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hypotenuse."},{"text":"Acute-angled triangles (containing only angles below 90°).","correct":false,"explanation":"Confusing an acute angle with a right angle makes this tempting. Standard Pythagoras' theorem requires a right-angled triangle containing an angle of exactly 90°."},{"text":"Obtuse-angled triangles (containing an angle above 90°).","correct":false,"explanation":"An obtuse triangle has no right angle, so the standard Pythagorean relationship does not apply. The required triangle is right-angled, with one angle equal to 90°."},{"text":"Equilateral triangles (containing three 60° angles).","correct":false,"explanation":"Equilateral triangles have three equal 60° angles rather than a right angle. Pythagoras' theorem requires a right-angled triangle containing a 90° angle."}]},{"id":"52515c36-58ba-c8c7-95d5-73c2daf94c02","cardType":"cloze","points":null,"front":"The hypotenuse is the [ ] side and is always opposite the right angle.","back":"The hypotenuse is the longest side and is always opposite the right angle.","mcqOptions":[{"text":"longest","correct":true,"explanation":"The hypotenuse lies opposite the 90° angle and is always the longest side of a right-angled triangle."},{"text":"shortest","correct":false,"explanation":"Mistaking the hypotenuse for one of the perpendicular sides reverses the length relationship. The hypotenuse is the longest side because it lies opposite the largest angle."},{"text":"vertical","correct":false,"explanation":"A triangle can be rotated, so the hypotenuse is not identified by its orientation. It is the longest side and lies opposite the right angle."},{"text":"horizontal","correct":false,"explanation":"Orientation does not determine which side is the hypotenuse, since diagrams may be rotated. The hypotenuse is the longest side opposite the right angle."}]},{"id":"744f8bb0-883c-65e7-c7ac-15b9523400f9","cardType":"cloze","points":null,"front":"The hypotenuse is the longest side and is always [ ] the right angle.","back":"The hypotenuse is the longest side and is always opposite the right angle.","mcqOptions":[{"text":"opposite","correct":true,"explanation":"The hypotenuse is identified as the side directly opposite the 90° angle, and it is consequently the longest side of the right-angled triangle."},{"text":"adjacent to","correct":false,"explanation":"The two shorter sides are adjacent to the right angle, which can cause this confusion. The hypotenuse is the side opposite the right angle."},{"text":"parallel to","correct":false,"explanation":"Sides and angles are not described as parallel to one another. The hypotenuse is identified by being opposite the right angle."},{"text":"beneath","correct":false,"explanation":"A diagram's orientation is irrelevant, so the hypotenuse need not appear beneath the right angle. It is always the side opposite the right angle."}]},{"id":"04cc53b5-5eaa-b96a-1164-889eeccf33e7","cardType":"cloze","points":null,"front":"In Pythagoras' theorem, the formula is $\\text{[...]}$, where $c$ represents the hypotenuse.","back":"In Pythagoras' theorem, the formula is $\\boxed{a^2 + b^2 = c^2}$, where $c$ represents the hypotenuse.","mcqOptions":[{"text":"$$a^2 + b^2 = c^2$","correct":true,"explanation":"In a right-angled triangle, the sum of the squares of the two shorter sides equals the square of the hypotenuse, represented by $c$."},{"text":"$$a + b = c$","correct":false,"explanation":"Adding the side lengths directly omits the essential squaring operation. Pythagoras' theorem states that $a^2 + b^2 = c^2$ when $c$ is the hypotenuse."},{"text":"$$a^2 - b^2 = c^2$","correct":false,"explanation":"Subtraction is used only after rearranging to find a shorter side. With $c$ as the hypotenuse, the original formula is $a^2 + b^2 = c^2$."},{"text":"$$a^2 + b^2 = c$","correct":false,"explanation":"Failing to square the hypotenuse produces an inconsistent relationship. The squares of the shorter sides add to the square of the hypotenuse: $a^2 + b^2 = c^2$."}]},{"id":"747debab-2218-a6da-85fa-4869fc6c1c93","cardType":"cloze","points":null,"front":"In Pythagoras' theorem, the formula is $a^2 + b^2 = c^2$, where $c$ represents the [ ].","back":"In Pythagoras' theorem, the formula is $a^2 + b^2 = c^2$, where $c$ represents the hypotenuse.","mcqOptions":[{"text":"hypotenuse","correct":true,"explanation":"In the standard Pythagorean formula, $c$ denotes the hypotenuse, which is the longest side and lies opposite the right angle."},{"text":"shortest side","correct":false,"explanation":"The letters $a$ and $b$ represent the two shorter sides, while $c$ represents the hypotenuse - the longest side opposite the right angle."},{"text":"right angle","correct":false,"explanation":"The variable $c$ represents a side length, not an angle. In the standard formula, it denotes the hypotenuse opposite the right angle."},{"text":"triangle area","correct":false,"explanation":"Pythagoras' theorem relates squared side lengths rather than directly calculating area. The variable $c$ represents the hypotenuse of the right-angled triangle."}]},{"id":"604096dd-382e-929b-b8aa-4e05c186cb26","cardType":"basic","points":2,"front":"How is the hypotenuse calculated if the two shorter sides are known?","back":"

