{"version":3,"file":"3G22-I8D10B83.js","sources":["../../src/modules/mathgraph.js","../../src/exercices/3e/3G22.js"],"sourcesContent":["import { loadMG32 } from './loaders.js'\n\n// la liste des exos mtg\nlet listeExos = []\nconst minSize = 100\nconst maxSize = 2000\n// nb max de figures dans le dom\nconst maxFigs = 100\n\n// l'appli mg32, si elle a été chargée\nlet mtgApp\n\nfunction setList (exos) {\n  if (!Array.isArray(exos)) return console.error(Error('Il faut passer une liste d’exercices'))\n  // on vérifie que les exos sont correctement définis, on laisse des trous (avec null) pour conserver les n° d'index qu'on nous a filé\n  listeExos = exos.map((exo, i) => {\n    try {\n      if (exo.typeExercice !== 'MG32') throw Error('pas un exercice MG32')\n      if (!Array.isArray(exo.dimensionsDivMg32) || exo.dimensionsDivMg32.length !== 2) throw Error('propriété dimensionsDivMg32 invalide')\n      exo.dimensionsDivMg32.forEach(dim => {\n        if (typeof dim !== 'number' || dim < minSize || dim > maxSize) throw Error(`dimensionsDivMg32 doit contenir deux number entre ${minSize} et ${maxSize}`)\n      })\n      if (typeof exo.MG32codeBase64 !== 'string' || !exo.MG32codeBase64) throw Error('propriété MG32codeBase64 invalide')\n      return exo\n    } catch (error) {\n      error.message = `Pb avec l’exercice d’index ${i} : ${error.message}`\n      console.error(error, exo)\n      return null\n    }\n  })\n  // reset mtgApp s'il avait déjà été chargé\n  if (mtgApp) {\n    mtgApp.docs.forEach(({ idDoc }) => mtgApp.removeDoc(idDoc))\n    // check à priori inutile, mais si ça déconne parce que la syntaxe a changé autant le savoir tout de suite\n    if (mtgApp.docs.length) console.error(Error('Après purge des docs mathgraph il en reste encore'), mtgApp.docs)\n  }\n}\n\n// on pourrait exporter cette fct, mais pour le moment on reste très lié à l'histoire d'index dans la liste,\n// on la garde donc privée ici\nasync function mg32Display (container, exo) {\n  const { dimensionsDivMg32: [width, height], MG32codeBase64, mg32Editable, MG32codeBase64corr } = exo\n  // on cherche d'abord dans la notre liste\n  let indexExo = listeExos.indexOf(exo)\n  if (indexExo === -1) {\n    console.warn('exo pas trouvé dans notre liste', exo, listeExos)\n    // on essaie de le deduire de l'id du conteneur (qui n'en a pas forcément)\n    const chunks = /^MG32div([0-9]+)$/.exec(container.id)\n    if (chunks) {\n      indexExo = Number(chunks[1])\n    } else {\n      // on cherche le premier dispo pour notre svg\n      indexExo = listeExos.length\n      while (document.getElementById(`MG32div${indexExo}`) && indexExo < maxFigs) indexExo++\n      if (indexExo === maxFigs) throw Error(`Trop de figures mathgraph dans le dom, ${maxFigs} max`)\n    }\n  }\n  const idDoc = `MG32svg${indexExo}`\n  const svgOptions = {\n    width,\n    height,\n    idSvg: idDoc\n  }\n  const mtgOptions = {\n    fig: MG32codeBase64,\n    isEditable: Boolean(mg32Editable)\n  }\n\n  // si on avait déjà chargé mtgApp, faut virer le doc avant de le remettre\n  if (mtgApp) mtgApp.removeDoc(idDoc)\n  mtgApp = await loadMG32(container, svgOptions, mtgOptions)\n  const idDocCorr = MG32codeBase64corr ? `MG32svgcorr${indexExo}` : ''\n  if (idDocCorr) {\n    container = document.getElementById(`MG32divcorr${indexExo}`)\n    if (container) {\n      svgOptions.idSvg = idDocCorr\n      mtgOptions.fig = MG32codeBase64corr\n      mtgOptions.isEditable = false\n      await loadMG32(container, svgOptions, mtgOptions)\n    } else {\n      console.error(Error(`Pas trouvé d’élément #MG32divcorr${indexExo} dans le dom, abandon de l'affichage de la correction`))\n    }\n  }\n\n  // et si y'a un modifier on l'exécute\n  /* avant c'était une string a exécuter en global, avec mtgApp mis au préalable dans window.mtg32App\n\n  let code_pour_modifier_la_figure = listeObjetsExercice[numero_figure].MG32code_pour_modifier_la_figure\n  if (window.mtg32App.docs.length === 1) {\n    code_pour_modifier_la_figure = code_pour_modifier_la_figure.replace('display', 'updateDisplay')\n  }\n  // eslint-disable-next-line no-new-func\n  const modification = new Function('numero_figure', code_pour_modifier_la_figure)\n  modification(numero_figure)\n  */\n  if (typeof exo.mg32init === 'function') await exo.mg32init(mtgApp, idDoc, idDocCorr)\n  return mtgApp\n}\n\n/**\n * Affiche toutes les figures\n */\nexport async function mg32DisplayAll (exos) {\n  // on passe ça à setList, ça sert de contrôle d'intégrité, mais c'est pas vraiment nécessaire,\n  // on pourrait