SACE Essential Mathematics Mastery Pack

1. Recognising plane shapes and classifying solids

Assessment status: school-assessed only. Begin by separating two-dimensional shapes from three-dimensional solids. A plane shape has length and width but no thickness: triangles, quadrilaterals, circles and composite polygons are examples. A solid occupies space and is described by its flat or curved surfaces. A prism has two congruent, parallel end faces joined by rectangular faces; it is named from its end face. A pyramid has one base and triangular faces meeting at an apex. Cylinders, cones and spheres have curved surfaces, so the word face is usually reserved for their flat surfaces.

Worked example 1. A camping tent has a triangular cross-section and a constant length. Model it as a triangular prism. Each triangular end contributes one face and the three long rectangles contribute three more, so F = 5. The two triangles contribute 3 + 3 edges and three joining edges run along the tent, so E = 9. There are 3 vertices at each end, so V = 6. Check Euler’s relation for a convex polyhedron: V − E + F = 6 − 9 + 5 = 2.

Worked example 2. A square-based display pyramid has one square base and four triangular side faces: F = 5. The square supplies four base edges and four sloping edges meet at the apex: E = 8. Four base corners plus the apex give V = 5. Again, V − E + F = 5 − 8 + 5 = 2. By contrast, a cube has 6 square faces, 12 edges and 8 vertices, and 8 − 12 + 6 = 2. Naming the defining faces prevents confusing a triangular prism with a triangular pyramid.

2. Faces, edges, vertices and cross-sections

Assessment status: school-assessed only. Count systematically rather than relying on what is visible in a perspective sketch. An edge is where two flat faces meet, and a vertex is a corner where edges meet. Hidden edges still count. For any convex polyhedron, Euler’s relation V − E + F = 2 provides a strong check. A cross-section is the plane shape exposed by a straight cut through a solid; its shape depends on the direction and position of the cut.

Worked example 1. An octagonal prism has two octagonal ends and eight rectangular side faces, so F = 2 + 8 = 10. It has 8 edges on each end plus 8 joining edges, so E = 8 + 8 + 8 = 24. It has 8 vertices on each end, so V = 16. The check gives 16 − 24 + 10 = 2. More generally, an n-gonal prism has F = n + 2, E = 3n and V = 2n.

Worked example 2. A pentagonal pyramid has one pentagonal base and five triangular sides, so F = 6. It has five base edges and five sloping edges, so E = 10, while five base vertices and one apex give V = 6. The Euler check is 6 − 10 + 6 = 2. If a plane cuts a cube parallel to a face, the cross-section is a square. If it cuts through three suitably chosen edges near one corner, the cross-section is a triangle. A horizontal cut through a cone is a circle, while a vertical cut through its axis is an isosceles triangle.

3. Reading and testing a net

Assessment status: school-assessed only. A net is a connected arrangement of plane regions that folds to form a solid without gaps or overlaps. To test a proposed net, identify a base face, imagine each adjacent face rotating about its shared edge, and track which free edges must meet. Equal edge lengths must be paired. Two faces that would occupy the same position make the net invalid even if the correct number of faces is shown.

Worked example 1. A closed rectangular prism measuring 8 cm by 5 cm by 3 cm requires six rectangles: two 8 × 5, two 8 × 3 and two 5 × 3. Suppose an 8 × 5 rectangle is used as the base. Attach an 8 × 3 rectangle to each 8 cm edge and a 5 × 3 rectangle to each 5 cm edge. The remaining 8 × 5 rectangle can attach to the outer 8 cm edge of either 8 × 3 side. Its perimeter before folding is not the issue; the matching shared-edge lengths determine whether it can close.

Worked example 2. A triangular prism with triangular ends of side lengths 3 cm, 4 cm and 5 cm and prism length 10 cm needs two congruent 3–4–5 triangles and rectangles 10 × 3, 10 × 4 and 10 × 5. Arrange the three rectangles in a strip along their 10 cm edges, then attach one triangle at each end so corresponding 3, 4 and 5 cm edges meet. The rectangular areas are 30 + 40 + 50 = 120 cm² and the two triangles add 2(½ × 3 × 4) = 12 cm², so the complete net has area 132 cm².