Square both sides
Add the squares
Square root the result

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

Square both sides
Add the squares
Square root the result

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

Square both sides
Subtract the squares
Square root the result

","correct":false,"explanation":"Subtraction is used when the hypotenuse is already known and a shorter side is missing. To find the hypotenuse, square the shorter sides, add, then square root."},{"text":"

Add both sides
Square the total
Square root the result

","correct":false,"explanation":"Adding the lengths before squaring introduces an unwanted $2ab$ term. Each shorter side must be squared separately, the squares added, and the result square-rooted."},{"text":"

Square root both sides
Add the results
Square the total

","correct":false,"explanation":"Reversing the operations does not follow $c^2=a^2+b^2$. Square each known shorter side, add those squares, then take the square root to obtain $c$."}]},{"id":"a1f4b6d1-5bfd-d4d3-19b6-79f4fb1b3a98","cardType":"basic","points":2,"front":"How is a shorter side calculated if the hypotenuse and one other side are known?","back":"

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

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

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

","correct":true,"explanation":"Rearranging $a^2+b^2=c^2$ gives a missing shorter side squared as $c^2$ minus the known shorter side squared. Square-rooting then gives the side length."},{"text":"

Square both sides
Add the smaller square to the hypotenuse square
Square root the result

","correct":false,"explanation":"Addition finds a hypotenuse when both shorter sides are known. To find a shorter side, subtract the known shorter-side square from the hypotenuse square, then square root."},{"text":"

Square both 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 is longest. Subtract the smaller square from the hypotenuse square before taking the square root."},{"text":"