directement appeler mg32Display avec le n° d'index\n  setList(exos)\n  return Promise.all(listeExos.map((exo, i) => {\n    const elt = document.getElementById(`MG32div${i}`)\n    return elt ? mg32Display(elt, exo) : null\n  }))\n}\n","import { fractionSimplifiee, texFractionReduite } from '../../lib/outils/deprecatedFractions.js'\nimport { texteExposant } from '../../lib/outils/ecritures'\nimport { katexPopup2 } from '../../lib/format/message.js'\nimport { numAlpha, sp } from '../../lib/outils/outilString.js'\nimport { pgcd } from '../../lib/outils/primalite.js'\nimport { texNombre } from '../../lib/outils/texNombre.js'\nimport Exercice from '../Exercice.js'\nimport Decimal from 'decimal.js'\nimport { context } from '../../modules/context.js'\nimport { listeQuestionsToContenu, randint } from '../../modules/outils.js'\nimport { mg32DisplayAll } from '../../modules/mathgraph.js'\nexport const titre = 'Know the effects of expansions/reductions on areas and volumes'\n\n/**\n* Problèmes calculs d'aire et de volumes utilisant l'effet d'une réduction sur les aires et les volumes\n* @author Jean-Claude Lhote\n*/\nexport const uuid = '960f9'\nexport const ref = '3G22'\nexport default function AgrandissementReduction () {\n  Exercice.call(this) // Héritage de la classe Exercice()\n  this.nbQuestions = 1\n  this.nbQuestionsModifiable = false\n  context.isHtml ? this.spacingCorr = 3.5 : this.spacingCorr = 1.5\n  context.isHtml ? this.spacing = 3 : this.spacing = 2\n  this.nbCols = 1\n  this.nbColsCorr = 1\n  this.quatrieme = false\n  this.sup = 1 //\n  this.sup2 = 1\n\n  this.nouvelleVersion = function (numeroExercice) {\n    this.listeQuestions = []\n    this.listeCorrections = []\n    let texte, texteCorr, r, r2, h1, h2, h3, c, c2, kprime\n    const pi = Decimal.acos(-1)\n    this.typeExercice = 'MG32'\n    this.dimensionsDivMg32 = [600, 700]\n    let codeBase64\n    let choix\n    if (this.sup === 1) {\n      choix = randint(1, 3)\n    } else if (this.sup === 2) {\n      choix = randint(4, 5)\n    } else {\n      choix = randint(1, 5)\n    }\n    switch (choix) {\n      case 1: { // calcul de l'aire de base, du volume d'une pyramide à base carrée. puis, calcul de la section, du volume de la petite pyramide et du volume du tronc\n        c = new Decimal(randint(30, 60)).div(10)\n        h1 = new Decimal(randint(12, 20)).div(2)\n        h2 = new Decimal(randint(3, h1.floor().sub(1).toNumber()))\n        if (this.sup2 < 3) {\n          if (this.sup2 === 1) {\n            // on veut un coefficient de réduction décimal à 1 chiffre après la virgule\n            while (!h2.div(h1).mul(10).eq(h2.div(h1).mul(10).round())) {\n              c = new Decimal(randint(30, 60)).div(10)\n              h1 = new Decimal(randint(12, 20)).div(2)\n              h2 = new Decimal(randint(3, h1.floor().sub(1).toNumber()))\n            }\n          } else {\n            // coefficient qui peut être décimal avec plus d'un chiffre ou rationnel non décimal.\n            while (h2.div(h1).mul(10).eq(h2.div(h1).mul(10).round())) {\n              c = new Decimal(randint(30, 60)).div(10)\n              h1 = new Decimal(randint(12, 20)).div(2)\n              h2 = new Decimal(randint(3, h1.floor().sub(1)))\n            }\n          }\n        }\n        codeBase64 = 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       if (!context.isHtml) {\n          texte = '\\\\begin{minipage}{0.65 \\\\linewidth} \\n\\t'\n        } else {\n          texte = ''\n        }\n        texte += `SABCD is a pyramid with a square base of height $SO${sp()} = ${sp()}${texNombre(h1, 1)}${sp()}$cm and base side $${texNombre(c, 1)}${sp()}$cm.<br>`\n        texte += ` Point O' is located on the height [SO] at ${texNombre(h2, 0)}${sp()}cm of S.`\n        texte += '<br>A plane parallel to the face ABCD passing through O\\' coupe la pyramide en formant la section A\\'B\\'C\\'D\\'.<br>'\n        if (!context.isHtml) {\n          texte += 'The figure is not in full size.<br>'\n        }\n        texte += numAlpha(0) + ' Calculate the' + katexPopup2(numeroExercice, 1, 'base area of the pyramid', 'Formula: Area of a square of side c', `$Area=\\\\text{c}$${texteExposant(2)}`) + '.<br>'\n        texte += numAlpha(1) + ' Calculate the' + katexPopup2(numeroExercice + 5, 1, 'volume of the pyramid', 'Formula: volume of a pyramid with base area $B$ and height h', '$Volume= \\\\dfrac{B \\\\times \\\\text{h}}{3}$') + ' SABCD.<br>'\n        texte += numAlpha(2) + ' Deduce the area of the' + katexPopup2(numeroExercice + 10, 1, 'section', 'Definition: plane section of a solid', `The section of a solid by a plane is a plane figure.