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3D shape. Because area is two-dimensional, it is expressed in square units, such as cm²."},{"text":"Surface area is the total area of all visible faces of a 3D shape, measured in square units.","correct":false,"explanation":"A perspective drawing may hide exterior faces such as the back or bottom. Surface area includes all outside faces, not only visible ones, and is measured in square units."},{"text":"Surface area is the total area of all outside faces of a 3D shape, measured in cubic units.","correct":false,"explanation":"Cubic units measure volume, which is the space inside a solid. Surface area covers all outside faces and must therefore be measured in square units."},{"text":"Surface area is the total area of all inside faces of a 3D shape, measured in linear units.","correct":false,"explanation":"This confuses surface area with interior surfaces and length measurement. Surface area totals all outside faces and uses square units because each face has two dimensions."}]},{"id":"5127a3cb-6601-444e-a362-c144d7f24b94","cardType":"basic","points":1,"front":"What does a net help you do for surface area?","back":"Add the areas of all outside faces.","mcqOptions":[{"text":"Add the areas of all outside faces.","correct":true,"explanation":"A net lays every exterior face flat, making each face easy to identify and measure. Adding their individual areas gives the total surface area."},{"text":"Add the areas of only visible faces.","correct":false,"explanation":"Counting only visible faces omits exterior faces hidden in a 3D drawing. A net helps you add the areas of all outside faces."},{"text":"Multiply the areas of all outside faces.","correct":false,"explanation":"Surface area is a total, so the separate face areas must be added rather than multiplied. A net displays all outside faces for this addition."},{"text":"Add the lengths of all outside edges.","correct":false,"explanation":"Adding edge lengths calculates a total length, not an area. A net is used to find and add the areas of all outside faces."}]},{"id":"a00018f9-24c7-44a1-9a87-07252321bc10","cardType":"basic","points":3,"front":"For a cuboid, which matching face pairs are included?","back":"

Front/back
Top/bottom
Left/right

","mcqOptions":[{"text":"

Front/back
Top/bottom
Left/right

","correct":true,"explanation":"Opposite faces of a cuboid are congruent rectangles. The front matches the back, the top matches the bottom, and the left matches the right."},{"text":"

Front/top
Back/bottom
Left/right

","correct":false,"explanation":"Front and top faces are adjacent and usually have different dimensions, as are back and bottom. The matching pairs are front/back, top/bottom and left/right."},{"text":"

Front/left
Back/right
Top/bottom

","correct":false,"explanation":"This pairs adjacent side faces rather than opposite congruent faces. A cuboid’s matching pairs are front/back, top/bottom and left/right."},{"text":"

