13.2 Shapes, Perimeter, Area and Volume
Key Takeaways
- MAT shape items expect remembered formulae for rectangle, trapezium, circle, cuboid, pyramid, cone, and sphere; QL usually prints a cuboid recipe.
- Trapezium area is (1/2)(a + b)h with a and b the parallel sides; treating the figure as a rectangle or a triangle is a designed trap.
- Cone and pyramid volumes use one-third of base area times perpendicular height, not the slant height or a slanted edge.
- Composite perimeters and surfaces omit internal edges and faces; similar figures scale lengths by k, areas by k², and volumes by k³.
- MAT mixes geometry with algebra: an unknown length in a formula becomes an equation you solve by hand, still without a calculator.
13.2 Shapes, Perimeter, Area and Volume
Quick Answer: MAT asks you to use properties of 2D and 3D shapes and to compute perimeter, area, and volume, including circles, rectangles, trapezia, spheres, cones, pyramids, composite solids, and scale factors. Formulae are not supplied the way Quantitative Literacy supplies a cuboid recipe. You also mix geometry with algebra: an unknown length in a trapezium, or a similar-solid ratio. The paper is calculator-free.
MAT geometry is not QL measurement
QL Shape, dimension and space is campus measurement: a noticeboard perimeter, a crate volume, a map scale, with the formula usually printed in the item. MAT Test Content asks you to apply properties of 2D and 3D shapes and to work perimeter, area, and volume as mathematical objects. Independent OpenExamPrep examples below are original. They are not official NBT questions, and they are not QL cuboid packing.
A QL crate is 40 cm by 25 cm by 20 cm and the volume is 20 000 cm³. A MAT item is more likely to say a trapezium has parallel sides x and x + 6 and height x, give the area 54, and ask for x — geometry and a quadratic in one stem. Treat the two papers as different jobs, even when both mention a box.
Properties you actually use
Rectangle. Opposite sides equal and parallel; four right angles; diagonals equal and bisect each other. Perimeter 2(l + w). Area l w.
Parallelogram. Opposite sides equal and parallel; area base × perpendicular height, not a slanted edge.
Rhombus. All sides equal; diagonals are perpendicular bisectors. Area (1/2) d₁ d₂.
Trapezium (South African school usage: exactly one pair of parallel sides). Area (1/2)(a + b) h, where a and b are the parallel sides and h is the perpendicular distance between them.
Circle. Radius r, diameter 2r. Circumference 2π r (or π d). Area π r². A tangent is perpendicular to the radius at the point of contact — that property is enough for a right triangle in this chapter; full circle theorems sit in the next chapter.
Cuboid. Volume l w h. Surface area 2(lw + lh + wh). The three face-diagonals are √(l² + w²), √(l² + h²), √(w² + h²); the space diagonal is √(l² + w² + h²). Those lengths are properties, not QL packing counts.
Pyramid. Volume (1/3) × (base area) × (perpendicular height). The slant height and the slant edge are not the perpendicular height.
Cone. Volume (1/3) π r² h. Slant height l = √(r² + h²). Curved surface area π r l. Total surface area π r l + π r².
Sphere. Volume (4/3) π r³. Surface area 4π r².
Learn the formulae. MAT does not promise to print them.
Rectangle, trapezium, circle — exact values
Rectangle. A courtyard is 15 m by 8 m. Perimeter 2(15 + 8) = 46 m. Area 120 m². A diagonal is √(225 + 64) = √289 = 17 m. That 8-15-17 triple is the same Pythagorean family as 3-4-5. Reporting 17 as a perimeter mixes the diagonal with the fence. Reporting 120 m as a length mixes area with perimeter.
Trapezium. Parallel sides 9 and 15, perpendicular height 8.
Area = (1/2)(9 + 15) × 8 = 12 × 8 = 96.
Using 9 × 15 = 135 treats it as a rectangle that it is not. Using (9 + 15) × 8 = 192 forgets the 1/2. Using (1/2)(9)(15) = 67.5 is a triangle formula with the wrong pair. Using height 6 by accident yields 72, another common distractor if a non-parallel side was copied as h.
Circle. Radius 6. Circumference 12π. Area 36π. If a stem gives diameter 6, the radius is 3, area 9π — four times smaller than 36π, because area grows with r². Name radius versus diameter before you substitute. The same error QL trains is still lethal on MAT, except now the formula is not printed in the stem.
Algebra inside a trapezium
A trapezium has parallel sides x and x + 6, and the height is x. The area is 54.
(1/2)(x + x + 6) x = 54
(1/2)(2x + 6) x = 54
(x + 3) x = 54
x² + 3x − 54 = 0
(x + 9)(x − 6) = 0
x = 6 (length is positive).
The parallel sides are 6 and 12, height 6. Check: (1/2)(18)(6) = 54. An option x = 9 is the discarded root with the sign flipped. An option x = 3 treats (1/2)(2x + 6) as if the 54 were 27. An option x = 54 / 6 = 9 also appears if someone divides the area by one parallel side and ignores the trapezium formula. This is the MAT standard: a shape property plus an equation, not a three-number product.
A second algebraic cuboid, still not QL packing: square ends of side x and length x + 1, volume 48.