<br>In the case of a section of a pyramid by a plane parallel to its base, this section is a polygon which is a reduction of the base.<br>In a reduction of coefficient k, the areas are multiplied by k${texteExposant(2)} and the volumes are multiplied by k${texteExposant(3)}.`) + ` A'B'C'D' knowing that SO'${sp()} = ${sp()}${texNombre(h2, 0)}${sp()}cm.<br>`\n        texte += numAlpha(3) + ' Calculate the volume of the pyramid SA\\'B\\'C\\'D\\'.<br>'\n        texte += numAlpha(4) + ' Calculate the volume of the pyramid\\'s trunk (part of the pyramid located between the base and the section).'\n        if (context.isHtml) {\n          texte += '<br>Point O can be moved and the viewing angle can be changed &#x3C6;'\n        } else {\n          texte += `\\n\\t \\\\end{minipage} \\n\\t \\\\begin{minipage}{0.35 \\\\linewidth} \\n\\t \\\\begin{tikzpicture}[scale=0.8] \\n\\t\n          \\\\definecolor{hhhhhh}{rgb}{0,0,0}\n          \\\\definecolor{phphph}{rgb}{0.5,0.5,0.5}\n          \\\\definecolor{dpdpdp}{rgb}{0.85,0.85,0.85}\n          \\\\definecolor{ofofof}{rgb}{0.5,0.5,0.5}\n          \\\\definecolor{ffhhhh}{rgb}{1,0,0}\n          \\\\clip (9.38,0) rectangle (0,14.47);\n          \\\\fill[color=black] (4.578,2.328) circle (0.063);\n          \\\\draw [color=black , dotted, line width = 0.4](6.773,0.916)--(8.236,3.174);\n          \\\\draw [color=black , dotted, line width = 0.4](8.236,3.174)--(2.384,3.739);\n          \\\\draw [color=black , dotted, line width = 0.4](2.384,3.739)--(0.921,1.481);\n          \\\\draw [color=black , dotted, line width = 0.4](0.921,1.481)--(6.773,0.916);\n          \\\\draw [color=black , dotted, line width = 0.4](4.578,13.458)--(0.921,1.481);\n          \\\\draw [color=black , dotted, line width = 0.4](4.578,13.458)--(6.773,0.916);\n          \\\\draw [color=black , dotted, line width = 0.4](4.578,13.458)--(8.236,3.174);\n          \\\\draw [color=black , dotted, line width = 0.4](4.578,13.458)--(2.384,3.739);\n          \\\\draw [color=black , line width = 0.8](0.921,1.481)--(6.773,0.916);\n          \\\\draw [color=black , line width = 0.8](6.773,0.916)--(8.236,3.174);\n          \\\\draw [color=black , line width = 0.8](4.578,13.458)--(0.921,1.481);\n          \\\\draw [color=black , line width = 0.8](4.578,13.458)--(6.773,0.916);\n          \\\\draw [color=black , line width = 0.8](4.578,13.458)--(8.236,3.174);\n          \\\\fill[color=black] (4.578,6.501) circle (0.063);\n          \\\\draw [color=black , dotted, line width = 0.4](4.578,2.328)--(8.236,3.174);\n          \\\\draw [color=dpdpdp , dotted, line width = 0.4](0.921,1.481)--(6.773,0.916)--(8.236,3.174)--(2.384,3.739)--(0.921,1.481)--cycle;\n          \\\\fill [color = ofofof, opacity = 0.2](0.921,1.481)--(6.773,0.916)--(8.236,3.174)--(2.384,3.739)--(0.921,1.481)--cycle;\n          \\\\node at (3.891, 14.198) [align=left,below right ,black,,font= \\\\sf \\\\fontsize {0.469cm} {0.586cm} \\\\selectfont] {\\\\textbf{S}};\n          \\\\draw [color=black , dotted, line width = 0.4](4.578,2.328)--(5.065,2.44);\n          \\\\fill[color=black] (5.065,6.614) circle (0.063);\n          \\\\draw [color=black , dotted, line width = 0.4](4.578,6.501)--(5.065,6.614);\n          \\\\draw [color=black , dotted, line width = 0.4](2.384,3.739)--(6.773,0.916);\n          \\\\draw [color=black , dotted, line width = 0.4](0.921,1.481)--(8.236,3.174);\n          \\\\draw [color=black , dotted, line width = 0.4](4.578,2.789)--(4.578,2.328);\n          \\\\draw [color=black , dotted, line width = 0.4](5.065,2.44)--(5.065,2.902);\n          \\\\draw [color=black , dotted, line width = 0.4](5.065,2.902)--(4.578,2.789);\n          \\\\draw [color=black , dotted, line width = 0.4](4.578,2.789)--(4.578,2.328)--(5.065,2.44)--(5.065,2.902)--(4.578,2.789)--cycle;\n          \\\\draw [color=black , dotted, line width = 0.4](4.578,6.963)--(4.578,6.501);\n          \\\\draw [color=black , dotted, line width = 0.4](5.065,6.614)--(5.065,7.075);\n          \\\\draw [color=black , dotted, line width = 0.4](5.065,7.075)--(4.578,6.963);\n          \\\\draw [color=black , dotted, line width = 0.4](4.578,6.963)--(4.578,6.501)--(5.065,6.614)--(5.065,7.075)--(4.578,6.963)--cycle;\n          \\\\draw [color=black , dotted, line width = 0.4](2.292,5.972)--(5.95,5.619)--(6.864,7.031)--(3.207,7.383)--(2.292,5.972)--cycle;\n          \\\\fill [color = ffhhhh, opacity = 0.2](2.292,5.972)--(5.95,5.619)--(6.864,7.031)--(3.207,7.383)--(2.292,5.972)--cycle;\n          \\\\draw [color=ffhhhh , dotted, line width = 0.4](2.292,5.972)--(6.864,7.031);\n          \\\\draw [color=ffhhhh , dotted, line width = 0.4](3.207,7.383)--(5.95,5.619);\n          \\\\draw [color=ffhhhh , line width = 0.8](2.292,5.972)--(5.95,5.619);\n          \\\\draw [color=ffhhhh , line width = 0.8](5.95,5.619)--(6.864,7.031);\n          \\\\node at (0.305, 1.74) [align=left,below right ,black,,font= \\\\sf \\\\fontsize {0.469cm} {0.586cm} \\\\selectfont] {A};\n          \\\\node at (6.742, 1.057) [align=left,below right ,black,,font= \\\\sf \\\\fontsize {0.469cm} {0.586cm} \\\\selectfont] {B};\n          \\\\node at (8.32, 3.571) [align=left,below right ,black,,font= \\\\sf \\\\fontsize {0.469cm} {0.586cm} \\\\selectfont] {C};\n          \\\\node at (1.914, 4.255) [align=left,below right ,black,,font= \\\\sf \\\\fontsize {0.469cm} {0.586cm} \\\\selectfont] {D};\n          \\\\node at (1.544, 6.407) [align=left,below right ,black,,font= \\\\sf \\\\fontsize {0.469cm} {0.586cm} \\\\selectfont] {A'};\n          \\\\node at (6.205, 5.667) [align=left,below right ,black,,font= \\\\sf \\\\fontsize {0.469cm} {0.586cm} \\\\selectfont] {B'};\n          \\\\node at (7.112, 7.564) [align=left,below right ,black,,font= \\\\sf \\\\fontsize {0.469cm} {0.586cm} \\\\selectfont] {C'};\n          \\\\node at (2.201, 7.992) [align=left,below right ,black,,font= \\\\sf \\\\fontsize {0.469cm} {0.586cm} \\\\selectfont] {D'};\n          \\\\node at (3.935, 5.466) [align=left,below right ,black,,font= \\\\sf \\\\fontsize {0.469cm} {0.586cm} \\\\selectfont] {\\\\textbf{O'}};\n          \\\\node at (4.247, 2.292) [align=left,below right ,black,,font= \\\\sf \\\\fontsize {0.469cm} {0.586cm} \\\\selectfont] {\\\\textbf{O}};\n          \\\\draw [color=black , dotted, line width = 0.4](4.578,2.328)--(4.578,13.458);\n          \\\\end{tikzpicture} \\n\\t \\\\end{minipage}`\n        }\n        texteCorr = numAlpha(0) + `The base area of the pyramid is: $${texNombre(c, 1)}^2$ cm${texteExposant(2)} $= ${texNombre(c * c)}$ cm${texteExposant(2)}.<br>`\n        texteCorr += numAlpha(1) + ` The volume of the pyramid is: $\\\\dfrac{A_\\\\text{base} \\\\times \\\\text{height}}{3}$ cm${texteExposant(3)} $= \\\\dfrac{${texNombre(c * c)}\\\\times ${texNombre(h1, 1)}}{3}$ cm${texteExposant(3)} $\\\\approx ${texNombre(c.mul(c).mul(h1).div(3), 3)}$ cm${texteExposant(3)}.<br>`\n        texteCorr += numAlpha(2) + ` The section is a reduction of the coefficient base $\\\\dfrac{${texNombre(h2, 0)}}{${texNombre(h1, 1)}}`\n        if (!Number.isInteger(h1) || pgcd(h2, h1) > 1) {\n          texteCorr += `=${texFractionReduite(h2.mul(10), h1.mul(10))}$.<br>`\n        } else {\n          texteCorr += '.$<br>'\n        }\n        texteCorr += `In a reduction of coefficient k, the areas are multiplied by k${texteExposant(2)}.<br>`\n        texteCorr += `So its area is $\\\\left(${texFractionReduite(h2.mul(10), h1.mul(10))}\\\\right)^2 \\\\times ${texNombre(c * c)}$ cm${texteExposant(2)} $=${texFractionReduite(h2.mul(c).mul(10).pow(2), h1.mul(10).pow(2))}$ cm${texteExposant(2)} $\\\\approx ${texNombre(h2.mul(c).div(h1).pow(2), 2)}$ cm${texteExposant(2)}.<br>`\n        texteCorr += numAlpha(3) + ` In a reduction of coefficient k, the volumes are multiplied by k ${texteExposant(3)}.<br>`\n        texteCorr += `So the volume of the pyramid SA'B'C'D' is: $\\\\left(${texFractionReduite(h2.mul(10), h1.mul(10))}\\\\right)^3 \\\\times \\\\dfrac{${texNombre(c * c * h1)}}{3}$ cm${texteExposant(3)} $\\\\approx ${texNombre(h2.pow(3).mul(c.pow(2)).div(h1.pow(2)).div(3), 3)}$ cm${texteExposant(3)}. <br>`\n        texteCorr += numAlpha(4) + ' The volume of the trunk of the pyramid is: '\n        texteCorr += `$V_\\\\text{SABCD} - V_\\\\text{SA'B'C'D'}$<br>Or: <br>$${texNombre(h1.mul(c).mul(c).div(3), 3)}$ cm${texteExposant(3)}$ - ${texNombre(h2.pow(3).mul(c).mul(c).div(h1.pow(2)).div(3), 3)}$ cm${texteExposant(3)}$ \\\\approx ${texNombre(h1.sub(h2.pow(3).div(h1.pow(2))).mul(c.pow(2)).div(3), 2)}$ cm${texteExposant(3)}.<br>`\n        texteCorr += `Which represents $${texFractionReduite(h1.pow(3).sub(h2.pow(3)).mul(1000), h1.pow(3).mul(1000))}$ of the volume of SABCD.