Front/bottom
Top/right
Back/left

","correct":false,"explanation":"Each listed pair contains adjacent faces, which do not generally share the same dimensions. Matching cuboid faces are the three opposite pairs: front/back, top/bottom and left/right."}]},{"id":"32346afe-7e44-476d-a58a-b3f5c2fd2a2c","cardType":"cloze","points":null,"front":"For a cuboid, surface area = [ ].","back":"For a cuboid, surface area = $2(lw+lh+wh)$.","mcqOptions":[{"text":"$$2(lw+lh+wh)$","correct":true,"explanation":"A cuboid has two faces of each rectangular area: $lw$, $lh$ and $wh$. Adding both of every pair gives $2(lw+lh+wh)$."},{"text":"$$lw+lh+wh$","correct":false,"explanation":"This includes only one face from each matching pair, so it gives half the surface area. Doubling the sum gives $2(lw+lh+wh)$."},{"text":"$$2(l+w+h)$","correct":false,"explanation":"This adds lengths and resembles a perimeter calculation, so it cannot produce square units. Multiplying dimension pairs gives the correct surface area $2(lw+lh+wh)$."},{"text":"$$lwh$","correct":false,"explanation":"Multiplying all three dimensions calculates the cuboid’s volume in cubic units. Surface area is found from the six faces using $2(lw+lh+wh)$."}]},{"id":"e9b78333-f96a-456d-8dd4-91e9a104a06f","cardType":"basic","points":1,"front":"What is missed if you only add the visible faces in a 3D diagram?","back":"Hidden outside faces, such as the back or bottom.","mcqOptions":[{"text":"Hidden outside faces, such as the back or bottom.","correct":true,"explanation":"A 3D drawing often conceals exterior faces behind or beneath the solid. These hidden outside faces still contribute to the total surface area."},{"text":"Visible outside faces, such as the front or top.","correct":false,"explanation":"The visible faces are precisely the ones already included by this method. What gets missed are hidden outside faces, commonly the back or bottom."},{"text":"Interior measurements, such as the length or width.","correct":false,"explanation":"Length and width are dimensions used to calculate face areas, not overlooked faces. Adding only visible faces misses hidden outside faces such as the back or bottom."},{"text":"Hidden inside faces, such as the centre or middle.","correct":false,"explanation":"An ordinary solid does not have faces at its centre or middle. The omitted parts are hidden outside faces, such as the back or bottom."}]},{"id":"69d4d6ea-8857-4cb9-9be4-0eac02b91566","cardType":"cloze","points":null,"front":"Triangle area = [ ].","back":"Triangle area = $\\frac{1}{2}\\times \\text{base}\\times \\text{height}$.","mcqOptions":[{"text":"$$\\frac{1}{2}\\times \\text{base}\\times \\text{height}$","correct":true,"explanation":"A triangle with a given base and perpendicular height has half the area of the corresponding rectangle, so its area is $\\frac{1}{2}\\times \\text{base}\\times \\text{height}$."},{"text":"$$\\text{base}\\times \\text{height}$","correct":false,"explanation":"This calculates the area of the corresponding rectangle and makes the triangle’s area twice too large. The triangle requires the factor $\\frac{1}{2}$."},{"text":"$$\\frac{1}{2}\\times(\\text{base}+\\text{height})$","correct":false,"explanation":"Adding base and height produces a length rather than an area. The two perpendicular measurements must be multiplied, then halved: $\\frac{1}{2}\\times \\text{base}\\times \\text{height}$."},{"text":"$$\\frac{1}{2}\\times \\text{base}\\times \\text{side}$","correct":false,"explanation":"Using a sloping side instead of the perpendicular height is a common mistake. Triangle area uses the base and its perpendicular height, multiplied and halved."}]},{"id":"9e936e4d-d15a-46cf-99f5-74cc30bba4d0","cardType":"cloze","points":null,"front":"Trapezium area = [ ].","back":"Trapezium area = $\\frac{1}{2}\\times(\\text{sum of parallel sides})\\times \\text{height}$.","mcqOptions":[{"text":"$$\\frac{1}{2}\\times(\\text{sum of parallel sides})\\times \\text{height}$","correct":true,"explanation":"A trapezium’s area equals the average length of its two parallel sides multiplied by the perpendicular height, giving half their sum times the height."},{"text":"$$(\\text{sum of parallel sides})\\times \\text{height}$","correct":false,"explanation":"This uses the sum rather than the average of the parallel sides, making the area twice too large. The formula needs the factor $\\frac{1}{2}$."},{"text":"$$\\frac{1}{2}\\times(\\text{product of parallel sides})\\times \\text{height}$","correct":false,"explanation":"The parallel side lengths must be added, not multiplied; multiplying them also gives the wrong dimensions. Halve their sum and multiply by the perpendicular height."},{"text":"$$\\frac{1}{2}\\times(\\text{sum of non-parallel sides})\\times \\text{height}$","correct":false,"explanation":"The area formula uses the two parallel sides because their average determines the trapezium’s effective width. The non-parallel sides are not used in this formula."}]},{"id":"66cded8d-129f-4b52-868f-db0332a3ad61","cardType":"basic","points":1,"front":"In a prism, what happens to the cross-section along its length?","back":"It stays the same all the way through.","mcqOptions":[{"text":"It stays the same all the way through.","correct":true,"explanation":"A prism has a constant cross-section perpendicular to its length. Its cross-section remains the same shape and size at every point along the prism."},{"text":"It gets steadily larger all the way through.","correct":false,"explanation":"A solid that steadily widens does not have the constant cross-section required of a prism. In a prism, the cross-section stays the same throughout its length."},{"text":"It gets steadily smaller all the way through.","correct":false,"explanation":"A tapering solid, such as a pyramid or cone, has changing cross-sections and is not a prism. A prism’s cross-section stays the same throughout."},{"text":"It changes shape and size all the way through.","correct":false,"explanation":"Changing shape or size contradicts the defining property of a prism. Its cross-section must stay the same shape and size all the way through."}]},{"id":"01fdaf62-1065-45b9-9101-b129ac2b95f1","cardType":"cloze","points":null,"front":"For any prism, surface area = [ ], where $P$ is the [ ].","back":"For any prism, surface area = $2A+PL$, where $P$ is the perimeter of the cross-section.","mcqOptions":[{"text":"For any prism, surface area = $2A+PL$, where $P$ is the perimeter of the cross-section.","correct":true,"explanation":"The two identical cross-sections have total area $2A$. The rectangular side faces have combined area $PL$, because their total width is the cross-section’s perimeter."},{"text":"For any prism, surface area = $A+PL$, where $P$ is the perimeter of the cross-section.","correct":false,"explanation":"Using $A$ counts only one end face. A prism has two identical cross-sections, so their combined area is $2A$ and the correct formula is $2A+PL$."},{"text":"For any prism, surface area = $2A+PL$, where $P$ is the area of the cross-section.","correct":false,"explanation":"This confuses perimeter with area. In $PL$, $P$ must be the perimeter of the cross-section so that multiplying it by length gives the total side area."},{"text":"For any prism, surface area = $2A+2PL$, where $P$ is the area of the cross-section.","correct":false,"explanation":"This both doubles the side area and misidentifies $P$. The side rectangles total $PL$ once, and $P$ represents the perimeter of the cross-section, giving $2A+PL$."}]},{"id":"6c54e620-c64b-4abd-97a1-3b666db32cbb","cardType":"basic","points":2,"front":"Why does $PL$ give the side-rectangle area of a prism?","back":"

