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units","correct":false,"explanation":"Linear units measure one-dimensional quantities such as length or perimeter. Area measures two-dimensional space, so the correct term is square units."},{"text":"cubic units","correct":false,"explanation":"Cubic units measure three-dimensional volume rather than area. Because area covers a two-dimensional surface, it is measured in square units."},{"text":"standard units","correct":false,"explanation":"Standard units is too general and does not identify the required dimension. Area is specifically measured in square units, as shown by cm² and m²."}]},{"id":"13c0c858-f3d7-40dc-8b66-f8c3adcd5e33","cardType":"basic","points":1,"front":"What is the area formula for a rectangle with base $b$ and height $h$?","back":"$A=bh$","mcqOptions":[{"text":"$A=bh$","correct":true,"explanation":"A rectangle contains $b$ rows of height $h$, or equivalently has area equal to its base multiplied by its perpendicular height."},{"text":"$A=b+h$","correct":false,"explanation":"Adding the base and height does not calculate the space inside the rectangle. Its area is found by multiplying these dimensions, giving $A=bh$."},{"text":"$A=\\frac{1}{2}bh$","correct":false,"explanation":"The factor of one-half belongs to the triangle formula, since a triangle is half a corresponding rectangle. A rectangle’s area is $A=bh$."},{"text":"$A=b^2h^2$","correct":false,"explanation":"Squaring both dimensions incorrectly multiplies the rectangle’s true area by another factor of $bh$. The correct formula is simply $A=bh$."}]},{"id":"8dff76dd-5c12-4b04-b1d3-ddb61dd1964c","cardType":"basic","points":1,"front":"What is the area formula for a triangle with base $b$ and height $h$?","back":"$A=\\frac{1}{2}bh$","mcqOptions":[{"text":"$A=\\frac{1}{2}bh$","correct":true,"explanation":"A triangle occupies half the area of a rectangle or parallelogram with the same base and perpendicular height, so its area is $A=\\frac{1}{2}bh$."},{"text":"$A=bh$","correct":false,"explanation":"Multiplying base by height gives the area of the corresponding rectangle or parallelogram. A triangle occupies half that area, so $A=\\frac{1}{2}bh$."},{"text":"$A=\\frac{1}{2}(b+h)$","correct":false,"explanation":"Adding base and height does not measure the triangle’s two-dimensional area. Multiply the base by the perpendicular height and halve the result: $A=\\frac{1}{2}bh$."},{"text":"$A=\\frac{1}{4}bh$","correct":false,"explanation":"Halving both dimensions introduces a factor of one-quarter instead of one-half. Only the product $bh$ is halved, giving $A=\\frac{1}{2}bh$."}]},{"id":"e67aad80-ebdf-4003-8df5-205118157c73","cardType":"basic","points":1,"front":"What is the area formula for a trapezium with parallel sides $a$, $b$ and height $h$?","back":"$A=\\frac{1}{2}(a+b)h$","mcqOptions":[{"text":"$A=\\frac{1}{2}(a+b)h$","correct":true,"explanation":"The mean length of the two parallel sides is $\\frac{a+b}{2}$. Multiplying this by the perpendicular height gives $A=\\frac{1}{2}(a+b)h$."},{"text":"$A=\\frac{1}{2}(a-b)h$","correct":false,"explanation":"This uses the difference between the parallel sides rather than their mean total. A trapezium’s area is $A=\\frac{1}{2}(a+b)h$."},{"text":"$A=(a+b)h$","correct":false,"explanation":"Omitting the factor of one-half doubles the area by using the sum rather than the mean of the parallel sides. The correct formula is $A=\\frac{1}{2}(a+b)h$."},{"text":"$A=\\frac{1}{2}abh$","correct":false,"explanation":"Multiplying the parallel sides together treats them as perpendicular dimensions and gives incorrect units after also multiplying by height. Use $A=\\frac{1}{2}(a+b)h$."}]},{"id":"a277dc68-03cd-4ce1-b735-c01f1faee42b","cardType":"cloze","points":null,"front":"For triangles and trapezia, the height must be [ ] to the [ ].","back":"For triangles and trapezia, the height must be perpendicular to the base.","mcqOptions":[{"text":"For triangles and trapezia, the height must be perpendicular to the base.","correct":true,"explanation":"Area formulas require the shortest separation from the base to the opposite vertex or parallel side. This distance meets the chosen base at a right angle."},{"text":"For triangles and trapezia, the height must be parallel to the base.","correct":false,"explanation":"A line parallel to the base does not measure the separation from it. The height must be perpendicular to the base, meeting it at a right angle."