`\n        this.MG32codeBase64 = codeBase64\n        this.mg32init = (mtg32App, idDoc) => {\n          mtg32App.giveFormula2(idDoc, 'c', c.toString())\n          mtg32App.giveFormula2(idDoc, 'h1', h1.toString())\n          mtg32App.giveFormula2(idDoc, 'h2', h2.toString())\n          // le 2e argument à true sert à recalculer les rand() éventuels contenus dans les formules,\n          // à priori inutile ici (car on initialise une figure), mais ça mange pas de pain\n          mtg32App.calculate(idDoc, true)\n          return mtg32App.display(idDoc)\n        }\n        break\n      }\n      case 2: // calcul de l'aire de base, du volume d'un cône. puis, calcul de la section, du volume du cône réduit et du volume du tronc\n        r = new Decimal(randint(12, 35)).div(10)\n        h1 = new Decimal(randint(12, 20)).div(2)\n        h2 = new Decimal(3, Math.floor(h1) - 1)\n        if (this.sup2 < 3) {\n          if (this.sup2 === 1) { // coefficient de réduction décimal\n            while (!h2.div(h1).mul(10).eq(h2.div(h1).mul(10).round())) {\n              r = new Decimal(randint(12, 35)).div(10)\n              h1 = new Decimal(randint(12, 20)).div(2)\n              h2 = new Decimal(3, Math.floor(h1) - 1)\n            }\n          } else { // coefficient de réduction rationnel\n            while (h2.div(h1).mul(10).eq(h2.div(h1).mul(10).round())) {\n              r = new Decimal(randint(12, 35)).div(10)\n              h1 = new Decimal(randint(12, 20)).div(2)\n              h2 = new Decimal(3, Math.floor(h1) - 1)\n            }\n          }\n        }\n        codeBase64 = 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       if (!context.isHtml) { texte = '\\\\begin{minipage}{0.7 \\\\linewidth} \\n\\t' } else { texte = '' }\n        texte += `A cone has radius $${texNombre(r, 1)}$${sp()}cm and height $SO${sp()} = ${sp()}${texNombre(h1, 1)}${sp()}$cm.<br>The point O' is located on the height [SO] to ${texNombre(h2, 0)}${sp()}cm of S.<br>`\n        texte += 'A plane parallel to the base passing through O\\' coupe le cône.<br>'\n        texte += 'We thus obtain a section' + katexPopup2(numeroExercice, 1, 'similar', 'Definition: Similar Figures', 'Said of two figures/solids which have the same shapes, the same angles but not necessarily the same measurements.<br>Representations of the same figure at two different scales are similar figures.') + ' at the base and a reduced cone similar to the large cone.<br>'\n        if (!context.isHtml) { texte += 'The figure is not in full size.<br>' }\n        texte += numAlpha(0) + ' Calculate the' + katexPopup2(numeroExercice + 1, 1, 'base area of cone', 'Formula: Area of the disk with radius R', '$Area=\\\\pi \\\\times \\\\text{R}^{2}$') + '.<br>'\n        texte += numAlpha(1) + ' Calculate the' + katexPopup2(numeroExercice + 6, 1, 'cone volume', 'Formula: Volume of a cone of radius R and height h', '$Volume= \\\\dfrac{\\\\pi \\\\times \\\\text{R}^{2} \\\\times \\\\text{h}}{3}$') + '.<br>'\n        texte += numAlpha(2) + ' Deduce the area of the' + katexPopup2(numeroExercice + 11, 1, 'section', 'Definition: Plane section of a solid', `The section of a solid by a plane is a plane figure.<br>In the case of a section of a cone by a plane parallel to its base, this section is a disk which is a reduction of the base.<br>In a reduction of coefficient k, the areas are multiplied by k${texteExposant(2)} and the volumes are multiplied by k${texteExposant(3)}.`) + ` knowing that SO'${sp()} = ${sp()}${texNombre(h2, 0)}${sp()}cm.<br>`\n        texte += numAlpha(3) + ' Calculate the volume of the cone of height SO\\'.<br>'\n        texte += numAlpha(4) + ' Calculate the volume of the truncated cone (part of the cone located between the base and the section).'\n        if (context.isHtml) { texte += '<br>Point O can be moved and the viewing angle can be changed &#x3C6;' } else {\n          texte += `\\n\\t \\\\end{minipage} \\n\\t \\\\begin{minipage}{0.3 \\\\linewidth} \\n\\t \\\\begin{tikzpicture}[scale=0.7] \\n\\t\n          \\\\definecolor{hhhhhh}{rgb}{0,0,0}\n          \\\\definecolor{phphph}{rgb}{0.5,0.5,0.5}\n          \\\\definecolor{ofofof}{rgb}{0.5,0.5,0.5}\n          \\\\definecolor{ffffhh}{rgb}{1,1,0}\n          \\\\definecolor{enenen}{rgb}{0.9,0.9,0.9}\n          \\\\definecolor{dpdpdp}{rgb}{0.85,0.85,0.85}\n          \\\\definecolor{bdbdbd}{rgb}{0.74,0.74,0.74}\n          \\\\definecolor{alalal}{rgb}{0.64,0.64,0.64}\n          \\\\definecolor{ffhhhh}{rgb}{1,0,0}\n          \\\\clip (40.89,0) rectangle (0,22.58);\n          \\\\fill[color=black] (4.609,10.281) circle (0.063);\n          \\\\node at (3.828, 10.531) 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         \\\\node at (3.922, 22.559) [align=left,below right ,black,,font= \\\\sf \\\\fontsize {0.469cm} {0.586cm} \\\\selectfont] {S};\n          \\\\draw [color=black , dotted, line width = 