Class 8 · English Version · Mathematics
2026 Class 8 Mathematics · English Version · Chapter 3 · Chapter companion part
Choose area or volume for compound shapes, count overlaps once, calculate costs and explain all missing assumptions.
Class 8 English Version: part D of the Measurement companion. Eight sections, 30 written tasks and a notebook challenge. Selected companion coverage, not complete textbook coverage. Use all four parts for the mapped chapter methods; these fresh tasks do not reproduce every printed question. Drawings and explanations need guardian or teacher review.
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For a garden l by b with an outside uniform path of width w, outer dimensions are l + 2w and b + 2w. Path area = (l + 2w)(b + 2w) − lb. For a 20 m by 12 m garden and 2 m path, subtract 240 from 24 × 16 = 384 to get 144 m². A path on all sides adds width twice to each dimension.
For a uniform path inside the boundary, inner dimensions are l − 2w and b − 2w, provided each is non-negative. Area = lb − (l − 2w)(b − 2w). For a 30 m by 20 m plot and 2 m path, area = 600 − 26 × 16 = 184 m². State whether the path is inside or outside before using a formula.
Two perpendicular full-length roads, each width w, cross a rectangle l by b. Adding lw + bw counts the central w² twice, so subtract it once: A = lw + bw − w². For 40 m by 24 m and w = 2 m, area = 80 + 48 − 4 = 124 m². The same result follows by adding one full strip and two shortened arms.
If l and b are internal room dimensions, uniform wall thickness t extends outward; outer dimensions are l + 2t and b + 2t. Four-wall volume without openings = [(l + 2t)(b + 2t) − lb]h. A 6-by-4-m room, height 3 m, walls 0.2 m thick has wall volume (6.4 × 4.4 − 24) × 3 = 12.48 m³. Corners are included once. Paint area and masonry volume are different quantities.
For a rectangular open tank with internal l, b, h, inside lining area is lb + 2lh + 2bh: base plus four walls. A square tank with base side 4 m and depth 2 m needs 16 + 32 = 48 m². If volume and depth are known, first find base area V/h and then side length. State whether a top, doors, windows or openings are included.
Carpet and grass use area × Tk/m²; a fence uses perimeter × Tk/m. If length is 1.5 times breadth and area is 216 m², breadth is 12 m and length 18 m. The 60 m perimeter at Tk 5/m costs Tk 300. If reducing breadth by 3 m saves Tk 600 at a uniform area price, compare the change in area and cost before finding the original breadth.
Total door/window area divided by plank area is an ideal material-area estimate. Actual whole planks or mats must also fit, with cuts, joins and waste specified. For a 6-by-4-m floor and 3-by-2-m mats, four uncut mats tile it exactly in a 2-by-2 arrangement. For other shapes, rounding the area ratio up gives only a lower bound. Do not claim a guaranteed fit from the ratio alone. The source’s plank example needs this qualification.
For constant removal rate q in m³/s and initial water V in m³, time is V/q seconds. A 20-by-12-m pond with 2 m water depth has 480 m³; at 0.2 m³/s it takes 2400 s = 40 min, neglecting inflow and rate changes. The source creative question gives a removed volume without a time unit, so a unique time needs clarification. Correction to our earlier review: Exercise 22 DOES give a 1 m margin. Its 15 by 10 m floor leaves a 13 by 8 m region. Each stone has area 50 cm², not side 50 cm. The ideal area count is 20800; an actual whole-stone layout still needs shape and cutting conditions.
For two crossing roads of unequal widths u along length l and v along breadth b, count lu + bv − uv. With l = 30, b = 18, u = 2 and v = 3, area = 60 + 54 − 6 = 108 m². At Tk 12/m² cost is Tk 1296. This extension states which width belongs to which road.
The paths are uniform and rectangular; walls have a stated outward thickness. Openings, slopes, inflow and waste must be specified before including them.
Numeric tasks check the stated final value. Drawings, cutting plans and explanations require a guardian or teacher; a matching number alone is not an exam mark.
Our previous review incorrectly said Exercise 3(22) omitted the margin. The printed source gives 1 m on all sides of a 15 by 10 m floor. Subtract twice the margin from each dimension: 13 × 8 = 104 m² is covered. A stone area of 50 cm² is 0.005 m², giving 104/0.005 = 20800 area equivalents. This is not a 50 cm side length. Shape, cutting and a fit plan are still needed to guarantee an actual whole-stone count.