4. Constructing nets from dimensions

Assessment status: school-assessed only. Construct a net by listing every surface and its dimensions before drawing. Use a ruler and a consistent scale, mark fold lines, and add tabs only after the mathematical faces are correct. Tabs are manufacturing allowances, not faces of the ideal solid. A dimension must follow the same physical edge around the fold: a rectangle attached to a 6 cm edge must itself have a 6 cm joining side.

Worked example 1. Make a net for a square-based pyramid with base side 6 cm and slant height 5 cm. Draw a 6 cm square. On each side construct an isosceles triangle with base 6 cm and perpendicular height 5 cm. Each triangular face has area ½ × 6 × 5 = 15 cm², so the four sides total 60 cm²; the base is 6² = 36 cm². The net’s total area is 96 cm². The 5 cm value is the face’s slant height, not the perpendicular height of the solid.

Worked example 2. For a cylinder of radius 4 cm and height 11 cm, the curved surface unwraps to a rectangle whose height is 11 cm and whose length is the circumference 2πr = 8π ≈ 25.13 cm. Add two circles of radius 4 cm. The rectangle area is 11 × 8π = 88π cm² and the circles total 2π(4²) = 32π cm², giving 120π ≈ 376.99 cm². If a 1 cm glue tab is required along the rectangle, the manufactured rectangle becomes approximately 26.13 cm by 11 cm, but the mathematical curved surface remains 25.13 cm by 11 cm.

5. Interpreting perspective drawings

Assessment status: school-assessed only. A perspective drawing represents depth on a flat page; it is not normally drawn to scale. Parallel edges in the solid may appear slanted, and hidden edges are often dashed. Read labels and structural relationships rather than measuring the picture. An orthographic set—front, top and side views—shows true outlines from perpendicular directions and can remove ambiguity left by perspective.

Worked example 1. A storage box is labelled length 12 cm, width 7 cm and height 5 cm. Even if the drawn rear edges look shorter, opposite edges of the rectangular prism are equal. Its face pairs are 12 × 7, 12 × 5 and 7 × 5. Hence surface area = 2(84 + 60 + 35) = 358 cm². Its volume is 12 × 7 × 5 = 420 cm³. Measuring the sketch with a ruler would wrongly replace the stated dimensions.

Worked example 2. A stepped block is made from a 6 × 4 × 2 cm prism with a 2 × 4 × 2 cm prism stacked on one end. The total volume is 6×4×2 + 2×4×2 = 48 + 16 = 64 cm³. For surface area, adding the two separate surface areas gives 88 + 40 = 128 cm², but the contact rectangle 2 × 4 is counted twice and is internal. Subtract 2(8) to obtain 112 cm². A perspective diagram should show the step and dash only those edges hidden from the chosen viewpoint.

6. Connecting nets, solids and practical designs

Assessment status: school-assessed only. A practical design often combines identification, a net and material calculations. Decide whether the object is open or closed, distinguish internal seams from exposed surfaces, and include overlap or waste only when the context specifies it. The net should communicate which edges join; the numerical calculation should use the dimensions of the real object rather than apparent lengths in a drawing.

Worked example 1. An open-top gift box is 20 cm long, 12 cm wide and 6 cm deep. Its net contains a 20 × 12 base, two 20 × 6 sides and two 12 × 6 ends. Card area = 240 + 2(120) + 2(72) = 624 cm². If 10% extra card is allowed for tabs and waste, required card = 1.10 × 624 = 686.4 cm². A closed box would add a second 20 × 12 face, raising the ideal area to 864 cm².

Worked example 2. A party favour is a closed triangular prism with equilateral ends of side 6 cm and length 14 cm. Its three rectangles total 3(6 × 14) = 252 cm². One equilateral triangle has area (√3/4)6² = 9√3 ≈ 15.59 cm², so both ends add 31.18 cm². Total material before tabs is 283.18 cm². The net therefore needs a strip of three 14 × 6 rectangles plus two equilateral triangles. This final check links every calculated region to exactly one exposed face.

