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This correct method earns M1."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$A=\\htmlData{anchor=product}{\\color{@green}{13\\times 13}}$"},{"latex":"$$A=\\htmlData{anchor=area}{\\color{@green}{169\\text{ cm}^2}}$"}],"highlights":[{"color":"green","style":"text-color","anchor":"product"},{"color":"green","style":"text-color-with-background","anchor":"area"}],"connections":[{"to":"area","from":"product","color":"green","label":"calculate"}]},"order":2,"title":"Calculate the area","description":"Use the area formula for a square: multiply the side length by itself. The correct area earns A1."}]}],"generatedAt":"2026-07-19T19:17:03.604Z","modelVersion":"gpt-5.6-sol","explanationVersion":"v3"},"options":[]}],"options":[],"explanation":{"methods":[{"id":"7b6c2bbe-3f50-4ada-a9c4-d4bffdd8c4e1:two_step_arithmetic","label":"Two-Step Arithmetic Approach","steps":[{"type":"text","order":0,"title":"Find the side length","content":"A square has four equal sides. To find the length of one side, we divide the total perimeter by $4$.\n\n$$\\text{Side length} = 52 \\div 4 = 13\\text{ cm}$$\n\nThis calculation earns you the first mark (\\textbf{M1}) on the exam.","highlights":[{"text":"perimeter of 52 cm","location":"specification"}]},{"type":"text","order":1,"title":"Calculate the area","content":"Now that we know each side is $13\\text{ cm}$ long, we can find the area of the square. The area is found by squaring the side length (multiplying the side length by itself):\n\n$$\\text{Area} = 13 \\times 13 = 169\\text{ cm}^2$$\n\nCompleting this calculation correctly earns you the final accuracy mark (\\textbf{A1})."}]},{"id":"7b6c2bbe-3f50-4ada-a9c4-d4bffdd8c4e1:direct_formula","label":"Direct Perimeter-Area Formula","steps":[{"type":"text","order":0,"title":"Recall the direct formula","content":"To find the area $A$ of a square directly from its perimeter $P$, we can use the formula:\n\n$$A = \\left(\\frac{P}{4}\\right)^2$$\n\nThis formula works because a square has four equal sides, so dividing the perimeter by $4$ gives the side length $s = \\frac{P}{4}$. Squaring this value gives the area."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$P = \\htmlData{anchor=p-val}{\\color{@sky}{52}}\\text{ cm}$"},{"latex":"$$A = \\left(\\frac{\\htmlData{anchor=p-var}{\\color{@sky}{P}}}{4}\\right)^2$"},{"latex":"$$A = \\left(\\frac{\\color{@sky}{52}}{4}\\right)^2$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"p-val"},{"color":"sky","style":"text-color","anchor":"p-var"}],"connections":[{"to":"p-var","from":"p-val","color":"sky","label":"substitute","style":"solid"}]},"order":1,"title":"Substitute the perimeter","description":"We substitute the given perimeter into our area formula."},{"type":"text","order":2,"title":"Calculate the final area","content":"Now, let's evaluate the expression inside the brackets first, and then square it:\n\n1. Divide the perimeter by $4$ to find the side length:\n $$52 \\div 4 = 13\\text{ cm}$$\n *(This step earns you the method mark M1)*\n\n2. Square the side length to find the area:\n $$13^2 = 13 \\times 13 = 169\\text{ cm}^2$$\n *(This step earns you the final accuracy mark A1)*","calculator":[{"gdc":{"expression":"(52/4)^2","instructions":"Enter (52/4)^2 on the main calculation screen."},"ti84":{"expression":"(52/4)^2","instructions":"Type (52/4)^2 and press ENTER."},"desmos":{"expression":"\\left(\\frac{52}{4}\\right)^{2}","instructions":"Type (52/4)^2 into the input box to see the result."},"description":"You can quickly compute this calculation using any standard calculator."}]},{"type":"text","order":3,"title":"State the final answer","content":"The area of the square photo is **$169\\text{ cm}^2$**."