Subtract both sides
Square the remaining difference
Square root the result

","correct":false,"explanation":"Subtracting the original lengths ignores that Pythagoras relates their squares. Square both known sides first, subtract the smaller square from the hypotenuse square, then square root."}]},{"id":"dfdcce67-4af5-c3c4-534c-ae732cce27d3","cardType":"basic","points":1,"front":"When solving for a missing side, why would you use subtraction instead of addition?","back":"Because the hypotenuse (longest side) is already known.","mcqOptions":[{"text":"Because the hypotenuse (longest side) is already known.","correct":true,"explanation":"When the hypotenuse is known, rearranging $a^2+b^2=c^2$ to find a shorter side requires subtracting the known shorter-side square from the hypotenuse square."},{"text":"Because the shorter side (smallest side) is already known.","correct":false,"explanation":"Knowing one shorter side does not by itself determine the operation; the key is that the hypotenuse is known. Finding the other shorter side therefore requires subtraction."},{"text":"Because the right angle (90° angle) is already known.","correct":false,"explanation":"The right angle confirms that Pythagoras can be used, but it does not cause subtraction. Subtraction is used because the hypotenuse is already known."},{"text":"Because the triangle area (inside space) is already known.","correct":false,"explanation":"The triangle's area is not part of the Pythagorean side-length calculation. Subtraction is required when the hypotenuse is known and a shorter side must be found."}]},{"id":"6f07b426-b0b1-5c2a-a483-b3e369173c02","cardType":"cloze","points":null,"front":"To square a number like $x$, you [ ] ($x \\times x$).","back":"To square a number like $x$, you multiply it by itself ($x \\times x$).","mcqOptions":[{"text":"multiply it by itself","correct":true,"explanation":"Squaring means multiplying a number by itself once. Therefore, $x$ squared is $x \\times x$, written as $x^2$."},{"text":"multiply it by two","correct":false,"explanation":"Multiplying by two doubles the number rather than squaring it. To square $x$, multiply it by itself, giving $x \\times x = x^2$."},{"text":"add it to itself","correct":false,"explanation":"Adding a number to itself produces $2x$, which is the same as doubling it. Squaring requires multiplication by itself: $x \\times x = x^2$."},{"text":"divide it by itself","correct":false,"explanation":"Dividing a non-zero number by itself gives $1$, not its square. Squaring $x$ means multiplying it by itself to obtain $x^2$."}]},{"id":"c217c4c6-88a9-0597-b1d2-b859e148a2af","cardType":"basic","points":1,"front":"In a rectangle, what does the diagonal represent in relation to Pythagoras' theorem?","back":"The hypotenuse of a right-angled triangle.","mcqOptions":[{"text":"The hypotenuse of a right-angled triangle.","correct":true,"explanation":"A rectangle’s diagonal joins opposite corners and forms right-angled triangles with its sides. As the side opposite the right angle, it is the hypotenuse."},{"text":"The shortest side of a right-angled triangle.","correct":false,"explanation":"The diagonal is longer than either adjoining side, so it cannot be the shortest side. It lies opposite the right angle and is therefore the hypotenuse."},{"text":"The perpendicular height of a right-angled triangle.","correct":false,"explanation":"A rectangle’s side may act as the perpendicular height, but its diagonal does not. The diagonal is opposite the right angle, making it the hypotenuse."},{"text":"The horizontal base of a right-angled triangle.","correct":false,"explanation":"The horizontal edge of the rectangle may form the triangle’s base. The diagonal lies opposite the right angle, so it represents the hypotenuse."}]},{"id":"0f8af3e9-9914-e162-f25c-135141890dad","cardType":"basic","points":2,"front":"How do you create a right-angled triangle from an isosceles triangle to find its height?","back":"Draw a perpendicular height from the top, which bisects (halves) the base.","mcqOptions":[{"text":"Draw a perpendicular height from the top, which bisects (halves) the base.","correct":true,"explanation":"In an isosceles triangle, the perpendicular from the apex to the base also bisects the base. This produces two congruent right-angled triangles for using Pythagoras."},{"text":"Draw a parallel height from the top, which doubles the length of the base.","correct":false,"explanation":"A height must be perpendicular to the base, not parallel, and drawing it does not double the base. In an isosceles triangle, it bisects the base."},{"text":"Draw a perpendicular height from the top, which bisects (halves) an equal side.","correct":false,"explanation":"The apex height in an isosceles triangle bisects the base, not either equal sloping side. This perpendicular creates two right-angled triangles suitable for Pythagoras."},{"text":"Draw a diagonal line from the top, which trisects (thirds) the base.","correct":false,"explanation":"The required construction is a perpendicular height, and symmetry makes it halve rather than trisect the base. This creates two congruent right-angled triangles."}]},{"id":"e371b1f1-8e4f-242c-6232-ac5d00c64e71","cardType":"basic","points":1,"front":"What is the general strategy for a trapezium problem using Pythagoras?","back":"Draw in a height line to create a right-angled triangle.","mcqOptions":[{"text":"Draw in a height line to create a right-angled triangle.","correct":true,"explanation":"A perpendicular height between the parallel sides creates a right-angled triangle. Its known dimensions can then be connected using Pythagoras’ theorem."},{"text":"Draw in a diagonal line to create an equilateral triangle.","correct":false,"explanation":"A diagonal does not generally create an equilateral triangle in a trapezium. Drawing a perpendicular height creates the right-angled triangle needed for Pythagoras."},{"text":"Draw in a midpoint line to create an isosceles triangle.","correct":false,"explanation":"A midpoint line does not necessarily produce a right angle or an isosceles triangle. A perpendicular height is drawn to form a right-angled triangle."},{"text":"Draw in a parallel line to create an obtuse-angled triangle.","correct":false,"explanation":"Pythagoras applies directly to right-angled triangles, not obtuse-angled triangles. A height perpendicular to the parallel sides creates the necessary right angle."}]},{"id":"d363f546-1a2f-f524-7d80-1ba344598dec","cardType":"basic","points":1,"front":"What is the first step when a question involves two joined right-angled triangles?","back":"Find the length of the shared side first.","mcqOptions":[{"text":"Find the length of the shared side first.","correct":true,"explanation":"The shared side links the two right-angled triangles. Calculating it from the triangle with sufficient information provides the missing length needed for the second application of Pythagoras."