0.4](4.609,21.776)--(8.987,10.281);\n          \\\\draw [color=ofofof , dotted, line width = 0.4](4.609,17.178)--(6.36,17.178);\n          \\\\draw [color=black , dotted, line width = 0.4](4.609,10.281)--(8.987,10.281);\n          \\\\draw [color=black , dotted, line width = 0.4](4.609,10.281)--(4.609,21.776);\n          \\\\draw [color=black , dotted, line width = 0.4](4.609,17.546)--(4.609,17.178);\n          \\\\draw [color=black , dotted, line width = 0.4](4.609,17.178)--(5.009,17.178);\n          \\\\draw [color=black , dotted, line width = 0.4](5.009,17.178)--(5.009,17.546);\n          \\\\draw [color=black , dotted, line width = 0.4](5.009,17.546)--(4.609,17.546);\n          \\\\draw [color=black , dotted, line width = 0.4](4.609,17.546)--(4.609,17.178)--(5.009,17.178)--(5.009,17.546)--(4.609,17.546)--cycle;\n          \\\\draw [color=black , dotted, line width = 0.4](4.609,10.649)--(4.609,10.281);\n          \\\\draw [color=black , dotted, line width = 0.4](4.609,10.281)--(5.009,10.281);\n          \\\\draw [color=black , dotted, line width = 0.4](5.009,10.281)--(5.009,10.649);\n          \\\\draw [color=black , dotted, line width = 0.4](5.009,10.649)--(4.609,10.649);\n          \\\\draw [color=black , dotted, line width = 0.4](4.609,10.649)--(4.609,10.281)--(5.009,10.281)--(5.009,10.649)--(4.609,10.649)--cycle;\n          \\\\end{tikzpicture} \\n\\t \\\\end{minipage}`\n        }\n\n        texteCorr = numAlpha(0) + `The base area of the cone is: $\\\\pi \\\\times R^2$ cm${texteExposant(2)} $= \\\\pi \\\\times ${texNombre(r, 1)}^2$ cm${texteExposant(2)} $= ${texNombre(r.pow(2), 2)}\\\\pi$ cm${texteExposant(2)} $\\\\approx ${texNombre(r.pow(2).mul(pi).toDP(2), 2)}$ cm${texteExposant(2)}. <br>`\n        texteCorr += numAlpha(1) + ` The volume of the cone is $\\\\dfrac{A_\\\\text{base}}{3}\\\\times \\\\text{height}$ cm${texteExposant(3)} $= \\\\dfrac{${texNombre(r.pow(2), 2)}\\\\pi}{3} \\\\times ${texNombre(h1, 1)} $ cm${texteExposant(3)} $= \\\\dfrac{${texNombre(r.pow(2).mul(h1), 3)}}{3}\\\\pi$ cm${texteExposant(3)} $\\\\approx ${texNombre(r.pow(2).mul(h1).mul(pi).div(3), 2)}$ cm${texteExposant(3)}.<br>`\n        texteCorr += numAlpha(2) + ` The section is a reduction of the coefficient base $\\\\dfrac{${texNombre(h2, 1)}}{${texNombre(h1, 1)}}`\n        if (!Number.isInteger(h1) || pgcd(h2, h1) > 1) { texteCorr += `=${texFractionReduite(h2.mul(10), h1.mul(10))}$.<br>` } else { texteCorr += '.$<br>' }\n        texteCorr += `In a reduction of coefficient k, the areas are multiplied by k${texteExposant(2)}.<br>`\n        texteCorr += `So its area is $\\\\left(${texFractionReduite(h2.mul(10), h1.mul(10))}\\\\right)^2 \\\\times ${texNombre(r.pow(2), 2)}\\\\pi$ cm${texteExposant(2)} $=${texFractionReduite(h2.mul(r).mul(10).pow(2), h1.mul(10).pow(2))}\\\\pi$ cm${texteExposant(2)} $\\\\approx${texNombre(h2.mul(r).div(h1).pow(2).mul(pi), 2)}$ cm${texteExposant(2)}.<br>`\n        texteCorr += numAlpha(3) + ` In a reduction of coefficient k, the volumes are multiplied by k${texteExposant(3)}.<br>`\n        texteCorr += `So the volume of the cone of height SO' is: $\\\\left(${texFractionReduite(h2.mul(10), h1.mul(10))}\\\\right)^3 \\\\times \\\\dfrac{${texNombre(r.mul(r).mul(h1), 3)}}{3}\\\\pi$ cm${texteExposant(3)} $\\\\approx ${texNombre(h2.pow(3).mul(r.pow(2)).mul(pi).div(h1.pow(2)).div(3), 2)}$ cm${texteExposant(3)} '.<br>`\n        texteCorr += numAlpha(4) + ' The volume of the truncated cone is: '\n        texteCorr += `$V_\\\\text{Cône} - V_\\\\text{CôneRéduit}$<br>Soit : <br>$\\\\dfrac{${texNombre(r.pow(2).mul(h1), 3)}}{3}\\\\pi$ cm${texteExposant(3)}$ - \\\\left(${texFractionReduite(h2.mul(10), h1.mul(10))}\\\\right)^3 \\\\times \\\\dfrac{${texNombre(r.pow(2).mul(h1), 3)}}{3}\\\\pi$ cm${texteExposant(3)} `\n        texteCorr += `$ = \\\\left(1-\\\\left(${texFractionReduite(h2.mul(10), h1.mul(10))}\\\\right)^3\\\\right)\\\\times \\\\dfrac{${texNombre(r.pow(2).mul(h1))}}{3}\\\\pi$ cm${texteExposant(3)} `\n        texteCorr += `$ = \\\\left(1-\\\\dfrac{${fractionSimplifiee(h2.mul(10), h1.mul(10))[0] ** 3}}{${fractionSimplifiee(h2.mul(10), h1.mul(10))[1] ** 3}}\\\\right)\\\\times \\\\dfrac{${texNombre(r.pow(2).mul(h1), 3)}}{3}\\\\pi$ cm${texteExposant(3)} `\n        texteCorr += `$ = \\\\dfrac{${fractionSimplifiee(h2.mul(10), h1.mul(10))[1] ** 3 - fractionSimplifiee(h2.mul(10), h1.mul(10))[0] ** 3}}{${fractionSimplifiee(h2.mul(10), h1.mul(10))[1] ** 3}}\\\\times \\\\dfrac{${texNombre(r.pow(2).mul(h1), 3)}}{3}\\\\pi$ cm${texteExposant(3)} `\n        texteCorr += `$ \\\\approx ${texNombre((fractionSimplifiee(h2.mul(10), h1.mul(10))[1] ** 3 - fractionSimplifiee(h2.mul(10), h1.mul(10))[0] ** 3) * r.pow(2).mul(h1).mul(pi).div(fractionSimplifiee(h2.mul(10), h1.mul(10))[1] ** 3 * 3), 3)}$ cm${texteExposant(3)}<br>`\n        this.MG32codeBase64 = codeBase64\n        this.mg32init = (mtg32App, idDoc) => {\n          mtg32App.giveFormula2(idDoc, 'r', r.toString())\n          mtg32App.giveFormula2(idDoc, 'h1', h1.toString())\n          mtg32App.giveFormula2(idDoc, 'h2', h2.toString())\n          mtg32App.calculate(idDoc, true)\n          return mtg32App.display(idDoc)\n        }\n        break\n      case 3: // calcul de l'aire de base, du volume d'une pyramide à base triangulaire. puis, calcul de la section, du volume de la petite pyramide et du volume du tronc\n        c = new Decimal(randint(30, 60)).div(10)\n        c2 = new Decimal(randint(30, 60)).div(10)\n        h1 = new Decimal(randint(12, 20) / 2)\n        h2 = new Decimal(randint(3, h1.floor().sub(1)))\n        if (this.sup2 < 3) {\n          if (this.sup2 === 1) { // coefficient de réduction décimal\n            while (!h2.div(h1).mul(10).eq(h2.div(h1).mul(10).round())) {\n              c = new Decimal(randint(30, 60)).div(10)\n              c2 = new Decimal(randint(30, 60)).div(10)\n              h1 = new Decimal(randint(12, 20) / 2)\n              h2 = new Decimal(randint(3, h1.floor().sub(1)))\n            }\n          } else { // coefficient de réduction rationnel\n            while (h2.div(h1).mul(10).eq(h2.div(h1).mul(10).round())) {\n              c = new Decimal(randint(30, 60)).div(10)\n              c2 = new Decimal(randint(30, 60)).div(10)\n              h1 = new Decimal(randint(12, 20) / 2)\n              h2 = new Decimal(randint(3, h1.floor().sub(1)))\n            }\n          }\n        }\n        codeBase64 = 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       if (!context.isHtml) { texte = '\\\\begin{minipage}{0.7 \\\\linewidth} \\n\\t' } else { texte = '' }\n        texte += `SOAB is a pyramid with a triangular base of height $SO${sp()} = ${sp()}${texNombre(h1, 1)}${sp()}$cm.<br> Its base is a right triangle OAB in O such that $OA${sp()} = ${sp()}${texNombre(c, 1)}${sp()}$ cm and $OB${sp()} = ${sp()}${texNombre(c2, 1)}${sp()}$ cm.<br>`\n        texte += ` Point O' is located on the height [SO] at ${texNombre(h2, 0)}${sp()}cm of S.`\n        texte += '<br>A plane parallel to the face OAB passing through O\\' coupe la pyramide en formant la section O\\'A\\'B\\'.<br>'\n        if (!context.isHtml) { texte += 'The figure is not in full size.<br>' }\n        texte += numAlpha(0) + ' Calculate the' + katexPopup2(numeroExercice, 1, 'base area of the pyramid', 'Formula: Area of a right triangle', '$Area=\\\\dfrac{\\\\text{c}\\\\times\\\\text{c\\'}}{2}$') + '.<br>'\n        texte += numAlpha(1) + ' Calculate the' + katexPopup2(numeroExercice + 5, 1, 'volume of the pyramid', 'Formula: volume of a pyramid with base area $B$ and height h', '$Volume= \\\\dfrac{B \\\\times \\\\text{h}}{3}$') + ' SOAB.<br>'\n        texte += numAlpha(2) + ' Deduce the area of the' + katexPopup2(numeroExercice + 10, 1, 'section', 'Definition: plane section of a solid', `The section of a solid by a plane is a plane figure.<br>In the case of a section of a pyramid by a plane parallel to its base, this section is a polygon which is a reduction of the base.<br>In a reduction of coefficient k, the areas are multiplied by k${texteExposant(2)} and the volumes are multiplied by k${texteExposant(3)}.`) + ` O'A'B' knowing that SO'${sp()} = ${sp()}${texNombre(h2, 0)}${sp()}cm.<br>`\n        texte += numAlpha(3) + ' Calculate the volume of the pyramid SO\\'A\\'B\\'.<br>'\n        texte += numAlpha(4) + ' Calculate the volume of the pyramid\\'s trunk (part of the pyramid located between the base and the section).'\n        if (context.isHtml) { texte += '<br>Point O can be moved and the viewing angle can be changed &#x3C6;' } else {\n          texte += `\\n\\t \\\\end{minipage} \\n\\t \\\\begin{minipage}{0.3 \\\\linewidth} \\n\\t \\\\begin{tikzpicture}[scale=0.8]\n          \\\\definecolor{hhhhhh}{rgb}{0,0,0}\n          \\\\definecolor{phphph}{rgb}{0.5,0.5,0.5}\n          \\\\definecolor{ofofof}{rgb}{0.5,0.5,0.5}\n          \\\\definecolor{ffhhhh}{rgb}{1,0,0}\n          \\\\definecolor{hhofhh}{rgb}{0,0.5,0}\n          \\\\definecolor{hhhhff}{rgb}{0,0,1}\n          \\\\clip (8.05,0) rectangle (0,12.08);\n          \\\\fill[color=black] (3.878,2.24) circle (0.054);\n          \\\\node at (3.341, 