7. Worked design: a triangular-prism display sleeve

Assessment status: school-assessed only. A workshop wants a closed triangular-prism sleeve. Each end is a right triangle whose perpendicular sides are 6 cm and 8 cm, and the prism is 20 cm long. First identify the solid from its structural definition: it has two congruent parallel triangular ends and three rectangular side faces, so it is a triangular prism. The ends do not meet at an apex, which rules out a triangular pyramid. Use Pythagoras to find the third end edge: sqrt(6²+8²)=10 cm. The three joining rectangles are therefore 20 by 6, 20 by 8 and 20 by 10 cm. A proposed 20 by 9 rectangle cannot attach along any complete end edge, even if a perspective sketch makes the diagonal appear about 9 cm.

Count the ideal mathematical faces before drawing: two triangles plus three rectangles give F=5. The triangular ends contribute six distinct corner positions, three at each end, so V=6. Each end has three perimeter edges, and three lengthwise edges join corresponding vertices, giving E=3+3+3=9. Euler’s convex-polyhedron check gives V−E+F=6−9+5=2. This equation is a check after careful counting; it does not justify overlooking the three edges hidden behind the front faces. A drawing viewed from one side may show only two of the lengthwise edges, yet the third still joins the matching rear vertices.

To construct one net, place the 20 by 8 rectangle centrally. Join the 20 by 6 rectangle along one of its 20 cm sides and the 20 by 10 rectangle along the other. The three long rectangles now form a strip that wraps around the triangular cross-section. Attach one 6–8–10 triangular end to an exposed 8 cm edge at each short end of the central rectangle, arranging them so they close opposite ends when folded. Every physical face appears exactly once. The rectangles must be attached along corresponding 20 cm fold lines; putting a triangle’s 6 cm edge against a 20 cm edge is impossible even if the shapes touch on paper.

How can this net be tested without manufacturing it? Label each face and each intended fold edge. Imagine the central rectangle flat, rotate the 20 by 6 and 20 by 10 faces upward around its long sides, then rotate the two triangular ends to close the openings. Each triangle has one edge of 6, one of 8 and one of 10 cm. Its 8 cm edge is attached to the central rectangle; the 6 and 10 cm free edges must meet the corresponding 6 and 10 cm widths of the other rectangles. If a triangle is reflected or placed at the same end as the other triangle, the faces can overlap or leave one end open. Check the final adjacency, not only the tally of five faces.

The net’s mathematical surface area is useful for a material estimate even though the main lesson is geometric. Each triangle has area (1/2)(6)(8)=24 cm², so the two ends use 48 cm². The rectangles use 20(6+8+10)=480 cm². Total ideal area is 528 cm². Recompute by grouping one 20 cm length against the triangle perimeter 6+8+10=24 cm: lateral area 20·24=480 cm², then add the ends. The two routes agree. Glue tabs are extra pieces of cardboard, not sixth or seventh faces of the prism; a production allowance is added only when its width and location are supplied. If the sleeve is open at one end, one triangle is omitted and ideal area becomes 504 cm², while the physical object no longer has the same closed-polyhedron Euler count.

Suppose the workshop places a 1 cm-wide glue tab along only the free 20 cm edge of the 20 by 10 rectangle. The rectangular tab adds 20 cm² of material if it is not trimmed. A student who adds 1 cm to every rectangle width would invent two additional tabs and overestimate the blank. A student who counts the tab as a face would misclassify the solid. Record the manufacturing model separately: ideal closed faces 528 cm², specified tab 20 cm², blank before waste 548 cm². These figures answer different questions and should not be substituted for one another.

One final cross-section check ties the net to the solid. A cut perpendicular to the 20 cm axis exposes a 6–8–10 triangle congruent to the ends; a cut parallel to the axis through a suitable part of the prism can expose a rectangle. The right-triangle dimensions come from the true end face, not the slanted appearance of the perspective drawing. If the task provides front and top orthographic views, use their labelled true lengths to recover the end and lengthwise dimensions. The net, face count and cross-section should tell one consistent story about the same prism.

