Class 8 · English Version · Mathematics
2026 Class 8 Mathematics · English Version · Chapter 8 · Chapter companion part
Derive quadrilateral areas from triangles and calculate cube/cuboid surface area from labelled, matching faces.
Class 8 English Version: part C of the chapter 8 companion. Eight teaching sections, 30 written tasks and a notebook challenge. Selected companion coverage, not complete textbook coverage; use all five parts. Independent proofs, compass work, drawings and nets require guardian or teacher review.
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For convex ABCD, diagonal AC of length d separates two triangles. If B, D have perpendicular distances h₁, h₂ to line AC on opposite sides, area=dh₁/2+dh₂/2=d(h₁+h₂)/2. Use perpendicular distance, not a sloping side. The formula assumes the internal diagonal partitions the region without overlap.
For parallel bases a, b and perpendicular separation h, a trapezium has area (a+b)h/2. One derivation draws a diagonal: the two triangles have bases a, b and the same height h. Another places two congruent trapezia together as a parallelogram of base a+b and height h, then halves its area. Average of the two lengths means (a+b)/2, not another halving of an already averaged sum.
Rhombus diagonals p, q bisect each other at right angles. Taking p as a common triangle base gives two heights q/2, so total area=pq/2. Equivalently four right triangles each have legs p/2, q/2. For a general quadrilateral with nonperpendicular diagonals, pq/2 need not be its area. A parallelogram area is base×perpendicular height, not usually the product of adjacent side lengths.
A rectangular solid, or cuboid, has 6 rectangular faces,12 edges and8 vertices. Opposite faces have equal dimensions. A cube is the special cuboid whose edges all have equal length and whose faces are squares. A drawn solid is a perspective model: parallel depth edges appear slanted, and their page lengths are not measurements of the real solid.
An open net for a cuboid can use four side rectangles of widths l, b, l, b in a strip of height h, with two l-by-b end faces attached above and below one l-wide strip face. Fold the shared edges; verify all six faces close with no overlaps or missing face. Not every arrangement of six squares makes a cube net. Tabs for glue are additional material and are not part of the ideal surface area.
A cuboid with length l, breadth b, height h has two lb faces, two bh faces and two lh faces. Total surface area=2(lb+bh+lh), in square units. A cube of edge s has six s² faces, total6s². Volume lbh is a different quantity measured in cubic units. Label dimensions before inserting numbers to avoid interchanging a face area and a length.
An open-top l-by-b-by-h box has base lb and four walls2lh+2bh, so material area=lb+2lh+2bh, ignoring thickness, tabs and waste. Painting only four walls uses2h(l+b). A closed-box formula is wrong if the question explicitly excludes the top. State whether internal surfaces or both sides of a sheet are counted.
Convert all lengths to one unit before squaring or multiplying. Since1 m=100 cm,1m²=10,000cm². A 3-by-4 rectangle with a right triangle attached has area from the two regions; its outer perimeter excludes the shared boundary. If a diagonal gives a 3–4–5 right triangle, verify 3²+4²=5² before using that length; a later chapter develops the theorem fully.
An8×5×3 cuboid has paired face areas40,15,24 cm². Its closed surface area is2(40+15+24)=158 cm². Removing a 40 cm² top gives118 cm² for an ideal open-top box.
Plane area models use their stated dimensions. A perspective cuboid is a schematic view; use its dimension labels and matching net.
Ideal surface area excludes thickness, glue tabs and waste unless a task supplies them.
Let d be an internal diagonal and h₁, h₂ perpendicular heights on its opposite sides.
Area = dh₁/2 + dh₂/2 = d(h₁+h₂)/2 = 10×(3+4)/2 = 35 cm²
The sloping boundary lengths are not substitutes for these heights.
Use l=8, b=5, h=3 centimetres.
| Surface | Area expression | Area in cm² |
|---|---|---|
| Two caps | 2×8×5 | 80 |
| Two long walls | 2×8×3 | 48 |
| Two short walls | 2×5×3 | 30 |
| Closed total | 80+48+30 | 158 |
| Open-top total | 40+48+30 | 118 |
| Walls only | 48+30 | 78 |
Ideal material area excludes thickness, glue tabs and cutting waste. A face area is not volume.
Follow the matching dimensions before folding. Use manual stages or replay; nothing starts automatically.
Static frames remain readable. No autoplay; reduced motion disables timed playback, and leaving the page pauses it. This sequence explains face correspondence rather than simulating a physical paper fold.
Match each letter to its explanation. Read the quantities, units and relationships, then check the counts or equations. These are original teaching models.
Original labelled mathematical model, not copied source artwork. Verify stated conditions rather than judging shape from appearance.