}]}],"generatedAt":"2026-06-12T18:50:49.960Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"firstAttemptFullMarks":{"percent":72},"walkthroughVideo":{"url":"https://assets.mathsgenie.co.uk/walkthroughs/review/db6b2d7f-47bb-49c9-881c-6aac2ba22df0.mp4","durationSeconds":69,"posterUrl":"https://assets.mathsgenie.co.uk/walkthroughs/review/db6b2d7f-47bb-49c9-881c-6aac2ba22df0.jpg"}},{"id":"bebc0c3f-b5b9-40d5-97b4-8643e4373891","revisionId":"2485574b-e542-4a7b-8e96-81325fcbc878","questionNumber":null,"questionType":"LA","specification":"A plan for a garden patio, in the shape of a trapezium, is drawn on a centimetre grid.\n\n![A trapezium on a centimetre grid with parallel sides of length 3 cm and 7 cm, and a perpendicular height of 5 cm.](https://assets.mathsgenie.co.uk/question-image-redraws/sol-max-v1/bebc0c3f-b5b9-40d5-97b4-8643e4373891/image-1-5cfd2f423d6f0570372cfa8e.svg)","difficultyBand":"easy","tier":null,"totalMarks":2,"sourceTitle":null,"calculatorUse":"non_calc","isCalculation":true,"parts":[{"id":"201e2fd4-29a2-4839-9486-bb94cb76436c","label":"a","content":"Work out the area of the patio represented on the grid.","markscheme":"- Identifies dimensions from grid: $a = 3, b = 7, h = 5$ M1\n- Calculates area: $\\frac{1}{2}(3 + 7) \\times 5 = 25$ A1","marks":2,"order":0,"markingRubric":{"totalMarks":2,"points":[{"id":"p1","marks":1,"description":"Identifies dimensions $a=3, b=7, h=5$ from the grid","code":"M"},{"id":"p2","marks":1,"description":"Calculates the area correctly using the trapezium formula","code":"A"}],"finalAnswers":["25"],"generatedAt":"2026-05-06T13:58:48.577Z","modelVersion":"gemini-3-flash-preview"},"tableImages":[],"explanation":{"methods":[{"id":"trapezium_area","label":"Use the trapezium formula","steps":[{"type":"text","order":0,"title":"Read the dimensions","content":"The two **parallel** sides are $a=3\\text{ cm}$ and $b=7\\text{ cm}$. The perpendicular height is $h=5\\text{ cm}$; the right-angle symbol confirms that this is measured at $90^\\circ$ to the parallel sides.\n\nIdentifying these three dimensions earns **M1**.","highlights":[{"text":"A plan for a garden patio, in the shape of a trapezium, is drawn on a centimetre grid.","location":"specification"}]},{"type":"text","order":1,"title":"Choose the formula","content":"For a trapezium, use\n\n$$A=\\frac{1}{2}(a+b)h$$\n\nwhere $a$ and $b$ are the lengths of the parallel sides and $h$ is the perpendicular height."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$A=\\dfrac{1}{2}\\htmlData{anchor=formula_values}{\\color{@sky}{(a+b)h}}$"},{"latex":"$$A=\\dfrac{1}{2}\\htmlData{anchor=substituted_values}{\\color{@sky}{(3+7)\\times 5}}$"},{"latex":"$$A=\\dfrac{1}{2}\\times 10\\times 5$"},{"latex":"$$A=25\\text{ cm}^2$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"formula_values"},{"color":"sky","style":"text-color","anchor":"substituted_values"}],"connections":[{"to":"substituted_values","from":"formula_values","color":"sky","label":"substitute"}]},"order":2,"title":"Substitute and calculate","description":"Substitute the three measurements into the area formula. The calculation (1/2)(3 + 7) × 5 = 25 earns A1; because this is an area, give the answer as 25 cm²."}]}],"generatedAt":"2026-07-19T15:29:01.108Z","modelVersion":"gpt-5.6-sol","explanationVersion":"v3"},"options":[]}],"options":[],"explanation":{"methods":[{"id":"201e2fd4-29a2-4839-9486-bb94cb76436c:trapezium_formula","label":"Using Trapezium Formula","steps":[{"type":"text","order":0,"title":"Identify the parallel sides","content":"To find the area of the trapezium, we first identify the lengths of the two parallel sides, which we will call $a$ and $b$, by counting the grid squares:\n- The shorter parallel side (top) is $a = 3\\text{ cm}$.\n- The longer parallel side (bottom) is $b = 7\\text{ cm}$."},{"type":"text","order":1,"title":"Identify the height","content":"Next, find the perpendicular height ($h$) of the trapezium by counting the vertical grid squares between the two parallel sides:\n- The perpendicular height is $h = 5\\text{ cm}$."