$P$ is the total width around the cross-section.
$L$ is the prism length.

","mcqOptions":[{"text":"

$P$ is the total width around the cross-section.
$L$ is the prism length.

","correct":true,"explanation":"Unfolding the side faces produces rectangles whose combined width is the cross-section’s perimeter $P$. Their common length is $L$, so their total area is $P\\times L$."},{"text":"

$P$ is the total area of the cross-section.
$L$ is the prism length.

","correct":false,"explanation":"This confuses perimeter with area. $P$ is the total distance around the cross-section, and multiplying that combined width by the prism length $L$ gives the side area."},{"text":"

$P$ is the total width around the cross-section.
$L$ is the cross-section area.

","correct":false,"explanation":"This misidentifies $L$ as an area. $L$ is the prism’s length, while $P$ is the total width around its cross-section, so $PL$ measures area."},{"text":"

$P$ is the width of one side face.
$L$ is the prism length.

","correct":false,"explanation":"Using one face’s width would calculate only that face’s area. $P$ combines all widths around the cross-section, so $PL$ gives the area of every side rectangle."}]},{"id":"cfd3f5ad-4be7-44ed-b503-df345b31d186","cardType":"basic","points":2,"front":"Why does a triangular prism have 3 rectangular side faces?","back":"Each of the triangle’s 3 sides forms one rectangle.","mcqOptions":[{"text":"Each of the triangle’s 3 sides forms one rectangle.","correct":true,"explanation":"Each side of the triangular cross-section extends along the prism’s length to create one rectangular face. Since a triangle has three sides, there are three rectangles."},{"text":"Each of the triangle’s 2 sides forms one rectangle.","correct":false,"explanation":"This overlooks one side of the triangular cross-section. All three sides extend along the prism’s length, and each produces one rectangular side face."},{"text":"Each of the triangle’s 3 vertices forms one rectangle.","correct":false,"explanation":"This confuses vertices with edges. Rectangular faces are formed when the triangle’s three sides extend along the prism’s length, not when its vertices are counted."},{"text":"Each of the triangle’s 2 end faces forms one rectangle.","correct":false,"explanation":"The two end faces are triangles, not rectangles. The three rectangular side faces come from extending each of the triangle’s three sides along the prism’s length."}]},{"id":"94778d8f-38d2-4548-9e6e-987fd0cb49a5","cardType":"basic","points":1,"front":"In a trapezium prism, what is the trapezium area used for?","back":"The two identical end faces.","mcqOptions":[{"text":"The two identical end faces.","correct":true,"explanation":"The trapezium is the prism’s constant cross-section, appearing at both ends. Its area is therefore calculated twice when finding the prism’s total surface area."},{"text":"The four rectangular side faces.","correct":false,"explanation":"The side faces are rectangles formed by extending the trapezium’s sides along the prism. The trapezium area is used for the two identical end faces instead."},{"text":"The perimeter of the cross-section.","correct":false,"explanation":"Area and perimeter are different measurements. The trapezium area gives the two end-face areas, while its side lengths are added separately to find the cross-section’s perimeter."},{"text":"The length of the prism.","correct":false,"explanation":"The prism length is a separate linear measurement used to find rectangular side areas. The trapezium area represents each of the two identical end faces."}]},{"id":"eff30e23-c2d5-40a8-a0e5-300dc5575dcd","cardType":"basic","points":1,"front":"In a trapezium, when is the perpendicular height included in perimeter?","back":"Only if it is an outside side length.","mcqOptions":[{"text":"Only if it is an outside side length.","correct":true,"explanation":"Perimeter includes only lengths forming the outside boundary. A perpendicular height is included only when it is also an actual outside side of the trapezium."},{"text":"Always, because it is an important measurement.","correct":false,"explanation":"The height is needed for the trapezium’s area formula, but that does not automatically place it on the perimeter. It is included only when it is an outside side length."