A second classroom model helps distinguish prism counts from pyramid counts. A pentagonal prism has two pentagonal ends and five rectangular lateral faces, so F=7. It has five vertices at each end, V=10, and five perimeter edges at each end plus five joining edges, E=15; 10−15+7=2. A pentagonal pyramid instead has one pentagonal base and five triangles meeting at one apex, so F=6,V=6,E=10 and again 6−10+6=2. Both pass Euler’s check, so Euler alone cannot identify the solid. The number and arrangement of congruent parallel bases are decisive. A net for the pentagonal pyramid needs one pentagon and five triangles attached around its perimeter; substituting five rectangles would make a prism-like lateral strip and could not meet at a single apex.

For a pentagonal-prism cross-section perpendicular to its length, the exposed shape is a pentagon congruent to either end. A slice parallel to the length can give a rectangle if it cuts two parallel lengthwise edges. A perspective sketch may show one pentagonal end as a distorted slanted shape; that visual distortion does not change the five physical sides. If a student counts ten vertices but only twelve edges because three rear edges are hidden, Euler gives 10−12+7=5, signalling a count error. Restore the five connecting edges and all ten end-perimeter edges to obtain fifteen. The check tells the student to revisit structure, not to alter the face count to force an answer.

Return to the 6–8–10 triangular sleeve and test a proposed manufacturing shortcut: using one rectangle 20 by 24 in place of the three side rectangles. That rectangle has the correct combined lateral area 480 cm² and can be scored along its width at distances 6 cm and 14 cm from one edge, leaving a final 10 cm panel. Its two free 20 cm edges join to close the side wrap. Without the two score lines, it is only a rectangle and does not communicate which triangular edge belongs to which panel. The alternative net is valid when the crease widths 6,8,10 are retained and the triangular ends are attached to matching 6,8 or10 cm short edges. Area agreement alone does not replace this geometric matching.

8. Worked audit: selecting valid nets and reading hidden structure

Assessment status: school-assessed only. Consider a closed cuboid with dimensions 5 cm by 4 cm by 3 cm. A valid net must contain exactly six rectangles in three congruent pairs: two 5 by 4 faces, two 5 by 3 faces and two 4 by 3 faces. Its faces are not six arbitrary rectangles whose total area happens to be correct. The 5 by 4 faces must be opposite after folding, as must each other matching pair. Before folding, label them A and A′, B and B′, C and C′; then trace which physical edges meet. A net with both 5 by 4 rectangles forced onto the same side position fails even though its inventory is correct.

Compute the ideal surface area as 2(5·4+5·3+4·3)=2(20+15+12)=94 cm². A net’s summed face areas must equal 94 cm², but this is only a necessary check, not a sufficient one. Two rectangles can overlap when folded and still have the correct total. Conversely, a net missing a 4 by 3 face gives only 82 cm² and certainly cannot close the cuboid. The face-area check quickly rejects missing or duplicate faces; the fold-adjacency check decides whether the layout is physically possible.

To test a candidate layout, choose one 5 by 4 base. The faces joined to its two 5 cm edges must be 5 by 3; the faces joined to its two 4 cm edges must be 4 by 3. A second 5 by 4 face can close the top only if it is attached along a free edge of a side face and folds over the opening. If the candidate places the second 5 by 4 face immediately adjacent to the base along a 5 cm edge, it can still be a side in the flat drawing but cannot serve as a cuboid side of height 3 cm; the edge length alone is insufficient because the face's perpendicular dimension is 4 cm, not 3. Follow both dimensions around every fold.

For a convex cuboid, F=6, V=8 and E=12. Count edges by noting four on the bottom, four on the top and four vertical connectors; then 8−12+6=2. In a perspective view looking down on the front-right corner, the three rear or bottom edges may be hidden, but the solid still has twelve. Dashed lines conventionally indicate hidden edges, while a missing dashed line in a simplified sketch does not erase the edge from the object. Count from structure instead of counting only drawn strokes.