Match each letter to its explanation. Read the quantities, units and relationships, then check the counts or equations. These are original teaching models.
Original labelled mathematical model, not copied source artwork. Verify stated conditions rather than judging shape from appearance.
Match each letter to its explanation. Read the quantities, units and relationships, then check the counts or equations. These are original teaching models.
Original labelled mathematical model, not copied source artwork. Verify stated conditions rather than judging shape from appearance.
Match each letter to its explanation. Read the quantities, units and relationships, then check the counts or equations. These are original teaching models.
Original labelled mathematical model, not copied source artwork. Verify stated conditions rather than judging shape from appearance.
Match each letter to its explanation. Read the quantities, units and relationships, then check the counts or equations. These are original teaching models.
Original labelled mathematical model, not copied source artwork. Verify stated conditions rather than judging shape from appearance.
Show your working on paper. Check your answer, then choose Next.
35
10×(3+4)/2=35.
50
(12+8)×5/2=50.
12
Average base=area/height=60/5.
5
2×50/(12+8)=5.
48
12×8/2=48.
36
9×4=36; the sloping side is not used.
6
There are three opposite-face pairs.
12
Four edges in each of three dimension directions.
6
Every surface appears exactly once.
158
2×(40+15+24)=158.
121.5
6×4.5²=121.5.
118
40+2×24+2×15=118; no top.
78
2×3×(8+5)=78.
600
Each square metre is10,000 square centimetres.
18
12+4×3/2=18; the regions do not overlap.
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The diagonal divides the convex region into two triangles. Their areas are dh₁/2 and dh₂/2, whose sum is d(h₁+h₂)/2.
Draw both perpendicular heights and identify the internal diagonal.
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A diagonal gives two triangles with bases a, b and common height h; their sum is (a+b)h/2. Two congruent copies make a parallelogram of base a+b and height h, so one is half its area.
Give two labelled constructions, not two restatements of a formula.
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h is the shortest perpendicular separation of the parallel base lines. A sloping leg usually exceeds it; draw a right-angle height before calculating.
Identify the parallel sides and the foot of the perpendicular.
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Diagonals p, q meet perpendicularly and bisect. Two triangles with base p each have height q/2; total is2×(p×q/2)/2=pq/2.
Explain why q/2 is a perpendicular height.
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The formula is base times perpendicular height. A sloping adjacent side is a hypotenuse of a height triangle and is longer than its altitude unless perpendicular. A rectangle is the special case where the adjacent side is a height.
Use a labelled counterexample or the base9, height4 model.
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A cuboid has6 faces,12 edges,8 vertices. In a cube all12 edges have one length and all6 faces are equal squares; it still has the same counts.
Do not count a hidden edge as an absent edge.
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Use a strip of height3 with widths8,5,8,5. Attach an 8-by5 face above and another below an 8-wide face. There are two8×5, two8×3, two5×3 faces; folding the strip and two caps closes the box.
Check shared edge lengths and six distinct final faces.
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Folding around successive edges wraps a strip and makes faces overlap or leaves the top and bottom unclosed. Six correct face areas alone do not ensure the right adjacency and orientation.
Use a paper fold or distinct face directions to demonstrate failure.
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Add opposite pairs:lb+lb, bh+bh, lh+lh. Their total is2(lb+bh+lh). Each product is an area and all dimensions must use one length unit.
Show three pairs and square units.
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Each face is4.5²=20.25 cm²; six give121.5 cm². Volume is4.5³=91.125 cm³ and measures internal space instead.
Use distinct meanings and units for the two quantities.
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Closed area158 cm². Remove one8×5 top to get118 cm². Remove the base too to leave78 cm² of walls. These ideal areas ignore thickness, tabs and cutting waste.
Name exactly which faces are included.
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A1 m-by1 m square is100 cm by100 cm, containing100×100=10,000 unit squares. Both length factors must be converted.
Use a square model or multiplication of two converted dimensions.
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Area12+6=18 cm². The rectangle contributes the other three sides3+4+3=10; triangle contributes3+5=8, giving perimeter18 cm. The shared4 cm join is internal.
Draw the attachment and exclude its shared edge from perimeter.
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Convert0.08 m to8 cm. Then2(8×5+5×3+8×3)=158cm². Mixing0.08 with5 and3 without conversion would multiply incompatible numerical scales.
Write the conversion before substitution.
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Label six rectangles with the three matching opposite pairs. Every exposed face is counted once before folding, so their summed areas equal the ideal closed surface area. Exclude glue tabs or state their additional area separately.
Teacher review checks closure, no overlap and dimensions.
Make a labelled quadrilateral area diagram and a folding cuboid net.
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 8, printed pages 133–151. It does not replace the complete chapter or its practical teaching.
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