},{"text":"Only if it is an internal vertical line.","correct":false,"explanation":"Internal construction lines are not part of a shape’s perimeter. The perpendicular height is counted only when it lies on the trapezium’s outside boundary as a side."},{"text":"Only if it is a parallel side length.","correct":false,"explanation":"A perpendicular height cannot be one of the trapezium’s parallel sides. It contributes to perimeter only if it is an outside side length, whether vertical or sloping."}]},{"id":"6ba02f60-7bbd-4871-9c57-3836ecbfd79c","cardType":"basic","points":1,"front":"How do you find the area of an L-shaped cross-section?","back":"Split into rectangles and add their areas.","mcqOptions":[{"text":"Split into rectangles and add their areas.","correct":true,"explanation":"An L-shape can be divided into non-overlapping rectangles. Calculate each rectangle’s area using length multiplied by width, then add the results for the total cross-sectional area."},{"text":"Multiply the perimeter by the prism length.","correct":false,"explanation":"Perimeter multiplied by prism length gives the area of the prism’s rectangular side faces, not the cross-section. Split the L-shape into rectangles and add their areas."},{"text":"Treat it as one large rectangle without subtraction.","correct":false,"explanation":"This includes the missing corner and therefore overestimates the area. The L-shape should be split into rectangles and their non-overlapping areas added together."},{"text":"Add all the outside side lengths together.","correct":false,"explanation":"Adding outside side lengths finds the perimeter, not the area. To find the L-shaped cross-section’s area, split it into rectangles and add their individual areas."}]},{"id":"7fc2ed66-dfbc-43f0-bb85-49bd11bf8517","cardType":"basic","points":1,"front":"For an L-shaped cross-section, which lines are excluded from the perimeter?","back":"Internal split lines.","mcqOptions":[{"text":"Internal split lines.","correct":true,"explanation":"Split lines are drawn only to divide the L-shape into simpler rectangles. They lie inside the shape, so they are not part of its outside boundary or perimeter."},{"text":"Outside horizontal lines.","correct":false,"explanation":"Horizontal direction does not exclude a line from the perimeter. Every segment on the outside boundary is counted; only internal split lines used to divide the shape are excluded."},{"text":"Outside vertical lines.","correct":false,"explanation":"Vertical outside edges still form part of the boundary and must be counted. The lines excluded from the perimeter are internal split lines drawn within the L-shape."},{"text":"The shortest boundary lines.","correct":false,"explanation":"Boundary length does not determine whether a segment is included. All outside edges, however short, contribute to perimeter; only internal split lines are left out."}]},{"id":"b616781f-7c6e-4ec7-ba79-b3f5086f788b","cardType":"basic","points":1,"front":"To convert cm² to m², divide by what?","back":"10,000.","mcqOptions":[{"text":"10,000.","correct":true,"explanation":"Since $1\\text{ m}=100\\text{ cm}$, squaring the conversion gives $1\\text{ m}^2=10{,}000\\text{ cm}^2$. Therefore, divide a value in cm² by 10,000."},{"text":"100.","correct":false,"explanation":"Dividing by 100 uses the linear centimetre-to-metre conversion without squaring it. Area conversion requires $100^2=10{,}000$, so cm² must be divided by 10,000."},{"text":"1,000.","correct":false,"explanation":"This treats the conversion as three factors of ten, which does not match the area scale. Because $1\\text{ m}=100\\text{ cm}$, divide cm² by $100^2=10{,}000$."},{"text":"1,000,000.","correct":false,"explanation":"Dividing by 1,000,000 applies the cubic conversion used for cm³ to m³. For square units, square 100, so cm² is divided by 10,000."}]},{"id":"7656a2c8-7c5a-4fc3-8f25-bedd9964fb6f","cardType":"basic","points":1,"front":"A paint calculation gives 1.944 pots. How many pots are needed?","back":"2 pots.","mcqOptions":[{"text":"2 pots.","correct":true,"explanation":"Pots cannot usually be bought as a fraction, so the calculation must be rounded up. Two pots provide enough paint to cover the required surface area."