A cross-section depends on where and how the cut is made. A plane parallel to a 5 by 4 face through the interior produces a 5 by 4 rectangle. A plane parallel to a 5 by 3 face produces a 5 by 3 rectangle. A plane cutting through the cuboid at an angle can create a different polygon, so “the cuboid’s cross-section” is incomplete without a cutting plane. If a diagram shows a slice through the middle parallel to the base, its area is 20 cm²; do not infer 94 cm², which is the entire external surface area.

Orthographic views give another audit. A cuboid placed with its 5 by 4 face on the ground has top view 5 by 4, front view 5 by 3 and side view 4 by 3. These are two-dimensional plane shapes, not three separate solids. If an isometric picture looks narrower on the right, the true side width is still 4 cm because the labelled view specifies it. Measuring a slanted 3 cm edge on the page can produce a false physical length when the drawing is not to scale.

Now change the brief to an open-top storage tray with the same external base and wall dimensions. The ideal set of exposed cardboard faces has one 5 by 4 base, two 5 by 3 long walls and two 4 by 3 short walls. Its area is 20+30+24=74 cm², exactly 20 cm² below the closed cuboid. There are five panels, but the open tray is not a closed convex polyhedron, so using V−E+F=2 on an arbitrarily chosen panel count is inappropriate. If 2 cm-wide fold tabs are specified, calculate their areas by actual tab lengths and count each once; do not silently add a percentage or a second top panel.

For a practical decision, suppose sheets are 30 by 20 cm, giving 600 cm² each. Dividing 600 by 74 suggests at most eight ideal trays by area, since 8·74=592 cm². That arithmetic is only an upper bound: eight full nets may not fit on one sheet after layout, gaps and cutting clearance. Without a packing diagram, the defensible conclusion is “no more than eight by area”; it is not a guarantee of eight usable blanks. This distinction between numerical area and actual net arrangement is precisely why a design audit uses dimensions, adjacency and layout together.

A cylinder gives a useful contrast to the six-flat-face cuboid. A closed cylinder of radius 3 cm and height 10 cm has two circular plane end regions and one curved lateral surface. A lateral net is a rectangle whose one dimension is height 10 cm and whose other is the circular circumference 2pi r=6pi cm. Two radius-3 circles attach to close the ends. The ideal material area is 10·6pi+2(pi·3²)=60pi+18pi=78pi cm². A student who draws a 10 by 6 rectangle has confused circumference with diameter; its strip would be too short by the factor pi. A student who treats the circular ends as rectangles would not produce the required cross-section.

The cylinder is not a polyhedron with only flat polygonal faces, so the cuboid’s V−E+F count should not be applied naively to its curved surface. The word “face” is used differently in some elementary diagrams for a curved cylinder surface; state whether counting flat end faces or all boundary surfaces. A cut perpendicular to the cylinder axis exposes a radius-3 circle, while a cut through the axis gives a 6 by 10 rectangle. These cross-sections are not the same as the unwrapped lateral rectangle: both may involve 10 cm, but the through-axis section has width diameter 6 cm whereas the net strip has width circumference 6pi cm.

Suppose the cylindrical label overlaps by 1 cm along a vertical seam. Its printed rectangular blank must be 10 cm high and (6pi+1) cm wide, area 60pi+10 cm². This does not alter the ideal cylinder’s curved area 60pi cm², because the overlapping 1 cm lies on top of another part of the label. If the label also stops 0.5 cm short of both ends, its height is 9 cm and its blank area is 9(6pi+1) cm². The context must say whether the seam allowance is included and whether the label reaches the ends. This is a practical reason to keep geometric surface, net blank and manufacturing allowance as separate quantities.

A final reasonableness check compares the circle and strip contributions. The two end circles use 18pi cm², less than one third of the lateral 60pi cm² because the cylinder is relatively tall. If a proposed answer made the ends larger than the side for these dimensions, revisit whether radius 3 was mistaken for diameter 3 or whether the circumference was squared. The dimensional labels also differ: 6pi is a length used in the net width, while 78pi is an area for the full ideal surface.