},{"text":"For triangles and trapezia, the height must be perpendicular to the sloping side.","correct":false,"explanation":"The height is not generally measured perpendicular to a sloping side. It must be perpendicular to whichever side is being used as the base."},{"text":"For triangles and trapezia, the height must be parallel to the sloping side.","correct":false,"explanation":"Neither condition identifies the perpendicular distance required by the area formulas. The height must be perpendicular to the base, not parallel to a sloping side."}]},{"id":"2ea0bb6d-1125-4076-b381-e4225cf52433","cardType":"basic","points":2,"front":"When can a sloping side be used as the height of a triangle or trapezium?","back":"Only if it is perpendicular to the base.","mcqOptions":[{"text":"Only if it is perpendicular to the base.","correct":true,"explanation":"A height is the perpendicular distance from the base to the opposite vertex or parallel side. Therefore, a sloping side counts only when it meets the base at $90°$."},{"text":"Only if it is parallel to the base.","correct":false,"explanation":"Being parallel to the base prevents a side from measuring the required separation from it. The sloping side is a height only if it is perpendicular to the base."},{"text":"Only if it is the longest side.","correct":false,"explanation":"A side’s length does not determine whether it is a height. A sloping side can be used only when it is perpendicular to the chosen base."},{"text":"Only if it is labelled on the diagram.","correct":false,"explanation":"A side being labelled does not make it a perpendicular height. It can be used in the area formula only if it is perpendicular to the base."}]},{"id":"e953ad51-d69e-42ff-8c9d-ffd8bc534d3d","cardType":"cloze","points":null,"front":"Parallel lines stay [ ] and [ ].","back":"Parallel lines stay the same distance apart and never meet.","mcqOptions":[{"text":"Parallel lines stay the same distance apart and never meet.","correct":true,"explanation":"Parallel lines have the same direction and constant perpendicular separation. Consequently, they remain the same distance apart and do not intersect, however far they are extended."},{"text":"Parallel lines stay further and further apart and never meet.","correct":false,"explanation":"Lines whose separation continually increases are diverging rather than parallel. Parallel lines remain the same distance apart and never meet when extended."},{"text":"Parallel lines stay the same distance apart and eventually meet.","correct":false,"explanation":"Lines that eventually meet are not parallel, even if they appear nearly parallel in a short diagram. Parallel lines stay the same distance apart and never meet."},{"text":"Parallel lines stay closer and closer together and eventually meet.","correct":false,"explanation":"Lines that approach and meet are converging, not parallel. True parallel lines maintain a constant distance between them and never intersect."}]},{"id":"50a7fd04-2c71-420d-9370-e59f1b562610","cardType":"basic","points":1,"front":"If a diagram is labelled not accurately drawn, what should you use to find lengths?","back":"Use the labelled lengths only.","mcqOptions":[{"text":"Use the labelled lengths only.","correct":true,"explanation":"A diagram marked not accurately drawn may have misleading proportions or angles. Calculations must therefore use the stated measurements rather than lengths estimated from the picture."},{"text":"Use the visual proportions only.","correct":false,"explanation":"The picture’s proportions may be deliberately inaccurate, so visual estimates are unreliable. Use the labelled lengths only when calculating missing lengths or areas."},{"text":"Use the measured lengths only.","correct":false,"explanation":"Measuring the printed diagram assumes it has been drawn to scale, which the warning explicitly denies. Calculations should use the labelled lengths only."},{"text":"Use the grid square counts only.","correct":false,"explanation":"Counting grid squares is valid only when an accurate grid and scale are provided. For a diagram not accurately drawn, use the labelled lengths only."}]},{"id":"c313b960-ef3b-4838-944c-c4ddea33b079","cardType":"basic","points":3,"front":"For a compound shape made of joined pieces, what is the area strategy?","back":"

Split into simpler shapes.
