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It is also the longest side of any right-angled triangle."},{"text":"Directly next to the right angle.","correct":false,"explanation":"The two shorter sides are next to the right angle, so this confuses the hypotenuse with a perpendicular side. The hypotenuse is directly opposite the right angle."},{"text":"Directly opposite the smallest angle.","correct":false,"explanation":"The side opposite the smallest angle is the shortest side, not the hypotenuse. The hypotenuse is directly opposite the right angle."},{"text":"Directly opposite an acute angle.","correct":false,"explanation":"Each shorter side is opposite one of the triangle’s acute angles. The hypotenuse is specifically the side directly opposite the right angle."}]},{"id":"35e3a792-5333-43b8-b038-385f85fcb8fc","cardType":"basic","points":2,"front":"What are the two defining characteristics of a hypotenuse?","back":"

It is opposite the right angle.
It is the longest side.

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

It is opposite the right angle.
It is the longest side.

","correct":true,"explanation":"In every right-angled triangle, the hypotenuse lies opposite the right angle. Since the greatest angle faces the greatest side, it is also the longest side."},{"text":"

It is next to the right angle.
It is the longest side.

","correct":false,"explanation":"The sides next to the right angle are the two shorter perpendicular sides. The hypotenuse is the longest side, but it must also be opposite the right angle."},{"text":"

It is opposite the right angle.
It is the shortest side.

","correct":false,"explanation":"Although the hypotenuse is opposite the right angle, it cannot be the shortest side. The largest angle is opposite the longest side, so the hypotenuse is longest."},{"text":"

It is opposite the smallest angle.
It is an equal side.

","correct":false,"explanation":"This confuses the hypotenuse with properties that may describe other sides. A hypotenuse is always opposite the right angle and is always the triangle’s longest side."}]},{"id":"e195aa46-be48-4f30-b350-b634a1aaa96b","cardType":"cloze","points":null,"front":"Pythagoras' theorem states that $\\text{[...]}$, where $c$ represents the hypotenuse.","back":"Pythagoras' theorem states that $\\boxed{c^2 = a^2 + b^2}$, where $c$ represents the hypotenuse.","mcqOptions":[{"text":"$$c^2 = a^2 + b^2$","correct":true,"explanation":"Pythagoras’ theorem says that the square of the hypotenuse equals the sum of the squares of the two shorter sides."},{"text":"$$c = a^2 + b^2$","correct":false,"explanation":"This omits the square on the hypotenuse and mixes lengths with squared lengths. The correct relationship is $c^2 = a^2 + b^2$."},{"text":"$$c^2 = a + b$","correct":false,"explanation":"This squares only the hypotenuse and incorrectly adds the unsquared shorter sides. Pythagoras requires all three side lengths to be squared: $c^2 = a^2 + b^2$."},{"text":"$$c^2 = a^2 - b^2$","correct":false,"explanation":"Subtraction is used only after rearranging to find a shorter side. In the original theorem, the shorter-side squares are added: $c^2 = a^2 + b^2$."}]},{"id":"16e6a99c-624b-4f8f-8cf8-a6645d628329","cardType":"cloze","points":null,"front":"Pythagoras' theorem states that $c^2 = a^2 + b^2$, where $\\text{[...]}$ represents the hypotenuse.","back":"Pythagoras' theorem states that $c^2 = a^2 + b^2$, where $\\boxed{c}$ represents the hypotenuse.","mcqOptions":[{"text":"$$c$","correct":true,"explanation":"In the standard form $c^2 = a^2 + b^2$, $c$ is assigned to the hypotenuse, while $a$ and $b$ represent the two shorter sides."},{"text":"$$a$","correct":false,"explanation":"In the stated formula, $a$ represents one of the two shorter sides. The hypotenuse is represented by $c$, whose square is isolated on the left."},{"text":"$$b$","correct":false,"explanation":"In the stated formula, $b$ represents one of the two shorter sides. The letter $c$ represents the hypotenuse and therefore the longest side."},{"text":"$$a + b$","correct":false,"explanation":"Adding $a$ and $b$ does not name a side and generally does not give the hypotenuse. In this notation, the hypotenuse is represented by $c$."}]},{"id":"91124281-d66d-4fc1-8755-086f972889a7","cardType":"basic","points":2,"front":"Outline the steps to find the hypotenuse when given two shorter sides.","back":"

Square both shorter sides.
Add the squares together.
Square root the result.