},{"text":"1 pot.","correct":false,"explanation":"Rounding down leaves only one pot, which contains less paint than the 1.944 pots required. The answer must be rounded up to 2 pots."},{"text":"3 pots.","correct":false,"explanation":"Three pots would provide enough paint, but it is more than necessary. Since 1.944 rounds up to the next whole pot, the minimum required is 2 pots."},{"text":"1.944 pots.","correct":false,"explanation":"Although this is the calculated amount, pots are needed as whole items rather than a fraction. Rounding up gives the minimum practical answer of 2 pots."}]},{"id":"e28e208d-d4dd-4d7b-a2ad-d5a9ff23f8af","cardType":"basic","points":1,"front":"If a cylinder’s diameter is 0.6 m, what radius is used?","back":"$0.3\\text{ m}$.","mcqOptions":[{"text":"$0.3\\text{ m}$.","correct":true,"explanation":"The radius is half the diameter, so dividing $0.6\\text{ m}$ by $2$ gives $0.3\\text{ m}$."},{"text":"$0.6\\text{ m}$.","correct":false,"explanation":"This uses the diameter unchanged instead of halving it. The radius is half of $0.6\\text{ m}$, giving $0.3\\text{ m}$."},{"text":"$1.2\\text{ m}$.","correct":false,"explanation":"This doubles the diameter rather than dividing it by two, confusing the relationship between radius and diameter. The radius should be $0.3\\text{ m}$."},{"text":"$0.15\\text{ m}$.","correct":false,"explanation":"This halves the diameter twice, instead of halving it once. Since radius equals diameter divided by $2$, the correct value is $0.3\\text{ m}$."}]},{"id":"f324c0f7-6edd-41fc-b61a-c6a176c1e0f0","cardType":"cloze","points":null,"front":"For a closed cylinder, surface area = [ ].","back":"For a closed cylinder, surface area = $2\\pi r^2+2\\pi rh$.","mcqOptions":[{"text":"$$2\\pi r^2+2\\pi rh$","correct":true,"explanation":"A closed cylinder has two circular ends, contributing $2\\pi r^2$, and one curved surface, contributing $2\\pi rh$. Adding these gives the total surface area."},{"text":"$$2\\pi r^2+\\pi rh$","correct":false,"explanation":"The curved surface has circumference $2\\pi r$ multiplied by height $h$, so its area is $2\\pi rh$, not half that amount. The complete formula is $2\\pi r^2+2\\pi rh$."},{"text":"$$\\pi r^2+2\\pi rh$","correct":false,"explanation":"This counts only one circular end, but a closed cylinder has two. The two ends contribute $2\\pi r^2$, making the full formula $2\\pi r^2+2\\pi rh$."},{"text":"$$2\\pi r^2+2\\pi h$","correct":false,"explanation":"The curved area must include the radius in the circumference, giving $2\\pi rh$. Omitting $r$ loses a required length factor, so the correct formula is $2\\pi r^2+2\\pi rh$."}]},{"id":"686e31d7-ca7d-4dae-a35b-f56529e3ec92","cardType":"basic","points":2,"front":"Why is the curved side area of a cylinder $2\\pi rh$?","back":"Circumference $2\\pi r$ multiplied by height $h$.","mcqOptions":[{"text":"Circumference $2\\pi r$ multiplied by height $h$.","correct":true,"explanation":"Unwrapping the curved surface produces a rectangle. Its width is the cylinder’s circumference, $2\\pi r$, and its height is $h$, so the area is $2\\pi rh$."},{"text":"Diameter $2r$ multiplied by height $h$.","correct":false,"explanation":"The rectangle formed by unwrapping the curved surface has width equal to the full circumference, not the diameter. Using circumference $2\\pi r$ gives the correct area $2\\pi rh$."},{"text":"Area $\\pi r^2$ multiplied by height $h$.","correct":false,"explanation":"Multiplying the circular area by height gives a volume-related expression, not the area of the curved surface. The curved side unwraps into a rectangle with width $2\\pi r$ and height $h$."},{"text":"Circumference $2\\pi r$ multiplied by radius $h$.","correct":false,"explanation":"The second dimension of the unwrapped rectangle is the cylinder’s height, not its radius. Therefore the curved area uses circumference $2\\pi r$ multiplied by height $h$."}]}]}],"examCountdown":{"examDate":"2027-05-14","paperLabel":"Paper 1"},"heading":"Surface Area","title":"Surface Area Flashcards","topbar":{"backHref":"/igcse/maths/edexcel/flashcards"},"topicSidebar":"$L37"}]}],"topicId":"f60dbe70-8097-444b-b77f-25a3c947c8a9","walled":"$L38"}],"$L39","$L3a"]