Find each area separately.
Add the areas.

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

Split into simpler shapes.
Find each area separately.
Add the areas.

","correct":true,"explanation":"Splitting a compound shape into familiar shapes allows each area to be calculated using a standard formula. Because the pieces are joined without overlapping, their areas are then added."},{"text":"

Split into simpler shapes.
Find each perimeter separately.
Add the perimeters.

","correct":false,"explanation":"This confuses area with perimeter. The correct strategy is to split the compound shape, find the area of each simpler shape separately, and then add those areas."},{"text":"

Split into simpler shapes.
Find each area separately.
Multiply the areas.

","correct":false,"explanation":"Multiplying the separate areas does not give the area of the combined shape. After splitting and finding each piece’s area, the areas must be added together."},{"text":"

Measure the outside edges.
Find their total length.
Square the result.

","correct":false,"explanation":"Adding outside edges calculates perimeter, and squaring that result does not calculate area. Instead, split the shape into simpler pieces, calculate their areas, and add them."}]},{"id":"273de04e-9ddd-401d-82e1-fb05b58c626f","cardType":"cloze","points":null,"front":"For right-angled compound shapes, splitting along a [ ] or [ ] line usually creates rectangles.","back":"For right-angled compound shapes, splitting along a horizontal or vertical line usually creates rectangles.","mcqOptions":[{"text":"For right-angled compound shapes, splitting along a horizontal or vertical line usually creates rectangles.","correct":true,"explanation":"Horizontal and vertical splitting lines follow the directions of the perpendicular sides. Extending these lines across a right-angled compound shape usually divides it into rectangles."},{"text":"For right-angled compound shapes, splitting along a diagonal or vertical line usually creates rectangles.","correct":false,"explanation":"A diagonal split generally creates triangles or other non-rectangular pieces. Horizontal or vertical lines align with the right-angled sides and usually create rectangles."},{"text":"For right-angled compound shapes, splitting along a horizontal or diagonal line usually creates rectangles.","correct":false,"explanation":"The diagonal line is the misconception here because it cuts across the perpendicular grid of the shape. Horizontal and vertical splits are the pair that usually create rectangles."},{"text":"For right-angled compound shapes, splitting along a diagonal or curved line usually creates rectangles.","correct":false,"explanation":"Diagonal and curved lines do not normally produce rectangles, whose sides must meet at right angles. Horizontal and vertical splitting lines follow the shape’s perpendicular edges."}]},{"id":"48f944f8-767a-4009-897d-e2e040cccaa2","cardType":"basic","points":2,"front":"Why should you not multiply total width by total height for an L-shape?","back":"It includes space not in the shape.","mcqOptions":[{"text":"It includes space not in the shape.","correct":true,"explanation":"Total width multiplied by total height gives the area of the complete surrounding rectangle. An L-shape has a missing section, so this calculation includes space outside the shape."},{"text":"It excludes space inside the shape.","correct":false,"explanation":"The bounding rectangle does not exclude part of the L-shape; it includes too much. Total width times total height counts the missing corner, which is not in the shape."},{"text":"It measures the perimeter of the shape.","correct":false,"explanation":"Multiplying width by height calculates rectangular area, not perimeter. The problem is that the resulting rectangle includes the L-shape’s missing section, so its area is too large."