ring area(25−9)π=16πm². If2 m is interpreted as the outer radius, the geometry contradicts inner 3 m. Use and state the width reading.
Identify exactly where the labelled segment lies.
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R=7 m. Areaπ(49−16)=33π≈103.62 m². Cost=103.62×50=5181 BDT, under the stated approximation and no extra waste.
Keep radius, width, area and price units distinct.
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Each has radius 4 cm and arc length 4πcm. Four arcs give16πcm≈50.24 cm ifπ≈3.14. The four square sides are internal joins, so they do not add32 cm to the outer boundary.
Trace the exterior boundary once.
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The perimeter is 16πcm. The region area is square64 plus four half-disks totalling32π, so 64+32πcm². The attached disks lie outside different square sides and meet only at corner points.
Use area for filled regions and arc length for the boundary.
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Outer disk64π minus two circles2π minus rectangle3 gives 62π−3 cm². The result assumes disjoint cutouts fully inside; overlaps must not be subtracted twice.
Explain the assumptions and why the source oval-looking drawing does not override its rectangular dimensions.
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The fixed7 cm side is the axis, height7 cm. The perpendicular3 cm width sweeps a radius 3 cm, while its opposite7 cm side sweeps the curved wall as a generator.
Do not confuse revolution of a radius-width rectangle with rolling an18.84 cm-wide wall sheet.
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A generator cut opens the wall to a rectangle of height h and width2πr, the base circumference. Their product2πrh is the curved area. The straight diameter 2r would be too short for the unrolled width.
Label both rectangle dimensions and their units.
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Use a rectangle7 cm high and 6πcm wide, plus two separate disks of radius 3 cm. The rectangle wraps once, and each disk circumference matches a rim. This is a cut layout, not a cuboid-style net with straight-edged cap hinges.
Explain the curved joins and exclude tabs/thickness unless given.
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The wall contributes2πrh. Two caps add2πr² for closed total2πr(h+r). One cap addsπr² for open-top total2πrh+πr². No caps leaves only the wall.
Name the number of included circular faces.
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Wall 131.88 cm²; each cap 28.26 cm²; one-base 160.14 cm²; closed 188.40 cm². Differences of 28.26 or56.52 correspond to one or two actual caps.
Check area differences against the component faces.
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Convert0.03 m to 3 cm, then 2×3.14×3×7=131.88 cm². Alternatively convert7 cm to 0.07 m and obtain0.013188 m², which equals131.88 cm².
Convert before multiplying and check squared-unit conversion.
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Radius and height are both4 cm. Closed area 2π×4×(4+4)=64πcm². The generator has length 4 cm; the base diameter is 8 cm.
Keep the axis side, radius and diameter distinct.
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State whether a supplied length is radius, diameter or width; mark all included regions/faces; choose or retain π; convert units; calculate area or length as requested; then multiply by the matching price rate. Verify existence and cutout overlap assumptions.
Apply the checklist to one ring and one cylinder example.
Build one circular-region budget and one cylinder cut layout.
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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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