V = x²(x + 1) = 48. Testing small integers: x = 3 gives 9 × 4 = 36, too small; x = 4 gives 16 × 5 = 80, too big. So x is not a whole number here — stop and rearrange x³ + x² − 48 = 0 only if the options require it, or notice that a different stem would choose numbers that factor. The point is the setup: volume is not l + w + h, and the unknown sits inside a product. Independent OpenExamPrep examples keep at least one length as x so the item cannot be answered by multiplying three given edges.
Sphere, cone, pyramid — original numbers
Sphere, r = 3. Volume = (4/3)π(27) = 36π. Surface area = 4π(9) = 36π. The two 36π results are not interchangeable: one is cubic units, one is square units. A stem that asks for volume and offers 36π still requires you to know you were not asked for surface area — the number coinciding is a designed trap. Using (4/3)π r² = 12π forgets to cube the radius.
Cone, r = 5, h = 12. Then l = √(25 + 144) = √169 = 13.
Volume = (1/3)π(25)(12) = 100π.
Curved surface area = π(5)(13) = 65π.
Total surface area = 65π + 25π = 90π.
Forgetting the 1/3 gives 300π. Copying 65π into a volume option uses the wrong measure. Copying 90π uses total surface area. The 5-12-13 triple is the right triangle that produces the slant height; MAT will not caption slant height. Using 13 as the perpendicular height in the volume formula would give (1/3)π(25)(13), which is not 100π.
Square pyramid, base side 6, perpendicular height 4.
Base area 36. Volume = (1/3)(36)(4) = 48.
If someone uses the slant edge √34 from Section 13.1 as the height, the volume becomes (1/3)(36)√34, which is not 48. Perpendicular height to the base plane is the only height in V = (1/3)Ah. Forgetting the 1/3 gives 144, the prism with the same base and height.
Composite figures
A running-track enclosure is a rectangle 10 by 6 with a semicircle of diameter 6 attached to one width, forming a capsule end.
Area = 10 × 6 + (1/2)π(3)² = 60 + (9π)/2.
Perimeter (outer path): two lengths of 10, the far width 6, and the semicircle arc π r = 3π, so 26 + 3π. The diameter of the semicircle is internal to the figure, so it is not part of the perimeter. Adding the diameter produces 32 + 3π. Using a full-circle circumference 6π instead of the semicircle arc doubles the curved part. Using area 60 + 9π treats the end as a full disk.
A solid is a cylinder of radius 3 and length 8 with a hemisphere of radius 3 on each end (a full sphere plus the cylinder).
Volume = π(9)(8) + (4/3)π(27) = 72π + 36π = 108π.
External surface: cylinder curved area 2π r h = 2π(3)(8) = 48π, plus the two hemisphere outer surfaces which together are one full sphere 4π r² = 36π, total 84π. You do not include the two circular joins; they are internal. Adding 2 × π r² for those joins inflates the surface. Composite MAT items are won or lost on what is internal.
Scale factor
If two similar figures have corresponding lengths in the ratio k, then
- lengths scale by k
- areas (and surface areas) scale by k²
- volumes scale by k³
| Linear k | Area k² | Volume k³ |
|---|---|---|
| 1 | 1 | 1 |
| 2 | 4 | 8 |
| 3 | 9 | 27 |
| 2/3 | 4/9 | 8/27 |
Worked example. Two similar cones. Linear scale small : large = 2 : 3. The smaller volume is 16π. The larger volume is 16π × (27/8) = 54π.
Using 16π × (3/2) = 24π applies the length ratio to volume. Using 16π × (9/4) = 36π applies the area ratio to volume. Using 16π × 2 = 32π doubles as if k were 2 with no ratio to 3. Using 16π × 8/27 shrinks instead of enlarges.
A plan uses scale 1 : 200. A courtyard is 4 cm by 3 cm on the plan. Actual sides 800 cm and 600 cm, i.e. 8 m by 6 m, area 48 m². Area scale factor 200² = 40 000; 12 cm² × 40 000 = 480 000 cm² = 48 m². Linear scale is not area scale. That is the same k versus k² distinction as the cones, written as a map. QL often gives 1 cm : 20 m and a single product; MAT expects you to square 200 yourself.
Method
- Name the shape and write the formula from memory.
- Separate perimeter (length), area (square units), and volume (cubic units).
- For a cone or pyramid, mark perpendicular height versus slant.
- For a composite, subtract or omit internal faces.
- For similar figures, raise k to the dimension of the quantity.
- If a length is x, form the equation and discard negative roots.
Traps
- Multiplying three cuboid edges as if that were MAT’s whole job, the QL pattern
- Using a slant edge as perpendicular height
- Leaving an internal diameter in a composite perimeter
- Scaling volume by k or by k²
- Using diameter in a formula that needs radius
- Using a calculator
Independent OpenExamPrep MAT shape work is that mix of a remembered formula, a composite decision, and an algebraic unknown — not a QL crate with the product already framed.
A trapezium has parallel sides 9 and 15 and perpendicular height 8. Its area is
Two similar cones have linear scale small to large equal to 2 : 3. If the smaller volume is 16π, the larger volume is
A cone has radius 5 and perpendicular height 12. Its volume is