},{"text":"It counts the missing space twice.","correct":false,"explanation":"The missing section is included once, not counted twice, when total width is multiplied by total height. The correct approach splits the L-shape or subtracts the missing area."}]},{"id":"f46f5fe5-0dae-4f51-a383-1765644e44b4","cardType":"basic","points":1,"front":"In the L-shape example, the bottom length is 13 cm and the top length is 6 cm. What is the missing right width?","back":"$13-6=7$ cm","mcqOptions":[{"text":"$13-6=7$ cm","correct":true,"explanation":"The full bottom length consists of the 6 cm top section and the missing right width. Subtracting gives $13-6=7$, so the missing width is 7 cm."},{"text":"$13+6=19$ cm","correct":false,"explanation":"Adding treats the two given lengths as separate adjoining sections, but 6 cm is already part of the total 13 cm. Subtraction gives $13-6=7$ cm."},{"text":"$13-6=8$ cm","correct":false,"explanation":"The correct operation has been chosen, but the subtraction contains an arithmetic error. Since $13-6=7$, the missing right width is 7 cm, not 8 cm."},{"text":"$13-6=6$ cm","correct":false,"explanation":"This repeats the known top length instead of calculating the difference. The missing section completes 6 cm to 13 cm, so its width is $13-6=7$ cm."}]},{"id":"4e005f1b-6e78-49a7-b6ae-20c88fb199b9","cardType":"basic","points":2,"front":"An L-shape is split into rectangles $6\\text{ cm}\\times8\\text{ cm}$ and $7\\text{ cm}\\times5\\text{ cm}$. What is its total area?","back":"83 cm²","mcqOptions":[{"text":"83 cm²","correct":true,"explanation":"The rectangle areas are $6\\times8=48$ cm² and $7\\times5=35$ cm². Adding the non-overlapping pieces gives $48+35=83$ cm²."},{"text":"65 cm²","correct":false,"explanation":"This can result from using 5 cm as the height of both rectangles: $6\\times5+7\\times5=65$. The correct areas are 48 cm² and 35 cm², totalling 83 cm²."},{"text":"90 cm²","correct":false,"explanation":"This can result from incorrectly using 6 cm as the second rectangle’s height. Its height is 5 cm, so the total is $6\\times8+7\\times5=83$ cm²."},{"text":"104 cm²","correct":false,"explanation":"This is the area of a $13\\text{ cm}\\times8\\text{ cm}$ bounding rectangle, which includes the missing part of the L-shape. Adding the two given rectangles gives 83 cm²."}]},{"id":"b855b6a4-3434-41bf-9c15-3fb7d06c4467","cardType":"basic","points":1,"front":"In a right-angled triangle, which two sides are the base and height?","back":"The two sides that meet at the right angle.","mcqOptions":[{"text":"The two sides that meet at the right angle.","correct":true,"explanation":"The base and perpendicular height must meet at 90°. In a right-angled triangle, the two sides forming the right angle therefore provide the base and height."},{"text":"The two sides that are equal in length.","correct":false,"explanation":"Equal sides identify an isosceles triangle but do not necessarily give a perpendicular base and height. In a right-angled triangle, use the two sides meeting at the right angle."},{"text":"The two sides that meet at an acute angle.","correct":false,"explanation":"Sides meeting at an acute angle are not perpendicular, so they cannot directly serve as base and height. Use the two sides that meet at the 90° angle."},{"text":"The two sides that include the hypotenuse.","correct":false,"explanation":"The hypotenuse does not meet either adjacent side at 90°, so a perpendicular height would need to be constructed. The simpler base and height meet at the right angle."