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$a = \\htmlData{anchor=a-val}{\\color{@sky}{3}}\\text{ cm}, \\quad b = \\htmlData{anchor=b-val}{\\color{@amber}{7}}\\text{ cm}, \\quad h = \\htmlData{anchor=h-val}{\\color{@purple}{5}}\\text{ cm}$"},{"latex":"$$Area = \\frac{1}{2}\\left(\\htmlData{anchor=a-sub}{\\color{@sky}{a}} + \\htmlData{anchor=b-sub}{\\color{@amber}{b}}\\right)\\htmlData{anchor=h-sub}{\\color{@purple}{h}}$"},{"latex":"$$Area = \\frac{1}{2}\\left(\\color{@sky}{3} + \\color{@amber}{7}\\right) \\times \\color{@purple}{5}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"a-val"},{"color":"amber","style":"text-color","anchor":"b-val"},{"color":"purple","style":"text-color","anchor":"h-val"}],"connections":[{"to":"a-sub","from":"a-val","color":"sky","label":null,"style":"solid"},{"to":"b-sub","from":"b-val","color":"amber","label":null,"style":"solid"},{"to":"h-sub","from":"h-val","color":"purple","label":null,"style":"solid"}]},"order":2,"title":"Apply the trapezium formula","description":"Now, we substitute these values into the standard formula for the area of a trapezium: $$Area = \\frac{1}{2}(a + b)h$$"},{"type":"text","order":3,"title":"Calculate the final area","content":"Calculate the value step-by-step:\n\n$$\\begin{aligned} Area &= \\frac{1}{2}(3 + 7) \\times 5 \\\\ Area &= \\frac{1}{2}(10) \\times 5 \\\\ Area &= 5 \\times 5 \\\\ Area &= 25 \\text{ cm}^2 \\end{aligned}$$\n\nThis calculation earns **2 marks** in total: **1 mark** for correctly identifying the dimensions and writing down the correct calculation, and **1 mark** for the correct final answer of $25$."}]},{"id":"201e2fd4-29a2-4839-9486-bb94cb76436c:splitting_shapes","label":"Splitting into Simpler Shapes","steps":[{"type":"text","order":0,"title":"Identify the dimensions","content":"First, count the squares on the centimetre grid to find the dimensions of the trapezium:\n\n* **Top parallel side:** $3\\text{ cm}$\n* **Bottom parallel side:** $7\\text{ cm}$\n* **Perpendicular height:** $5\\text{ cm}$","highlights":[{"text":"3 cm","location":"specification"},{"text":"7 cm","location":"specification"},{"text":"5 cm","location":"specification"}]},{"type":"text","order":1,"title":"Split the trapezium","content":"We can split the trapezium into three simpler shapes:\n\n1. A **central rectangle** with a width of $3\\text{ cm}$ and a height of $5\\text{ cm}$.\n2. **Two right-angled triangles** on the sides.\n\nSince the total bottom width is $7\\text{ cm}$ and the rectangle takes up $3\\text{ cm}$, the remaining width for the two triangle bases is:\n\n$$7\\text{ cm} - 3\\text{ cm} = 4\\text{ cm}$$\n\nIf we split this symmetrically, each triangle has a base of $2\\text{ cm}$ and a height of $5\\text{ cm}$."},{"type":"hyper_latex","block":{"nodes":[{"text":"Area of the central rectangle:"},{"latex":"$$\\text{Area}_{\\text{rectangle}} = \\htmlData{anchor=rect-w}{\\color{@sky}{3}} \\times \\htmlData{anchor=rect-h}{\\color{@amber}{5}} = \\htmlData{anchor=rect-ans}{\\color{@purple}{15\\text{ cm}^2}}$"},{"text":"Area of the two side triangles:"},{"latex":"$$\\text{Area}_{\\text{two triangles}} = 2 \\times \\left( \\frac{1}{2} \\times \\htmlData{anchor=tri-b}{\\color{@teal}{2}} \\times \\htmlData{anchor=tri-h}{\\color{@amber}{5}} \\right) = \\htmlData{anchor=tri-ans}{\\color{@orange}{10\\text{ cm}^2}}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"rect-w"},{"color":"amber","style":"text-color","anchor":"rect-h"},{"color":"purple","style":"box","anchor":"rect-ans"},{"color":"teal","style":"text-color","anchor":"tri-b"},{"color":"amber","style":"text-color","anchor":"tri-h"},{"color":"orange","style":"box","anchor":"tri-ans"}],"connections":[{"to":"tri-h","from":"rect-h","color":"amber","label":"same height","style":"dashed"}]},"order":2,"title":"Calculate individual areas","description":"Calculate the area of the rectangle and the triangles separately using their base and height dimensions."