Exercise 3(18) has a 26 by 20 m floor and 4 by 2.5 m mats. The diagram supplies a complete uncut layout: four columns of eight mats give 32, and four rotated columns of five give 20. Their widths total 4 × 4 + 4 × 2.5 = 26 m, all reaching 20 m high. Thus 52 mats really fit, not just by an area estimate. At the stated Tk 27.50 each, cost is 52 × 27.50 = Tk 1430. The printed Tk 10400 is wrong.
Invented method guide, not measured data or construction prices.
| Task | Quantity | Rate or final unit |
|---|---|---|
| Fence boundary | Perimeter | metres × Tk/m |
| Grass, carpet or line a tank | Area | m² × Tk/m² |
| Masonry shell or tank contents | Volume | m³ |
| Air or material mass | Volume × stated density | m³ × kg/m³ or cm³ × g/cm³ |
| Pump water | Volume / removal rate | m³ / (m³/s) = seconds |
A 60 m boundary at Tk 5/m costs Tk 300. Using its 216 m² area instead would answer a different problem.
Use the stated inside or outside position and count overlaps once.
Outside path A = (l + 2w)(b + 2w) − lb Inside path A = lb − (l − 2w)(b − 2w) Crossed strips A = lu + bv − uv
The inside dimensions must be non-negative. A road on only one side is not the all-sides border formula.
State internal dimensions, uniform outward thickness and openings.
Four-wall V = [(l + 2t)(b + 2t) − lb]h Open-tank lining S = lb + 2lh + 2bh Opening volume = opening face area × wall thickness
The shell counts corners once. Subtract stated openings for masonry; do not subtract an area directly from a volume.
Cancellation works only when the named conditions stay unchanged.
Uniform area cost C = plb Saving from narrowing by d: ΔC = pld Breadth b = dC/ΔC Emptying time T = V/q
A removed volume without a time unit is not a removal rate. Real pumps may change rate; this is a specified constant-rate classroom model.
Exercise 3(18): use the source dimensions and price, with rotation allowed and no gaps.
Floor area = 26 × 20 = 520 m² One mat area = 4 × 2.5 = 10 m² Count = 32 + 20 = 52 Cost = 52 × 27.50 = Tk 1430
The printed count 52 is correct for this layout. The printed Tk 10400 cost is wrong. Both the count and the price multiplication must be checked.
Exercise 3(22B) explicitly gives a 1 m margin. This corrects our earlier missing-margin claim.
Covered area = (15 − 2 × 1)(10 − 2 × 1) = 104 m² Stone area = 50/10000 = 0.005 m² Ideal area count = 104/0.005 = 20800
Area equivalents are an ideal material budget. The source does not specify the stone shape, dimensions or cutting/fit, so do not promise a whole-piece arrangement. The floor margin does not reduce the room’s full air volume.
Match each letter to its explanation. Read the quantities, units and relationships, then check the counts or equations. These are original teaching models.
Original rectangular plans. Each panel has its own consistent scale, stated in drawing metadata; the two panels are not at one shared scale. Width labels apply to all four sides.
Match each letter to its explanation. Read the quantities, units and relationships, then check the counts or equations. These are original teaching models.
Original plan with exact rectangular proportions and uniform widths. The highlighted overlap is part of both roads, not an extra third road.
Match each letter to its explanation. Read the quantities, units and relationships, then check the counts or equations. These are original teaching models.
Original proportionate floor plan. Height is given as a label rather than drawn in this plan view. Uniform outward thickness and no openings are assumptions, not building specifications.
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 rectangular layout at one scale. Mats can rotate, touch edge to edge and have no cutting allowance or gap. Colour distinguishes orientation; the column counts and labels also explain it.
Match each letter to its explanation. Read the quantities, units and relationships, then check the counts or equations. These are original teaching models.
Original proportionate floor plan, not a stone-placement diagram. This corrects our earlier claim that the margin was missing; it does not invent the unprovided stone shape.
Show your working on paper. Check your answer, then choose Next.
144
24 × 16 − 20 × 12 = 144 m².
184
30 × 20 − 26 × 16 = 184 m².