}]},{"id":"6234f3c8-dd25-4967-ac11-654a222f43d2","cardType":"basic","points":2,"front":"A compound shape has a $3\\text{ cm}\\times10\\text{ cm}$ rectangle and a triangle with base 6 cm and height 4 cm. What is the total area?","back":"42 cm²","mcqOptions":[{"text":"42 cm²","correct":true,"explanation":"The rectangle has area $3\\times10=30$ cm². The triangle has area $\\frac{1}{2}\\times6\\times4=12$ cm², giving a total of $30+12=42$ cm²."},{"text":"54 cm²","correct":false,"explanation":"This uses $6\\times4=24$ cm² for the triangle and forgets to halve it. The triangle is 12 cm², so the total area is $30+12=42$ cm²."},{"text":"36 cm²","correct":false,"explanation":"This can result from dividing the triangle’s base-height product by 4 rather than 2. Its area is 12 cm², which added to 30 cm² gives 42 cm²."},{"text":"12 cm²","correct":false,"explanation":"This is only the triangle’s area, so the rectangle has been omitted. Adding the rectangle’s 30 cm² to the triangle’s 12 cm² gives 42 cm²."}]},{"id":"080101f2-aff7-434f-9b8d-9f21e10ad747","cardType":"basic","points":1,"front":"For a shaded region with a hole, what area calculation is used?","back":"Whole area - unshaded area","mcqOptions":[{"text":"Whole area - unshaded area","correct":true,"explanation":"The whole area includes both the shaded region and the hole. Subtracting the unshaded hole from the whole area leaves exactly the required shaded area."},{"text":"Whole area + unshaded area","correct":false,"explanation":"Adding the unshaded area counts the hole twice rather than removing it. The shaded region is found by subtracting the unshaded hole from the whole area."},{"text":"Whole area × unshaded area","correct":false,"explanation":"Multiplying two areas produces inappropriate squared-area units and does not remove the hole. The required calculation is whole area minus unshaded area."},{"text":"Whole area ÷ unshaded area","correct":false,"explanation":"Dividing the areas gives a ratio rather than the shaded area. Because the hole lies inside the whole shape, its unshaded area must be subtracted."}]},{"id":"63e7193a-a0f7-4c33-9dac-9a9379a17033","cardType":"cloze","points":null,"front":"If there are several identical holes, subtract [ ] of them, not just [ ].","back":"If there are several identical holes, subtract all of them, not just one.","mcqOptions":[{"text":"If there are several identical holes, subtract all of them, not just one.","correct":true,"explanation":"Every hole removes area from the compound shape. For identical holes, calculate one hole’s area, multiply by the number of holes, and subtract the resulting total."},{"text":"If there are several identical holes, subtract one of them, not just all.","correct":false,"explanation":"This reverses the required instruction. Every hole reduces the total area, so the area of all the holes must be subtracted, rather than the area of only one."},{"text":"If there are several identical holes, subtract half of them, not just one.","correct":false,"explanation":"Subtracting only half the holes leaves some removed regions included in the answer. The correct method is to subtract the combined area of all the holes."},{"text":"If there are several identical holes, subtract none of them, not just one.","correct":false,"explanation":"Ignoring the holes gives the area of the complete outer shape rather than the remaining material. The areas of all the holes must be subtracted."}]},{"id":"b635179b-bf2d-4e7f-ad1d-8962f416974d","cardType":"basic","points":2,"front":"A wall is $5\\text{ m}\\times3\\text{ m}$ and a door is $1\\text{ m}\\times2\\text{ m}$. What is the painted area?","back":"13 m²","mcqOptions":[{"text":"13 m²","correct":true,"explanation":"The wall has area 5 × 3 = 15 m² and the door has area 1 × 2 = 2 m². Therefore, the painted area is 15 - 2 = 13 m²."},{"text":"17 m²","correct":false,"explanation":"This comes from adding the door’s 2 m² area to the wall’s 15 m² area. The door is unpainted, so its area must be subtracted, giving 13 m²."