},{"text":"Find the length of the longest hypotenuse first.","correct":false,"explanation":"The longest hypotenuse often depends on an unknown length from the other triangle, so it cannot usually be found immediately. Calculate the shared side first."},{"text":"Find the size of the largest angle first.","correct":false,"explanation":"Pythagoras relates side lengths and does not require calculating an angle first. The shared side should be found because it supplies information for the second triangle."},{"text":"Find the area of the larger triangle first.","correct":false,"explanation":"Area is not normally needed to connect the two applications of Pythagoras. Finding the shared side first gives a required side length in the other triangle."}]},{"id":"00e5c7df-78dd-f374-39be-d30195500051","cardType":"basic","points":1,"front":"In worded problems, what angle is formed by a vertical wall and horizontal ground?","back":"90° (a right angle).","mcqOptions":[{"text":"90° (a right angle).","correct":true,"explanation":"Vertical and horizontal lines are perpendicular, so the angle where a wall meets level ground is $90°$. This creates the right-angled triangle required for Pythagoras."},{"text":"45° (an acute angle).","correct":false,"explanation":"A $45°$ angle may occur in an isosceles right-angled triangle, but not between vertical and horizontal lines. A wall and level ground meet at $90°$."},{"text":"180° (a straight angle).","correct":false,"explanation":"A $180°$ angle is formed by opposite rays along one straight line. Vertical and horizontal lines are perpendicular, so they meet at $90°$."},{"text":"360° (a full turn).","correct":false,"explanation":"A $360°$ angle represents one complete turn rather than the angle between two perpendicular lines. A vertical wall and horizontal ground form a $90°$ angle."}]},{"id":"940f9698-ba9f-ccf7-e167-e30f9c7662e4","cardType":"basic","points":1,"front":"A ladder leans against a wall. Which part of the triangle does the ladder represent?","back":"The hypotenuse.","mcqOptions":[{"text":"The hypotenuse.","correct":true,"explanation":"The wall and ground meet at the right angle, and the ladder lies opposite that angle. It is therefore the hypotenuse and the triangle’s longest side."},{"text":"The vertical height.","correct":false,"explanation":"The vertical height is the section of wall from the ground to the ladder’s top. The sloping ladder lies opposite the right angle, so it is the hypotenuse."},{"text":"The horizontal base.","correct":false,"explanation":"The horizontal base is the distance along the ground from the wall to the ladder’s foot. The ladder is the sloping hypotenuse."},{"text":"The right-angle vertex.","correct":false,"explanation":"The right-angle vertex is the point where the wall meets the ground, not a side of the triangle. The ladder forms the hypotenuse."}]},{"id":"9b8bc974-3a39-c423-287d-f17fc04e7b57","cardType":"basic","points":1,"front":"If a journey goes North then East, what is the geometric relationship between the two paths?","back":"They meet at a 90° angle.","mcqOptions":[{"text":"They meet at a 90° angle.","correct":true,"explanation":"North and East are perpendicular compass directions, so the two stages of the journey meet at $90°$. They form the shorter sides of a right-angled triangle."},{"text":"They meet at a 45° angle.","correct":false,"explanation":"A $45°$ direction would be North-East, halfway between North and East. Travelling North and then East involves a $90°$ turn."},{"text":"They meet at a 180° angle.","correct":false,"explanation":"A $180°$ angle would mean reversing direction along the same straight line, such as North then South. North and East are perpendicular and meet at $90°$."},{"text":"They meet at a 360° angle.","correct":false,"explanation":"A $360°$ angle is a complete turn and returns to the original direction. North and East differ by a quarter-turn, which is $90°$."}]},{"id":"eb63b5f8-0e49-36c9-c345-f8a6c6a61721","cardType":"basic","points":1,"front":"What must you check regarding units before starting a Pythagoras calculation?","back":"Ensure all side lengths are in the same units (e.g., all cm or all m).","mcqOptions":[{"text":"Ensure all side lengths are in the same units (e.g., all cm or all m).","correct":true,"explanation":"Pythagoras combines squared side lengths, so every measurement must first use the same unit. Otherwise, the numerical values cannot be substituted into the formula consistently."},{"text":"Ensure the two shorter side lengths are in different units (e.g., one cm and one m).","correct":false,"explanation":"Using different units makes the squared lengths incompatible. All side lengths must be converted into the same units, such as all centimetres or all metres, before calculating."},{"text":"Ensure all side lengths are in squared units (e.g., all cm² or all m²).","correct":false,"explanation":"Side lengths use linear units, not squared units; squaring occurs when applying the formula. The original measurements must all be in the same linear unit."},{"text":"Ensure only the hypotenuse is in standard units (e.g., cm rather than m).","correct":false,"explanation":"Converting only the hypotenuse leaves incompatible measurements in the calculation. Every side length involved must be expressed in the same unit before Pythagoras is applied."}]},{"id":"3f53b521-d705-cfd7-ff8e-92874e6a5f2e","cardType":"cloze","points":null,"front":"In the exam, you should keep [ ] until the final line before rounding.","back":"In the exam, you should keep full calculator values until the final line before rounding.","mcqOptions":[{"text":"full calculator values","correct":true,"explanation":"Keeping full calculator values avoids introducing intermediate rounding errors. Round only the final answer to the accuracy requested in the question."},{"text":"rounded calculator values","correct":false,"explanation":"Rounding during intermediate steps can accumulate errors and change the final answer. Keep full calculator values throughout, then round only on the final line."},{"text":"estimated calculator values","correct":false,"explanation":"Estimates are useful for checking reasonableness, but not for the main calculation. Keep full calculator values until the final line to preserve accuracy."},{"text":"truncated calculator values","correct":false,"explanation":"Truncating removes digits without rounding and reduces the calculation’s accuracy. Full calculator values should be retained until the final answer is rounded appropriately."}]},{"id":"9ebbfffb-6d36-ccf6-3eb9-f7450d5e5e35","cardType":"basic","points":2,"front":"What are the two initial steps recommended before writing the formula in an exam?","back":"

