Class 8 · English Version · Mathematics
2026 Class 8 Mathematics · English Version · Chapter 10 · Chapter companion part
Explain pi through measured ratios, choose approximations explicitly, and derive/use circumference and area with correct units.
Class 8 English Version:part C 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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For an ideal Euclidean circle, the ratio C/d is one fixed number called pi,π. A ratio compares quantities by division; C/d says how many diameter-lengths make one circumference. It has no unit because both lengths use the same unit. π≈3.14159… is irrational:its decimal neither terminates nor repeats. Neither3.14 nor22/7 is exactlyπ. A finite checked task states the approximation to use.
Wrap string or a flexible tape around a circular rim and measure a straight diameter through the centre, in the same unit. Divide measured C by d. Close but unequal experimental ratios reflect drawing and measurement limits, not different ideal values of π. The source table uses circumference/diameter although one activity sentence says radius. For radius the ratio is C/r=2π. Also22/7=3.142857…;3.142 is a truncation, while rounding it to 3 decimal places gives 3.143.
Circumference C=πd=2πr. For diameter 12 cm, use π≈3.14 to obtain37.68 cm. The source Example1 starts with diameter 10 but calls it radius 10 in its final sentence;31.4 cm uses diameter 10. If radius were10, the same approximation would give62.8 cm. Always label the supplied dimension before multiplying.
From a circumference find d=C/π and r=C/(2π). For C=88 cm and π≈22/7, r=14 cm under that convention. The formula contains division by 2 only when the requested quantity is radius. Keep exact symbolicπ when no numerical approximation is requested. A value computed with a stated approximation is not a measurement guaranteed exact in the real world.
Without slipping, distance=number of turns×circumference. For diameter 70 cm and π≈22/7, a turn covers220 cm; five turns cover1100 cm=11 m. Two loops of garden fencing likewise need twice the circumference, before any stated joining allowance. Cost=required length×price per length in matching units. The wheel’s radius is not the distance travelled in a turn.
Cut a disk into an even number of equal sectors and alternate their directions. Eight pieces make a scalloped strip;16 or32 have shallower curved bulges. The pieces preserve area exactly, but no finite strip is an exact rectangle. As the sectors become narrower, its limiting height isr and limiting long side is half the circumference,πr. This motivates area A=(πr)r=πr²; do not replace actual curved sectors by triangles and pretend their areas are equal.
A disk of radius r has area πr². For r=7 cm and π≈22/7, this gives 154 cm². Count fully enclosed grid squares for a lower bound and every square with interior overlap for an upper bound. Boundary-cut squares need partial areas for a closer estimate. For the radius 4 unit-grid model, the full/cut counts are supplied beside the diagram; the exact area 16π lies between the bounds. Counting only square centres is an estimate, not an exact area proof.
If A=81πcm², then r²=81 and the positive radius is 9 cm. If radius doubles, circumference doubles but area becomes 4 times as large because(2r)²=4r². A unit-radius disk has area π square units. Convert lengths before squaring:1 m²=10,000 cm². In written work, name whether a number is length, area, ratio or a coefficient of π.
For diameter 10 cm, radius 5 cm, C≈31.4 cm and A≈78.5 cm² whenπ≈3.14. The two results count different things and carry different units.
The worked31.4 cm answer follows diameter 10 cm. Radius10 cm would require a different circumference.
Curved sector arcs remain visible in our models; increasing their number illustrates the limiting argument.
Example readings are idealised arithmetic under π≈3.14, not collected observations.
| Radius (cm) | Diameter (cm) | Circumference (cm) | C/d |
|---|---|---|---|
| 2 | 4 | 12.56 | 3.14 |
| 3 | 6 | 18.84 | 3.14 |
| 5 | 10 | 31.40 | 3.14 |
Actual ruler/string ratios vary with measurement error. π is C/d; C/r=2π. The source’s22/7=3.142 entry truncates the number:22/7 rounds to 3.143 at 3 decimal places, whereas π rounds to 3.142.
Diameter10 cm means radius 5 cm. Use π≈3.14.
C=πd=3.14×10=31.4 cm A=πr²=3.14×5²=78.5 cm²
Source Example1 says diameter 10 initially but radius 10 in its conclusion. The31.4 result uses the diameter; a radius 10 would give62.8 cm.
Move through four stages manually. The pieces stay genuine sectors with equal total area.
No autoplay. Static stages, manual replay and reduced-motion support remain available.