},{"text":"14 m²","correct":false,"explanation":"This subtracts only 1 m², using one door dimension instead of its area. The door covers 1 × 2 = 2 m², so the painted area is 13 m²."},{"text":"12 m²","correct":false,"explanation":"This treats the door’s dimensions as an area by adding 1 + 2 and subtracting 3 m². Area requires multiplication, so subtracting 2 m² gives 13 m²."}]},{"id":"2fba455c-9793-4c0f-8266-6a7b954d3197","cardType":"basic","points":2,"front":"A triangle has base 14 cm and height 10 cm. A $4\\text{ cm}\\times3\\text{ cm}$ rectangle is removed. What is the shaded area?","back":"58 cm²","mcqOptions":[{"text":"58 cm²","correct":true,"explanation":"The triangle’s area is ½ × 14 × 10 = 70 cm². The removed rectangle has area 4 × 3 = 12 cm², leaving 70 - 12 = 58 cm²."},{"text":"82 cm²","correct":false,"explanation":"This adds the rectangle’s 12 cm² area to the triangle’s 70 cm² area. Because the rectangle is removed, its area must be subtracted, giving 58 cm²."},{"text":"66 cm²","correct":false,"explanation":"This subtracts the rectangle’s width of 4 cm as though it were an area. The rectangle’s area is 4 × 3 = 12 cm², so the shaded area is 58 cm²."},{"text":"128 cm²","correct":false,"explanation":"This calculates the triangle as 14 × 10, forgetting the factor of ½. The triangle’s area is 70 cm², and subtracting 12 cm² leaves 58 cm²."}]},{"id":"656c7b7d-20c8-4fd4-91c2-11601c70af56","cardType":"cloze","points":null,"front":"If a price is given per square metre, the area must be in [ ] before multiplying by the price.","back":"If a price is given per square metre, the area must be in m² before multiplying by the price.","mcqOptions":[{"text":"m²","correct":true,"explanation":"A price per square metre applies to an area measured in square metres. Converting the area to m² before multiplying ensures that the units match the given rate."},{"text":"cm²","correct":false,"explanation":"Square centimetres do not match a price quoted per square metre. The area must first be converted to m², otherwise the calculated cost will be incorrectly scaled."},{"text":"m","correct":false,"explanation":"Metres measure length rather than area, so they cannot be multiplied directly by a price per square metre. The required area unit is m²."},{"text":"m³","correct":false,"explanation":"Cubic metres measure volume, not area. A rate per square metre must be multiplied by an area expressed in m² to produce the correct price."}]},{"id":"e2002087-f9ce-4f93-9aad-95d77b52e7e9","cardType":"basic","points":3,"front":"A field is a $70\\text{ m}\\times90\\text{ m}$ rectangle plus a triangle with base 80 m and height 65 m. It sells for £4 per m². What is received?","back":"£35,600","mcqOptions":[{"text":"£35,600","correct":true,"explanation":"The rectangle’s area is 70 × 90 = 6,300 m² and the triangle’s area is ½ × 80 × 65 = 2,600 m². Multiplying 8,900 m² by £4 gives £35,600."},{"text":"£25,200","correct":false,"explanation":"This uses only the rectangle’s 6,300 m² area and omits the triangular section. Including the triangle gives 8,900 m², which sells for £35,600."},{"text":"£46,000","correct":false,"explanation":"This calculates the triangular area as 80 × 65, forgetting to divide by 2. Using 2,600 m² for the triangle gives a total sale price of £35,600."},{"text":"£8,900","correct":false,"explanation":"This is the field’s total area in square metres, not the amount received. Multiplying 8,900 m² by the price of £4 per m² gives £35,600."}]}]}],"examCountdown":{"examDate":"2027-05-14","paperLabel":"Paper 1"},"heading":"Area of Compound Shapes","title":"Area of Compound Shapes Flashcards","topbar":{"backHref":"/igcse/maths/edexcel/flashcards"},"topicSidebar":"$L38"}]}],"topicId":"0b31b976-faaa-4193-8f77-2e7e7a3e6cb8","walled":"$L39"}],"$L3a","$L3b"]