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

Square both shorter sides.
Add the squares together.
Square root the result.

","correct":true,"explanation":"Pythagoras gives $c^2 = a^2 + b^2$. Adding the squared shorter sides finds $c^2$, and taking the square root then gives the hypotenuse $c$."},{"text":"

Square both shorter sides.
Subtract the squares.
Square root the result.

","correct":false,"explanation":"Subtraction is used when finding a missing shorter side, not the hypotenuse. To find the hypotenuse, square both shorter sides, add the squares, then square root."},{"text":"

Add both shorter sides.
Square the total.
Square root the result.

","correct":false,"explanation":"This calculates the sum of the side lengths rather than applying Pythagoras. Square each shorter side separately, add those squares, and then square root the result."},{"text":"

Double both shorter sides.
Add the results together.
Square root the result.

","correct":false,"explanation":"Doubling a side is not the same as squaring it, a common operation error. Pythagoras requires squaring both shorter sides, adding, and then taking the square root."}]},{"id":"ae7122a7-4577-42f5-9355-87e8f26c6dff","cardType":"basic","points":2,"front":"Outline the steps to find a shorter side when given the hypotenuse and one side.","back":"

Square both known sides.
Subtract the smaller square from the hypotenuse square.
Square root the result.

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

Square both known sides.
Subtract the smaller square from the hypotenuse square.
Square root the result.

","correct":true,"explanation":"Rearranging Pythagoras gives a shorter-side square as the hypotenuse square minus the other side’s square. Taking the square root then gives the missing length."},{"text":"

Square both known sides.
Add the two squares together.
Square root the result.

","correct":false,"explanation":"Adding the squares is the method for finding a hypotenuse, so it would produce a length larger than the known hypotenuse. A shorter side requires subtraction before square rooting."},{"text":"

Subtract the two known sides.
Square the difference.
Square root the result.

","correct":false,"explanation":"This wrongly subtracts the lengths before squaring them. Pythagoras uses the squares of the sides, so square first, subtract the squares, and then square root."},{"text":"

Square both known sides.
Subtract the hypotenuse square from the smaller square.
Square root the result.