},{"type":"text","order":3,"title":"Sum the areas","content":"Add the areas of the shapes together to get the total area of the patio:\n\n$$\\text{Total Area} = 15\\text{ cm}^2 + 10\\text{ cm}^2 = 25\\text{ cm}^2$$"}]},{"id":"201e2fd4-29a2-4839-9486-bb94cb76436c:counting_squares","label":"Counting Grid Squares","steps":[{"type":"text","order":0,"title":"Identify the method","content":"We want to find the area of the patio represented on the grid. We will do this by counting the centimetre grid squares ($1\\text{ cm}^2$ each) that make up the shape."},{"type":"text","order":1,"title":"Count the full squares","content":"First, let's count the number of grid squares that are completely inside the trapezium.\n\nLooking at the central rectangular region of the shape:\n- The middle section has a width of $3\\text{ cm}$ (matching the top parallel side) and a height of $5\\text{ cm}$.\n- This central region contains exactly $$3 \\times 5 = 15\\text{ full squares}$$\n\nEach of these represents $1\\text{ cm}^2$, giving us an initial area of $15\\text{ cm}^2$."},{"type":"text","order":2,"title":"Analyze the partial squares","content":"Now, let's look at the triangular sections on either side of the central rectangle:\n- On the left side, there is a right-angled triangular section that has a width of $2\\text{ cm}$ and a height of $5\\text{ cm}$.\n- On the right side, there is another right-angled triangular section with a width of $2\\text{ cm}$ and a height of $5\\text{ cm}$.\n\nLet's pair up the partial squares from these two matching sloped sides:\n- For every partial square on the left side, there is a corresponding partial square on the right side that completes it to make exactly $1$ full square.\n- For example, a square that is $\\frac{1}{5}$ filled on one side matches with a square that is $\\frac{4}{5}$ filled on the other side."},{"type":"text","order":3,"title":"Combine the partial squares","content":"Since both triangular sections are identical, we can combine them to form a single rectangle:\n- Combining the left and right sloped sections gives a rectangle with a width of $2\\text{ cm}$ and a height of $5\\text{ cm}$.\n- This combined region gives us: \n\n$$2 \\times 5 = 10\\text{ squares}$$"},{"type":"text","order":4,"title":"Calculate total area","content":"Now we add the full squares and the combined partial squares together:\n\n$$\\text{Total Area} = 15\\text{ (full squares)} + 10\\text{ (combined partial squares)} = 25\\text{ cm}^2$$\n\nThis gives us a final area of $25\\text{ cm}^2$."}]}],"generatedAt":"2026-06-13T04:15:56.372Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"firstAttemptFullMarks":{"percent":86},"walkthroughVideo":null}],"subjectCode":"gcse_maths__wjec","summaries":[{"id":"01be8c93-1a0d-42bc-806e-03e9b874ef5f","difficultyBand":"medium","calculatorUse":"non_calc","totalMarks":3,"questionType":"LA","isCalculation":true},{"id":"04166146-be94-4688-a5f4-216a10be292f","difficultyBand":"medium","calculatorUse":"non_calc","totalMarks":2,"questionType":"LA","isCalculation":true},{"id":"0fafeb26-4bc6-463d-a87d-b7c573e423a1","difficultyBand":"medium","calculatorUse":"non_calc","totalMarks":2,"questionType":"LA","isCalculation":false},{"id":"1bd40506-c128-49c5-97ca-4366f6603309","difficultyBand":"medium","calculatorUse":null,"totalMarks":2,"questionType":"LA","isCalculation":true},{"id":"3c2e6f35-38a3-4e22-b73e-a38851bfb3a5","difficultyBand":"hard","calculatorUse":"non_calc","totalMarks":2,"questionType":"LA","isCalculation":false},{"id":"7d21c775-4905-441e-a4ce-42c69e118e0d","difficultyBand":"medium","calculatorUse":"non_calc","totalMarks":2,"questionType":"LA","isCalculation":true},{"id":"abac78c3-d0ef-472c-8bdc-ee4e190ac2a8","difficultyBand":"easy","calculatorUse":"non_calc","totalMarks":2,"questionType":"LA","isCalculation":true},{"id":"b3db4d8a-8110-4824-a7d9-a28ec1b8b5ce","difficultyBand":"medium","calculatorUse":"non_calc","totalMarks":3,"questionType":"LA","