Mark the right angle
Label the hypotenuse

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

Mark the right angle
Label the hypotenuse

","correct":true,"explanation":"Marking the right angle confirms that Pythagoras applies, while labelling the hypotenuse identifies the side opposite that angle and prevents incorrect substitution."},{"text":"

Measure the right angle
Label the shortest side

","correct":false,"explanation":"A diagram may not be drawn to scale, so measuring is unreliable, and the shortest side has no special role. Mark the right angle and label the hypotenuse."},{"text":"

Mark the acute angles
Label the longest leg

","correct":false,"explanation":"The acute angles and longer leg do not determine the formula’s arrangement. Mark the right angle and label the hypotenuse, which is opposite it."},{"text":"

Measure the side lengths
Label the right angle

","correct":false,"explanation":"Measurements should come from the question rather than the drawing, which may not be to scale. Mark the right angle and label the hypotenuse first."}]},{"id":"dbc0c0e3-6c74-03a6-859f-155a7c66740b","cardType":"basic","points":1,"front":"What is the square root of a number?","back":"The positive number which, when multiplied by itself, produces the original number.","mcqOptions":[{"text":"The positive number which, when multiplied by itself, produces the original number.","correct":true,"explanation":"The square root reverses squaring: multiplying the positive square root by itself gives the original number. For example, $csqrt{49}=7$ because $7ctimes7=49$."},{"text":"The negative number which, when multiplied by itself, produces the original number.","correct":false,"explanation":"Although a negative number can have the same square, the square root symbol denotes the positive value. For example, $csqrt{49}=7$, not $-7$."},{"text":"The positive number which, when multiplied by two, produces the original number.","correct":false,"explanation":"Multiplying by two finds a relationship involving halves, not square roots. A square root is the positive number that produces the original number when multiplied by itself."},{"text":"The positive number which, when added to itself, produces the original number.","correct":false,"explanation":"Adding a number to itself finds half of the original number. A square root is found by identifying the positive number whose square equals the original."}]},{"id":"0d79faca-3d41-20c7-3e13-602efc8bca59","cardType":"basic","points":1,"front":"How do you find the side $EF$ if $EF^2 = 207$?","back":"Calculate $\\sqrt{207}$.","mcqOptions":[{"text":"Calculate $\\sqrt{207}$.","correct":true,"explanation":"Since $EF^2=207$, applying the square root to both sides gives $EF=csqrt{207}$. A length uses the positive square root."},{"text":"Calculate $207^2$.","correct":false,"explanation":"Squaring again applies the same operation rather than reversing it, producing a much larger value. Take the square root of $207$ to find $EF$."},{"text":"Calculate $cfrac{207}{2}$.","correct":false,"explanation":"Dividing by two reverses multiplication by two, not squaring. Since $EF$ has been squared, calculate $csqrt{207}$ to undo that operation."},{"text":"Calculate $csqrt{207^2}$.","correct":false,"explanation":"Squaring $207$ before taking its square root simply returns $207$, not the required side length. Apply the square root directly to $207$."}]}]}],"examCountdown":{"examDate":"2027-05-14","paperLabel":"Paper 1"},"heading":"Pythagoras","title":"Pythagoras Flashcards","topbar":{"backHref":"/gcse/maths/ocr/flashcards"},"topicSidebar":"$L36"}]}],"topicId":"a2fb6550-47c7-4135-98a9-e6acb1096efc","walled":"$L37"}],"$L38","$L39"]