","correct":false,"explanation":"Reversing the subtraction produces a negative value because the hypotenuse square is larger. Subtract the smaller side’s square from the hypotenuse square, then square root."}]},{"id":"2983b91a-d134-41de-a283-d27f19619de9","cardType":"basic","points":1,"front":"How can you use the length of the hypotenuse to perform a quick sense check on your answer?","back":"The hypotenuse must be longer than either of the two shorter sides.","mcqOptions":[{"text":"The hypotenuse must be longer than either of the two shorter sides.","correct":true,"explanation":"The hypotenuse lies opposite the largest angle, which is the right angle, so it must be longer than each shorter side. Otherwise, the calculated answer is unreasonable."},{"text":"The hypotenuse must be shorter than either of the two shorter sides.","correct":false,"explanation":"This reverses the correct size relationship. The hypotenuse is opposite the largest angle, so it must be longer than each of the two shorter sides."},{"text":"The hypotenuse must be equal to either of the two shorter sides.","correct":false,"explanation":"Even in an isosceles right-angled triangle, only the two shorter sides are equal. The hypotenuse remains longer than either of them."},{"text":"The hypotenuse must be twice as long as either of the two shorter sides.","correct":false,"explanation":"There is no fixed doubling relationship between the hypotenuse and a shorter side. The reliable sense check is only that the hypotenuse must be longer than each shorter side."}]},{"id":"32278bdb-4dd9-4183-a1b2-8ca824eaecc0","cardType":"cloze","points":null,"front":"When finding a shorter side, you must calculate $\\text{[...]}$ rather than $(15 - 8)^2$ because Pythagoras uses the squares of the sides.","back":"When finding a shorter side, you must calculate $\\boxed{15^2 - 8^2}$ rather than $(15 - 8)^2$ because Pythagoras uses the squares of the sides.","mcqOptions":[{"text":"$$15^2 - 8^2$","correct":true,"explanation":"Pythagoras works with squared side lengths, so the correct rearrangement subtracts the known shorter-side square from the hypotenuse square: $15^2 - 8^2$."},{"text":"$$15^2 + 8^2$","correct":false,"explanation":"Adding the squares would calculate a new hypotenuse rather than a missing shorter side. Since $15$ is the hypotenuse, the required calculation is $15^2 - 8^2$."},{"text":"$$15 - 8^2$","correct":false,"explanation":"This squares only one of the two known lengths, mixing lengths with squared lengths. Both values must be squared before subtraction, giving $15^2 - 8^2$."},{"text":"$$(15 - 8)^2$","correct":false,"explanation":"Squaring the difference is not equivalent to subtracting the squares, because $(15 - 8)^2$ includes an extra cross term. Pythagoras requires $15^2 - 8^2$."}]},{"id":"4b73fe85-9d7f-4d0f-b2a4-3471beae0d16","cardType":"cloze","points":null,"front":"When finding a shorter side, you must calculate $15^2 - 8^2$ rather than $\\text{[...]}$ because Pythagoras uses the squares of the sides.","back":"When finding a shorter side, you must calculate $15^2 - 8^2$ rather than $\\boxed{(15 - 8)^2}$ because Pythagoras uses the squares of the sides.","mcqOptions":[{"text":"$$(15 - 8)^2$","correct":true,"explanation":"Pythagoras requires the difference between the squares of the hypotenuse and known shorter side. Therefore, the correct calculation is $15^2 - 8^2$, not the square of their difference."},{"text":"$$(15 + 8)^2$","correct":false,"explanation":"Adding the lengths before squaring confuses Pythagoras with a perimeter calculation. The incorrect alternative is being contrasted with $(15 - 8)^2$, while the required calculation is $15^2 - 8^2$."},{"text":"$$15 - 8$","correct":false,"explanation":"Subtracting the unsquared lengths ignores that Pythagoras relates squares. The incorrect alternative is $(15 - 8)^2$, whereas the shorter side is found using $15^2 - 8^2$."},{"text":"$$15^2 - 8$","correct":false,"explanation":"Squaring only the hypotenuse and then subtracting the other length applies the powers inconsistently. The required calculation squares both known side lengths: $15^2 - 8^2$."}]},{"id":"470e8001-841b-432f-89e8-605f5a298aa7","cardType":"basic","points":2,"front":"When calculating the height of an isosceles triangle, what must you do to the base first?","back":"Halve the base to create a right-angled triangle.","mcqOptions":[{"text":"Halve the base to create a right-angled triangle.","correct":true,"explanation":"The height of an isosceles triangle bisects its base and meets it at a right angle. Using half the base therefore creates a right-angled triangle for Pythagoras."