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answers correct to 1 decimal place.","markscheme":"- Uses ratio to set up Pythagoras: $(4x)^2 + (3x)^2 = 51^2$ M1\n- $16x^2 + 9x^2 = 2601 \\Rightarrow 25x^2 = 2601 \\Rightarrow x = 10.2$ M1\n- Length $= 4 \\times 10.2 = 40.8$, Width $= 3 \\times 10.2 = 30.6$ A1","marks":3,"order":0,"markingRubric":{"totalMarks":3,"points":[{"id":"p1","marks":1,"description":"Uses the ratio to set up Pythagoras: $(4x)^2 + (3x)^2 = 51^2$.","code":"M","acceptable":["$$(4x)^2 + (3x)^2 = 51^2$"]},{"id":"p2","marks":1,"description":"Solves the equation to obtain $x=10.2$.","code":"M","acceptable":["$$16x^2 + 9x^2 = 2601$","$$25x^2 = 2601$","$$x^2 = 104.04$","$$x = 10.2$"]},{"id":"p3","marks":1,"description":"Finds the length and the width correct to 1 decimal place.","code":"A","acceptable":["Length $=40.8$ and width $=30.6$","Length $=40.8$ inches and width $=30.6$ inches"]}],"finalAnswers":["Length = 40.8 inches, width = 30.6 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triangle","width":760,"height":500,"shapes":[{"id":"ab","type":"line","color":"black","points":[{"x":210,"y":90},{"x":585,"y":90}]},{"id":"bc","type":"line","color":"black","points":[{"x":585,"y":90},{"x":585,"y":190}]},{"id":"cd","type":"line","color":"black","points":[{"x":585,"y":190},{"x":210,"y":390}]},{"id":"ae","type":"line","color":"black","points":[{"x":210,"y":90},{"x":210,"y":190}]},{"id":"ed","type":"line","color":"black","points":[{"x":210,"y":190},{"x":210,"y":390}]},{"id":"ec","dash":"dashed","type":"line","color":"blue","points":[{"x":210,"y":190},{"x":585,"y":190}]},{"id":"right_a","type":"line","color":"grey","points":[{"x":210,"y":114},{"x":234,"y":114},{"x":234,"y":90}]},{"id":"right_e","type":"line","color":"blue","points":[{"x":210,"y":166},{"x":234,"y":166},{"x":234,"y":190}]},{"x":184,"y":62,"id":"label_a","size":"m","text":"A","type":"text"},{"x":598,"y":62,"id":"label_b","size":"m","text":"B","type":"text"},{"x":600,"y":194,"id":"label_c","size":"m","text":"C","type":"text"},{"x":184,"y":408,"id":"label_d","size":"m","text":"D","type":"text"},{"x":184,"y":180,"id":"label_e","size":"s","text":"E","type":"text","color":"blue"},{"x":370,"y":55,"id":"label_ab","gap":8,"size":"m","text":"15 cm","type":"text","anchorTo":"ab","orientation":"above"},{"x":600,"y":130,"id":"label_bc","gap":10,"size":"m","text":"BC","type":"text","anchorTo":"bc","orientation":"right"},{"x":380,"y":330,"id":"label_cd","gap":18,"size":"m","text":"17 cm","type":"text","anchorTo":"cd","orientation":"below"},{"x":370,"y":160,"id":"label_ec","gap":8,"size":"m","text":"15 cm","type":"text","color":"blue","anchorTo":"ec","orientation":"above"},{"x":180,"y":280,"id":"label_x","gap":12,"size":"m","text":"x","type":"text","color":"blue","anchorTo":"ed","orientation":"left"},{"id":"dim_ad","type":"line","color":"grey","points":[{"x":120,"y":90},{"x":120,"y":390}]},{"id":"tick_ad_top","type":"line","color":"grey","points":[{"x":110,"y":90},{"x":130,"y":90}]},{"id":"tick_ad_bottom","type":"line","color":"grey","points":[{"x":110,"y":390},{"x":130,"y":390}]},{"x":55,"y":230,"id":"label_ad","gap":12,"size":"m","text":"12 cm","type":"text","anchorTo":"dim_ad","orientation":"left"}],"caption":"Draw $EC$ parallel to $AB$. Since $AB$ is perpendicular to $AD$, $EC$ is also perpendicular to $AD$; and since $AD$ is parallel to $BC$, $ABCE$ is a rectangle. Therefore $EC=AB=15$ cm and $AE=BC$. Triangle $ECD$ is right-angled at $E$, with hypotenuse $17$ cm and $ED=x$. Constructing and identifying this right-angled triangle, with $EC=15$ cm, earns the first M1."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$x^2+15^2=17^2$"},{"latex":"$$x^2=17^2-15^2$"},{"latex":"$$x^2=64$"},{"latex":"$$x=8$"}]},"order":1,"title":"Apply Pythagoras","description":"Rearranging and evaluating gives 17² − 15² = 64, so x² = 64 and x = 8. Reaching these results earns the second M1."