124
40 × 2 + 24 × 2 − 2² = 124 m².
4
The crossing is 2 × 2 = 4 m².
12.48
[(6.4 × 4.4) − (6 × 4)] × 3 = 12.48 m³.
48
4² + 4 × 4 × 2 = 48 m²; no lid.
720
48 × 15 = 720.
12
1.5b² = 216, so positive b = 12 m.
60
2(18 + 12) = 60 m.
300
Perimeter 60 m × Tk 5/m = Tk 300.
4
Two along length and two along breadth give four mats.
340
4 × 85 = 340.
2400
20 × 12 × 2 / 0.2 = 2400 s.
40
2400/60 = 40 min.
12
Cost ratio 2400/600 = breadth/3, so breadth = 12 m.
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A path of width w occurs on both ends of each dimension. Outer area is (l + 2w)(b + 2w); subtract lb for the inner garden. Adding w only once omits one side in each direction.
Label inner dimensions, all four borders and outer dimensions.
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Inner dimensions 23 by 16 give area 368 m². Whole plot area is 450 m², so path area 82 m² and cost Tk 656.
Use inside subtraction and an area rate.
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Use 80 + 48 − 4 = 124 m², or one 80 m² strip plus two other arms of total length 22 m and width 2 m, adding 44 m².
Count the overlap once under both methods.
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Outer side is 20 m. Path area is 20² − 18² = 400 − 324 = 76 m².
Do not use perimeter × width without accounting for corners.
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Outer base is 6.4 × 4.4 = 28.16 m². Inner base is 24 m². Shell base area 4.16 m² times 3 m gives 12.48 m³.
Treat dimensions as internal and count corner material once.
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Door opening volume = 2 × 0.2 = 0.4 m³. Subtract from 12.48 to get 12.08 m³, under the uniform-through-wall model.
An opening’s area cannot be subtracted directly from a volume.
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V = 32 m³, base area = 16 m² and side = 4 m. Lining = 16 + 4 × 4 × 2 = 48 m², costing Tk 720.
Exclude a lid and use full internal depth as specified.
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Area is 216 m². From 1.5b² = 216 obtain b = 12 m and l = 18 m. Perimeter 60 m costs Tk 300.
Keep area and boundary rates separate and verify both dimensions.
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With unchanged length l and price p per m², full cost plb = 2400 and saving pl × 3 = 600. Divide to cancel pl: b/3 = 4, so b = 12 m.
State unchanged length and uniform rate; length alone is not determined.
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Shapes can leave waste or fail to tile without cuts. For example a 3-by-3 square cannot be tiled by two uncut 2-by-3 rectangles even though ceil(9/6) = 2. A cutting/joining plan or extra material assumption is needed.
Give a geometry reason, not only a rounded ratio.
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Door area 6 m² plus window area 3 m² = 9 m², or three 3 m² plank-area equivalents. This is an ideal area budget, not a guaranteed whole-plank cutting plan.
Preserve the qualification even when the area ratio is an integer.
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If 20 by 12 are the outside plot dimensions, inner water is 18 by 10 m. Bank area = 240 − 180 = 60 m²; volume = 180 × 2 = 360 m³; time = 360/0.2 = 1800 s = 30 min.
Explicitly state that the supplied outside dimensions belong to the plot; otherwise the result changes.
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A volume per operation or a removal rate with a time unit is needed, such as 0.1 m³ per second. Volume alone does not determine time.
Do not silently insert per second into the original question.
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If the room is l by b and margin w is left all round, usable area is (l − 2w)(b − 2w). Different w gives different usable area and layout, so stone size alone is insufficient.
Require the margin, shape and a layout/cutting convention before a finite count.
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Identify inside/outside dimensions and openings; convert units; choose perimeter, area, volume or rate-time relation; show every substitution; state cutting, density, flow and rounding assumptions; check units and plausibility.
A reviewer should be able to reproduce the result and identify any missing condition.
Create a short mixed-problem paper containing a path, a wall shell, a covering plan and a pump calculation.
Keep your work on paper or in a local file. This page does not upload it or automatically grade the project.
This original Kidemy practice follows the topic order of 2026 Class 8 Mathematics · English Version, chapter 3, printed pages 29–47. It does not replace the complete chapter or its practical teaching.
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