},{"text":"Double the base to create a right-angled triangle.","correct":false,"explanation":"Doubling the base produces a length outside the original triangle and does not form the required right-angled triangle. The base must be halved because the height bisects it."},{"text":"Square the base to create a right-angled triangle.","correct":false,"explanation":"Squaring the base is part of a later Pythagoras calculation, not the geometric construction. First halve the base because an isosceles triangle’s height divides it equally."},{"text":"Keep the whole base to create a right-angled triangle.","correct":false,"explanation":"Using the whole base overlooks that the height bisects an isosceles triangle. Halving the base creates one of the two congruent right-angled triangles needed for Pythagoras."}]},{"id":"94820e73-7ed8-4d90-9992-9417376f23e7","cardType":"cloze","points":null,"front":"In an isosceles triangle problem, the height splits the base into [ ].","back":"In an isosceles triangle problem, the height splits the base into two equal halves.","mcqOptions":[{"text":"two equal halves","correct":true,"explanation":"In an isosceles triangle, the perpendicular height from the apex is also a median. It bisects the base, producing two segments of exactly equal length."},{"text":"two unequal parts","correct":false,"explanation":"The height from the apex of an isosceles triangle bisects the base, so the resulting parts cannot be unequal. It creates two equal halves."},{"text":"three equal thirds","correct":false,"explanation":"A single height divides the base at one point, producing two sections rather than three. In an isosceles triangle, those two sections are equal halves."},{"text":"two equal sides","correct":false,"explanation":"The height divides the base into segments, not into sides of the triangle. Because it is drawn from the apex of an isosceles triangle, it creates two equal halves."}]},{"id":"9a79f496-57af-4478-9f0a-9b94820e0678","cardType":"basic","points":1,"front":"A ladder is leaning against a wall. Which part of the scenario represents the hypotenuse?","back":"The ladder itself.","mcqOptions":[{"text":"The ladder itself.","correct":true,"explanation":"The wall and ground meet at a right angle, making them the two shorter sides. The ladder lies opposite that right angle, so it is the hypotenuse."},{"text":"The wall itself.","correct":false,"explanation":"The wall forms one of the perpendicular shorter sides of the triangle, not the hypotenuse. The ladder is opposite the right angle and is therefore the hypotenuse."},{"text":"The ground itself.","correct":false,"explanation":"The ground forms one of the two sides meeting at the right angle, so it is a shorter side. The ladder opposite the right angle is the hypotenuse."},{"text":"The contact point itself.","correct":false,"explanation":"The contact point is a vertex rather than a side, so it cannot represent the hypotenuse. The ladder is the side opposite the right angle."}]},{"id":"1bd776e2-591a-40a3-9a2e-e9bec2eda1bd","cardType":"basic","points":1,"front":"What must be done if a question provides side lengths in both metres and centimetres?","back":"Convert all lengths to the same unit before squaring.","mcqOptions":[{"text":"Convert all lengths to the same unit before squaring.","correct":true,"explanation":"Pythagoras can only combine measurements expressed in the same unit. Converting before squaring ensures that every squared length uses a consistent scale."},{"text":"Convert all lengths to the same unit after squaring.","correct":false,"explanation":"Squaring measurements in different units produces incompatible square units, so they cannot be added or subtracted directly. Convert every length to the same unit before squaring."},{"text":"Use all lengths in the given units before squaring.","correct":false,"explanation":"Metres and centimetres represent different scales, so using the values unchanged gives an invalid calculation. All lengths must be converted to the same unit before squaring."},{"text":"Convert only the shortest length to metres before squaring.","correct":false,"explanation":"Converting only one selected side may still leave inconsistent units in the calculation. Every length must be expressed in the same unit before applying Pythagoras and squaring."}]},{"id":"de5c71dc-ab67-4961-b078-36fe6f5cfb0b","cardType":"cloze","points":null,"front":"To maintain accuracy in multi-step problems, keep the [ ] in your calculator until the final step.","back":"To maintain accuracy in multi-step problems, keep the full unrounded value in your calculator until the final step.","mcqOptions":[{"text":"full unrounded value","correct":true,"explanation":"Keeping the full calculator value prevents small rounding errors from accumulating across several operations. Rounding should be performed only once, when giving the final answer."