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$AE=12-8$"},{"latex":"$$AE=4$"},{"latex":"$$BC=AE$"},{"latex":"$$BC=4\\text{ cm}$"}]},"order":2,"title":"Subtract from the height","description":"The full height satisfies AD = AE + ED. Since ED is 8 cm, AE is 12 − 8 = 4 cm. Also, AE = BC because ABCE is a rectangle. This correct final length earns A1."}]}],"generatedAt":"2026-07-19T19:14:29.048Z","modelVersion":"gpt-5.6-sol","explanationVersion":"v3"},"options":[]}],"options":[],"explanation":{"methods":[{"id":"ba6e9f65-46d8-4250-9ec4-de68b48c17f4:pythagorean","label":"Using Pythagoras' Theorem","steps":[{"type":"text","order":0,"title":"Analyze the trapezium","content":"In the given trapezium $ABCD$:\n- $AB = 15\\text{ cm}$ is perpendicular to $AD$.\n- $AD = 12\\text{ cm}$ is parallel to $BC$.\n- $DC = 17\\text{ cm}$ is the slanted side.\n\nWe need to find the length of $BC$.","highlights":[{"text":"side AB = 15 cm","location":"specification"},{"text":"side AD = 12 cm","location":"specification"},{"text":"side DC = 17 cm","location":"specification"}]},{"type":"text","order":1,"title":"Construct a right-angled triangle","content":"To find $BC$, let's construct a right-angled triangle by drawing a line from $D$ that is perpendicular to the line of $BC$ (meeting $BC$ at a point $E$).\n\n- The length of this perpendicular height $DE$ is equal to the parallel side $AB$:\n $$DE = 15\\text{ cm}$$\n- This forms a right-angled triangle $DEC$ where:\n - The hypotenuse is $DC = 17\\text{ cm}$\n - One side is $DE = 15\\text{ cm}$\n - The other side is $EC$"},{"type":"text","order":2,"title":"Apply Pythagoras' theorem","content":"Using Pythagoras' theorem ($a^2 + b^2 = c^2$) in the right-angled triangle $DEC$:\n\n$$EC^2 + DE^2 = DC^2$$\n\nSubstitute the known values:\n\n$$\\begin{aligned}\nEC^2 + 15^2 &= 17^2 \\\\\nEC^2 + 225 &= 289 \\\\\nEC^2 &= 289 - 225 \\\\\nEC^2 &= 64 \\\\\nEC &= \\sqrt{64} = 8\\text{ cm}\n\\end{aligned}$$"},{"type":"text","order":3,"title":"Calculate the length of BC","content":"Since $AD = 12\\text{ cm}$, the total parallel height from the top line to the bottom is $12\\text{ cm}$ ($BE = AD = 12\\text{ cm}$).\n\nTo find the length of $BC$, we subtract the segment $EC$ from this parallel height:\n\n$$\\begin{aligned}\nBC &= BE - EC \\\\\nBC &= 12 - 8 \\\\\nBC &= 4\\text{ cm}\n\\end{aligned}$$"}]}],"generatedAt":"2026-06-12T19:30:49.154Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"firstAttemptFullMarks":null,"walkthroughVideo":{"url":"https://assets.mathsgenie.co.uk/walkthroughs/review/120deb03-d0dc-47e0-ab45-7a7f8f86d93b.mp4","durationSeconds":104,"posterUrl":"https://assets.mathsgenie.co.uk/walkthroughs/review/120deb03-d0dc-47e0-ab45-7a7f8f86d93b.jpg"}},{"id":"3afaa395-59f0-433c-9d5f-6d6024c29a62","revisionId":"4593e93b-08a6-45b7-8456-a76e82e7fdb5","questionNumber":null,"questionType":"LA","specification":"","difficultyBand":"hard","tier":null,"totalMarks":3,"sourceTitle":null,"calculatorUse":null,"isCalculation":true,"parts":[{"id":"9a3123f1-e5b6-4d1a-8bdb-e32cde3aaa62","label":null,"content":"A television has a diagonal length of $50$ inches.\n\nThe ratio of the length of the television to the width of the television is $4:3$\n\nCalculate the length and the width of the television. \nGive your answers in exact form.","markscheme":"Uses ratio to set up Pythagoras: $(4x)^2 + (3x)^2 = 50^2$ M1\n$16x^2 + 9x^2 = 2500 \\Rightarrow 25x^2 = 2500 \\Rightarrow x = 10$ M1\nLength $= 4 \\times 10 = 40$, Width $= 3 \\times 10 = 30$ A1","marks":3,"order":0,"markingRubric":{"totalMarks":3,"points":[{"id":"p1","marks":1,"description":"Uses ratio to set up Pythagoras: $(4x)^2 + (3x)^2 = 50^2$","code":"M","acceptable":["$$(4x)^2 + (3x)^2 = 50^2$"]},{"id":"p2","marks":1,"description":"Solves $16x^2 + 9x^2 = 2500$ to obtain $x=10$","code":"M","acceptable":["$$16x^2 + 9x^2 = 2500 \\Rightarrow 25x^2 = 2500 \\Rightarrow x = 10$"]},{"id":"p3","marks":1,"description":"Obtains the exact length and width","code":"A","acceptable":["Length $=40$, width $=30$","$$40$ inches and $30$ inches"]}],"finalAnswers":["Length $=40$ inches, width $=30$ 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Pythagoras

Question 8

Pythagoras Questions

Pythagoras

Question 8

Pythagoras Questions