isCalculation":true},{"id":"b9dc3aab-e5e1-4de9-8e37-5f1ede282010","difficultyBand":"medium","calculatorUse":"non_calc","totalMarks":2,"questionType":"LA","isCalculation":false},{"id":"bf814d0a-30ed-43d5-a5d9-8cbf6ebecd0c","difficultyBand":"medium","calculatorUse":"non_calc","totalMarks":2,"questionType":"LA","isCalculation":true},{"id":"c22d2f4c-67fb-4321-ad92-22100cbb0e36","difficultyBand":"medium","calculatorUse":"non_calc","totalMarks":2,"questionType":"LA","isCalculation":true},{"id":"f4ca9396-b5af-4307-b2da-dfcfcdb0e488","difficultyBand":"medium","calculatorUse":"non_calc","totalMarks":2,"questionType":"LA","isCalculation":false},{"id":"b455e793-ccf5-48ee-86a8-0254cd572a6e","difficultyBand":"easy","calculatorUse":"non_calc","totalMarks":2,"questionType":"LA","isCalculation":false},{"id":"00f209b2-e194-460a-b72f-8d72226bbb34","difficultyBand":"medium","calculatorUse":"non_calc","totalMarks":3,"questionType":"LA","isCalculation":true},{"id":"32530915-a67a-4823-bac7-18ecf671bcdc","difficultyBand":"easy","calculatorUse":null,"totalMarks":2,"questionType":"LA","isCalculation":true},{"id":"42cf4e2a-45fd-4acc-9f3a-cca0bc9a2a06","difficultyBand":"medium","calculatorUse":"non_calc","totalMarks":5,"questionType":"LA","isCalculation":true},{"id":"431fc7b7-00b2-4d98-9014-bbfc69423c50","difficultyBand":"medium","calculatorUse":"non_calc","totalMarks":2,"questionType":"LA","isCalculation":false},{"id":"4d85da44-7606-4a1d-a0d8-c92a790b16e9","difficultyBand":"medium","calculatorUse":"non_calc","totalMarks":3,"questionType":"LA","isCalculation":true},{"id":"66e634c3-9e2e-4afb-b9b4-c6e9ad011244","difficultyBand":"easy","calculatorUse":"non_calc","totalMarks":2,"questionType":"LA","isCalculation":true},{"id":"9f5dbab7-3e8b-463d-bc31-1097fe60dae9","difficultyBand":"medium","calculatorUse":"non_calc","totalMarks":2,"questionType":"LA","isCalculation":false},{"id":"b92ce3ef-bdf0-42c0-8a3f-26991716fa9e","difficultyBand":"medium","calculatorUse":"non_calc","totalMarks":2,"questionType":"LA","isCalculation":false},{"id":"bebc0c3f-b5b9-40d5-97b4-8643e4373891","difficultyBand":"easy","calculatorUse":"non_calc","totalMarks":2,"questionType":"LA","isCalculation":true},{"id":"cb6ef5fc-8d27-4e8f-9170-a8a43e116ca1","difficultyBand":"medium","calculatorUse":"non_calc","totalMarks":2,"questionType":"LA","isCalculation":true},{"id":"d36ef591-127c-434a-8132-a6c13cca7c42","difficultyBand":"medium","calculatorUse":"non_calc","totalMarks":2,"questionType":"LA","isCalculation":true},{"id":"e6f75db0-1957-421c-a34f-35253501d76b","difficultyBand":"medium","calculatorUse":"non_calc","totalMarks":2,"questionType":"LA","isCalculation":true},{"id":"ef444e44-505f-44e3-912c-f32aedfddd59","difficultyBand":"medium","calculatorUse":"non_calc","totalMarks":2,"questionType":"LA","isCalculation":false},{"id":"60177e3b-ba2e-4982-84b8-fc35d7371f57","difficultyBand":"medium","calculatorUse":"non_calc","totalMarks":2,"questionType":"LA","isCalculation":false},{"id":"0dec45ac-afe5-4d57-8578-23af1da54bbf","difficultyBand":"easy","calculatorUse":"non_calc","totalMarks":2,"questionType":"LA","isCalculation":true},{"id":"330e504a-a911-4430-9bcb-bc07f92f0b92","difficultyBand":"medium","calculatorUse":"non_calc","totalMarks":2,"questionType":"LA","isCalculation":true},{"id":"5343f430-1f80-4e21-b59c-3be96ae0ec2d","difficultyBand":"medium","calculatorUse":"non_calc","totalMarks":3,"questionType":"LA","isCalculation":true},{"id":"57e38bb8-e6f9-493b-b05d-dc12edf635b0","difficultyBand":"medium","calculatorUse":"non_calc","totalMarks":2,"questionType":"LA","isCalc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and Perimeter","topicId":"a318dc6d-8c6d-4372-5fbd-6b007306229d","topicSidebar":"$L3a","topicSlug":"grade-2-2-11-area-and-perimeter","topicTitle":"Area and Perimeter"}]}],"$L3b","$L3c"]}]

Area and Perimeter

Question 4

Area and Perimeter Questions

Area and Perimeter

Question 4

Area and Perimeter Questions