},{"text":"value rounded to 1 d.p.","correct":false,"explanation":"Using a rounded value during later calculations introduces premature-rounding error, which can alter the final result. Keep the full unrounded value until the final step."},{"text":"whole-number value","correct":false,"explanation":"Discarding the decimal part removes potentially important information and can produce a substantially inaccurate result. Keep the full unrounded value in the calculator until the final step."},{"text":"value rounded after each step","correct":false,"explanation":"Repeatedly rounding after each operation allows errors to accumulate throughout a multi-step calculation. Retain the full unrounded value and round only at the final step."}]},{"id":"c19f19a9-cfe2-4540-97b1-dfdf70f280a0","cardType":"cloze","points":null,"front":"To maintain accuracy in multi-step problems, keep the full unrounded value in your calculator until the [ ].","back":"To maintain accuracy in multi-step problems, keep the full unrounded value in your calculator until the final step.","mcqOptions":[{"text":"final step","correct":true,"explanation":"Rounding only at the final step preserves all available calculator precision during intermediate calculations. This prevents small rounding errors from accumulating and changing the final answer."},{"text":"first step","correct":false,"explanation":"Rounding at the first step carries an approximation through every later operation, allowing the error to grow. Keep the full unrounded value until the final step."},{"text":"squaring step","correct":false,"explanation":"Rounding immediately after squaring can distort subsequent subtraction or square-root calculations. The full calculator value should be retained until the final step."},{"text":"penultimate step","correct":false,"explanation":"The penultimate step is still an intermediate stage, so rounding there can affect the last calculation. Retain the full unrounded value until the final step."}]},{"id":"28f7cede-c3d2-490b-afa3-65a9843ac6f4","cardType":"basic","points":1,"front":"How is a right-angled triangle formed within a rectangle?","back":"By drawing a diagonal from one corner to the opposite corner.","mcqOptions":[{"text":"By drawing a diagonal from one corner to the opposite corner.","correct":true,"explanation":"A rectangle’s diagonal divides it into two congruent triangles. Each triangle contains one of the rectangle’s right angles, so both are right-angled triangles."},{"text":"By drawing a line from one corner to an adjacent corner.","correct":false,"explanation":"Joining adjacent corners simply traces one side of the rectangle and does not divide its interior into triangles. A diagonal must join opposite corners."},{"text":"By drawing a line between the midpoints of opposite sides.","correct":false,"explanation":"A line joining the midpoints of opposite sides divides the shape into smaller rectangles, not triangles. Joining one corner to the opposite corner creates right-angled triangles."},{"text":"By drawing a line from the centre to one corner.","correct":false,"explanation":"A line from the centre to one corner does not by itself enclose a triangle within the rectangle. A full diagonal between opposite corners creates two right-angled triangles."}]},{"id":"d3c37e53-986c-4897-83db-339603127902","cardType":"basic","points":2,"front":"What is the general strategy for multi-step Pythagoras questions involving two triangles?","back":"Use Pythagoras to find the shared side first, then use that side to solve the second triangle.","mcqOptions":[{"text":"Use Pythagoras to find the shared side first, then use that side to solve the second triangle.","correct":true,"explanation":"The shared side links the two right-angled triangles. Calculating it from the triangle with sufficient information provides the missing measurement needed to apply Pythagoras to the second triangle."},{"text":"Use Pythagoras on the larger triangle first, then use its hypotenuse to solve the smaller triangle.","correct":false,"explanation":"Starting with the larger triangle does not work unless it already has two known sides. The correct strategy is to calculate the shared side first, then use it in the second triangle."},{"text":"Add the known sides of both triangles first, then square the result to find the missing side.","correct":false,"explanation":"Pythagoras involves adding or subtracting squared side lengths within one right-angled triangle, not combining sides from both. Find the shared side first, then solve the second triangle."