Radius4 unit-grid model; squares touching only at a point have no positive overlap.
| Count or calculation | Meaning | Square units |
|---|---|---|
| 32 full cells | Certain covered area: lower bound | 32 |
| 28 cut cells | Count each as at most one | 28 |
| 32+28 | Upper bound | 60 |
| 16π | Exact ideal disk area | ≈50.27 |
A bound guarantees the area lies between32 and 60. Counting half of every cut cell gives an estimate46, which need not equal the exact area. The separate practice count40 full/20 cut is invented and clearly labelled.
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.
3.14
31.4/10=3.14; this measured ratio approximatesπ.
6.28
C/r=2π, so 2×3.14=6.28; using radius instead of diameter doubles the ratio.
37.68
C=3.14×12=37.68.
44
2×22/7×7=44 under the stated approximation.
14
88/(2×22/7)=14.
20
62.8/3.14=20.
11
One turn220 cm; five turns1100 cm=11 m.
1760
Circumference44 m, two loops88 m, cost88×20=1760, no extra allowance.
22.5
360/16=22.5.
154
22/7×49=154.
78.5
Radius5 cm;3.14×25=78.5.
9
r²=81; choose the positive root9.
4
Area is proportional to radius squared:2²=4.
1
A=π×1²=1π square units.
60
Count every cut square as at most 1:40+20=60. This is a stated count, not the separate radius 4 diagram.
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It is circumference divided by diameter, so it measures how many diameter-lengths fit one complete boundary. Both lengths use one unit, which cancels. The ideal ratio is constant;3.14 and 22/7 are approximations.
State numerato r, denominator and meaning rather than only quoting a decimal.
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Record circle label, radius, diameter, circumference and C/d with units on the lengths. Checkd≈2r. Repeat careful string/ruler measurements; ratios should approachπ within practical error. Do not label C/r asπ.
Use your own measurements and discuss the limits of the instruments.
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22/7=3.142857… exactly as a recurring decimal.3.142 cuts off digits; rounding22/7 to 3 decimal places gives 3.143.π itself differs from 22/7 and rounds to 3.142 at 3 decimal places.
Keep the rational approximation, the truncation and π itself separate.
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Its input diameter 10 cm gives 31.4 cm whenπ≈3.14. The conclusion mistakenly says radius 10 cm, which would instead give62.8 cm. Keep the stated diameter for the worked arithmetic and show the wording error.
Do not silently change the input dimension.
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d=88÷(22/7)=28 cm; r=14 cm. Reinsert2×22/7×14=88. Calling28 the radius would double the implied circumference.
Show both inverses and the check.
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Circumference220 cm is the distance per full revolution. Five turns give1100 cm=11 m. Slipping changes the travelled-distance relationship, so no-slip is a required assumption.
Convert centimetres to metres after calculating consistently.
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One loop44 m; two loops88 m; cost 1760 BDT. This excludes knots, waste and overlaps unless an allowance is supplied.
Name what each intermediate length counts.
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The outer edges remain circular arcs, so there are scallops and slightly slanted end edges. More, narrower sectors reduce those features. Area is preserved by moving the pieces; the rectangular dimensionsπr andr describe the limiting shape.
Do not claim32 pieces make every boundary perfectly straight.
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Alternating narrow sectors creates a strip whose limiting long side is half the circumference,πr, and limiting height isr. The pieces preserve disk area; the limiting rectangle gives πr×r=πr².
Identify why half the boundary, not the whole circumference, forms a long side.
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r=10/2=5 cm. A=3.14×5²=78.5 cm². Radius must be squared and the result measures covered area, not boundary length.
Show the diameter-to-radius conversion.
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Count fully inside squares for the lower bound. Add every boundary-cut square for an upper bound, ignoring squares that touch only at an edge or point. Partial boundary fractions can improve the estimate; the exact curve area need not be an integer.
State a counting rule before comparing an estimate with πr².
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The area is at least40 and at most60 square units. Treating every cut square as one-half gives an estimate50, but their actual fractions need not all be halves. This count is not asserted for a particular radius unless specified.
Explain what a bound guarantees and what an estimate does not.
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C=2πr becomes3C. A=πr² becomes9A. One formula has one factor of length and the other has two.
Use formulas or a scaled diagram, not a guess that both triple.
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Divide by π to getr²=81; positive radius 9 cm. Theπ cancels exactly, so an approximate decimal for π is unnecessary.
Do not introduce rounding when it is not needed.
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Use π≈22/7. The calculation is exact arithmetic under the stated approximation, but the ideal circumference uses irrational π and a measured radius also has instrument error. State the convention or leave the result as a multiple of π.
Separate mathematical exactness, chosen approximation and physical measurement.
Measure three circles and explain an area rearrangement.
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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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