},{"text":"Use Pythagoras to find both hypotenuses separately, then subtract them to obtain the shared side.","correct":false,"explanation":"The triangles cannot be solved separately when the shared side is initially unknown. Calculate that shared side from the triangle with enough information, then use it in the other triangle."}]},{"id":"ae219725-02f1-4d34-82a2-706b402519bf","cardType":"cloze","points":null,"front":"If a rectangle's sides are in the ratio 16:9, they should be written algebraically as $\\text{[...]}$ and $9k$ before using Pythagoras.","back":"If a rectangle's sides are in the ratio 16:9, they should be written algebraically as $\\boxed{16k}$ and $9k$ before using Pythagoras.","mcqOptions":[{"text":"$$16k$","correct":true,"explanation":"Multiplying both parts of the ratio by the same scale factor gives side lengths $16k$ and $9k$. Their ratio remains $16:9$ for every positive value of $k$."},{"text":"$$9k$","correct":false,"explanation":"Using $9k$ for both sides would give a ratio of $1:1$, making the rectangle a square. The first side must be $16k$ so the ratio is $16:9$."},{"text":"$$7k$","correct":false,"explanation":"Subtracting the ratio parts, $16-9=7$, does not produce either side length. The ratio requires corresponding multiples $16k$ and $9k$, so the missing expression is $16k$."},{"text":"$$16+k$","correct":false,"explanation":"A ratio scale factor must multiply each ratio part, not be added to it. Therefore the sides are represented by $16k$ and $9k$, preserving the ratio $16:9$."}]},{"id":"460adca5-17f9-431f-bac9-327717944b06","cardType":"cloze","points":null,"front":"If a rectangle's sides are in the ratio 16:9, they should be written algebraically as $16k$ and $\\text{[...]}$ before using Pythagoras.","back":"If a rectangle's sides are in the ratio 16:9, they should be written algebraically as $16k$ and $\\boxed{9k}$ before using Pythagoras.","mcqOptions":[{"text":"$$9k$","correct":true,"explanation":"Both parts of the ratio are multiplied by the same scale factor $k$. Therefore side lengths $16k$ and $9k$ maintain the required ratio of $16:9$."},{"text":"$$16k$","correct":false,"explanation":"Writing both sides as $16k$ creates equal side lengths and therefore a ratio of $1:1$. The second side must be $9k$ to preserve the ratio $16:9$."},{"text":"$$7k$","correct":false,"explanation":"The difference between the ratio parts is $7$, but that is not the second side’s coefficient. The ratio parts must be scaled directly, giving $16k$ and $9k$."},{"text":"$$9+k$","correct":false,"explanation":"Adding $k$ to a ratio part does not apply a common scale factor. Multiplying both parts by $k$ gives $16k$ and $9k$, which retain the ratio $16:9$."}]},{"id":"f8d6d0df-f0ae-4939-8c86-7c567f9a2999","cardType":"basic","points":1,"front":"Why should you NOT identify the hypotenuse simply by looking for the side that \"looks\" longest?","back":"Diagrams may not be to scale; it must be identified as the side opposite the right angle.","mcqOptions":[{"text":"Diagrams may not be to scale; it must be identified as the side opposite the right angle.","correct":true,"explanation":"A drawing’s appearance is unreliable because its lengths may not be proportional. By definition, the hypotenuse is always the side directly opposite the right angle."},{"text":"Diagrams may not show every length; it must be identified as the side next to the right angle.","correct":false,"explanation":"The two sides next to the right angle are the shorter perpendicular sides, not the hypotenuse. The hypotenuse is identified as the side opposite the right angle."},{"text":"Diagrams may be drawn upside down; it must be identified as the side below the right angle.","correct":false,"explanation":"A triangle’s orientation does not determine which side is the hypotenuse. Whether the diagram is rotated or inverted, the hypotenuse is the side opposite the right angle."},{"text":"Diagrams may contain equal sides; it must be identified as the side touching the right angle.","correct":false,"explanation":"The sides touching the right angle are the two perpendicular shorter sides. The hypotenuse does not touch the right-angle vertex; it lies opposite that angle."}]},{"id":"bcdbddb2-ca6f-45c3-b883-9de5c753228c","cardType":"cloze","points":null,"front":"Rounding to 3 significant figures means keeping the first three [ ] digits and rounding the last one based on the following digit.","back":"Rounding to 3 significant figures means keeping the first three non-zero digits and rounding the last one based on the following digit.","mcqOptions":[{"text":"non-zero","correct":true,"explanation":"Significant-figure counting begins at the first non-zero digit, so leading zeros are ignored. Keep the first three significant digits, then inspect the fourth to decide whether to round."},{"text":"decimal","correct":false,"explanation":"Counting decimal digits describes decimal-place rounding, not significant-figure rounding. For significant figures, begin at the first non-zero digit and retain three significant digits before rounding."},{"text":"even","correct":false,"explanation":"Whether digits are odd or even has no role in deciding which digits to retain. Significant figures are counted from the first non-zero digit, followed by rounding from the next digit."},{"text":"final","correct":false,"explanation":"Significant figures are not simply the final three digits of a number. Counting begins at the first non-zero digit, and the next digit determines whether the third is rounded."}]},{"id":"1c698c53-0495-454d-9897-4b6812eaaba6","cardType":"basic","points":2,"front":"In a worded problem, what physical features often indicate the two shorter sides of a right-angled triangle?","back":"Features that meet at a 90° angle, such as a vertical wall and horizontal ground.","mcqOptions":[{"text":"Features that meet at a 90° angle, such as a vertical wall and horizontal ground.","correct":true,"explanation":"Perpendicular physical features form the two shorter sides of a right-angled triangle. A vertical wall and horizontal ground meet at $90°$, so the sloping side is the hypotenuse."},{"text":"Features that meet at a 45° angle, such as a sloping roof and diagonal ladder.","correct":false,"explanation":"An angle of $45°$ does not by itself identify the shorter sides of a right-angled triangle. Look for perpendicular features meeting at $90°$, such as a vertical wall and horizontal ground."},{"text":"Features that run in parallel directions, such as a vertical wall and upright pole.","correct":false,"explanation":"Parallel features do not meet, so they cannot form two sides at a right-angle vertex. The shorter sides are usually perpendicular features, such as a vertical wall and horizontal ground."},{"text":"Features that form a straight line, such as a horizontal road and level ground.","correct":false,"explanation":"Collinear features form an angle of $180°$, not a right-angled triangle. Look instead for features meeting at $90°$, especially a vertical feature paired with a horizontal one."}]},{"id":"3261c29a-8427-45e8-9e50-c9bdc556b3a7","cardType":"cloze","points":null,"front":"A [ ] reverses the process of squaring a number.","back":"A square root reverses the process of squaring a number.","mcqOptions":[{"text":"square root","correct":true,"explanation":"Taking a square root undoes squaring because $sqrt{x^2}=x$ for non-negative $x$. For example, squaring $5$ gives $25$, and the square root of $25$ is $5$."},{"text":"cube root","correct":false,"explanation":"A cube root reverses cubing, not squaring. Squaring is reversed by taking the square root, as shown by $5^2=25$ and $sqrt{25}=5$."},{"text":"reciprocal","correct":false,"explanation":"Taking a reciprocal reverses multiplication by a number through division; it does not undo an exponent of $2$. The operation that reverses squaring is taking the square root."},{"text":"logarithm","correct":false,"explanation":"A logarithm is used to find an unknown exponent and is not the direct inverse of squaring. Taking the square root reverses the operation of raising a number to power $2$."}]},{"id":"c3bbc556-19b6-4a48-b0df-c397491fe458","cardType":"basic","points":1,"front":"If the length and width of a rectangle are $a$ and $b$, what is the formula for the diagonal $d$?","back":"$$d = √(a^2 + b^2)$","mcqOptions":[{"text":"$$d = √(a^2 + b^2)$","correct":true,"explanation":"The rectangle’s length and width are perpendicular, while its diagonal is the hypotenuse. Pythagoras gives $d^2=a^2+b^2$, so taking the square root gives the stated formula."},{"text":"$$d = √(a^2 - b^2)$","correct":false,"explanation":"Subtraction is used when the hypotenuse and one shorter side are known. Here $a$ and $b$ are both shorter sides, so their squares must be added: $d=√(a^2+b^2)$."},{"text":"$$d = a^2 + b^2$","correct":false,"explanation":"Adding the squares gives $d^2$, not $d$. A square root is required after applying Pythagoras, so the diagonal is $d=√(a^2+b^2)$."},{"text":"$$d = √(a + b)$","correct":false,"explanation":"Pythagoras requires the side lengths to be squared before they are added. The correct formula is $d=√(a^2+b^2)$, not the square root of their ordinary sum."}]}]}],"examCountdown":{"examDate":"2027-05-14","paperLabel":"Paper 1"},"heading":"Pythagoras","title":"Pythagoras Flashcards","topbar":{"backHref":"/igcse/maths/edexcel/flashcards"},"topicSidebar":"$L38"}]}],"topicId":"33e6e1c3-2193-45d2-9479-a690d09d670e","walled":"$L39"}],"$L3a","$L3b"]