2.562) [align=left,inner sep = 0pt, outer sep = 0pt,below right,black,font= \\\\sf \\\\fontsize {0.376cm} {0.47cm} \\\\selectfont] {$\\\\text{O}$};\n          \\\\fill[color=black] (3.878,7.567) circle (0.054);\n          \\\\node at (3.663, 7.889) [align=left,inner sep = 0pt, outer sep = 0pt,below right,black,font= \\\\sf \\\\fontsize {0.376cm} {0.47cm} \\\\selectfont] {$\\\\text{O'}$};\n          \\\\draw [color=black , dotted, line width = 0.4](1.822,1.235)--(3.878,2.24);\n          \\\\draw [color=black , dotted, line width = 0.4](3.878,2.24)--(6.791,0.978);\n          \\\\draw [color=black , dotted, line width = 0.4](6.791,0.978)--(1.822,1.235);\n          \\\\draw [color=black , dotted, line width = 0.4](3.878,11.118)--(1.822,1.235);\n          \\\\draw [color=black , dotted, line width = 0.4](3.878,11.118)--(6.791,0.978);\n          \\\\draw [color=black , dotted, line width = 0.4](3.055,7.165)--(3.878,7.567);\n          \\\\draw [color=black , dotted, line width = 0.4](3.878,7.567)--(5.043,7.062);\n          \\\\draw [color=black , dotted, line width = 0.4](3.055,7.165)--(5.043,7.062);\n          \\\\draw [color=black , line width = 0.8](3.878,11.118)--(1.822,1.235);\n          \\\\draw [color=black , line width = 0.8](3.878,11.118)--(6.791,0.978);\n          \\\\draw [color=black , dotted, line width = 0.4](3.878,11.118)--(3.878,2.24);\n          \\\\draw [color=black , line width = 0.8](3.055,7.165)--(5.043,7.062);\n          \\\\node at (2.345, 7.43) [align=left,below right ,black,,font= \\\\sf \\\\fontsize {0.403cm} {0.503cm} \\\\selectfont] {\\\\textbf{A'}};\n          \\\\node at (5.278, 7.27) [align=left,below right ,black,,font= \\\\sf \\\\fontsize {0.403cm} {0.503cm} \\\\selectfont] {\\\\textbf{B'}};\n          \\\\node at (1.178, 1.584) [align=left,below right ,black,,font= \\\\sf \\\\fontsize {0.403cm} {0.503cm} \\\\selectfont] {\\\\textbf{A}};\n          \\\\node at (7.155, 1.145) [align=left,below right ,black,,font= \\\\sf \\\\fontsize {0.403cm} {0.503cm} \\\\selectfont] {\\\\textbf{B}};\n          \\\\draw [color=ofofof , dotted, line width = 0.4](3.878,2.24)--(4.242,2.083);\n          \\\\draw [color=ofofof , dotted, line width = 0.4](3.535,2.073)--(3.878,2.24);\n          \\\\draw [color=ofofof , dotted, line width = 0.4](3.535,2.073)--(3.899,1.915);\n          \\\\draw [color=ofofof , dotted, line width = 0.4](3.899,1.915)--(4.242,2.083);\n          \\\\draw [color=ofofof , dotted, line width = 0.4](4.242,2.083)--(3.878,2.24)--(3.535,2.073)--(3.899,1.915)--(4.242,2.083)--cycle;\n          \\\\draw [color=ofofof , dotted, line width = 0.4](3.878,7.567)--(4.242,7.409);\n          \\\\draw [color=ofofof , dotted, line width = 0.4](3.535,7.399)--(3.878,7.567);\n          \\\\draw [color=ofofof , dotted, line width = 0.4](3.535,7.399)--(3.899,7.241);\n          \\\\draw [color=ofofof , dotted, line width = 0.4](3.899,7.241)--(4.242,7.409);\n          \\\\draw [color=ofofof , dotted, line width = 0.4](4.242,7.409)--(3.878,7.567)--(3.535,7.399)--(3.899,7.241)--(4.242,7.409)--cycle;\n          \\\\draw [color=ofofof , dotted, line width = 0.4](3.878,2.684)--(3.878,2.24);\n          \\\\draw [color=ofofof , dotted, line width = 0.4](4.242,2.083)--(4.242,2.526);\n          \\\\draw [color=ofofof , dotted, line width = 0.4](4.242,2.526)--(3.878,2.684);\n          \\\\draw [color=ofofof , dotted, line width = 0.4](3.878,2.684)--(3.878,2.24)--(4.242,2.083)--(4.242,2.526)--(3.878,2.684)--cycle;\n          \\\\draw [color=ofofof , dotted, line width = 0.4](3.878,2.684)--(3.535,2.517);\n          \\\\draw [color=ofofof , dotted, line width = 0.4](3.535,2.517)--(3.535,2.073);\n          \\\\draw [color=ofofof , dotted, line width = 0.4](3.535,2.073)--(3.878,2.24)--(3.878,2.684)--(3.535,2.517)--(3.535,2.073)--cycle;\n          \\\\draw [color=ofofof , dotted, line width = 0.4](3.878,8.011)--(3.878,7.567);\n          \\\\draw [color=ofofof , dotted, line width = 0.4](4.242,7.409)--(4.242,7.853);\n          \\\\draw [color=ofofof , dotted, line width = 0.4](4.242,7.853)--(3.878,8.011);\n          \\\\draw [color=ofofof , dotted, line width = 0.4](3.878,8.011)--(3.878,7.567)--(4.242,7.409)--(4.242,7.853)--(3.878,8.011)--cycle;\n          \\\\draw [color=ofofof , dotted, line width = 0.4](3.878,8.011)--(3.535,7.843);\n          \\\\draw [color=ofofof , dotted, line width = 0.4](3.535,7.843)--(3.535,7.399);\n          \\\\draw [color=ofofof , dotted, line width = 0.4](3.535,7.399)--(3.878,7.567)--(3.878,8.011)--(3.535,7.843)--(3.535,7.399)--cycle;\n          \\\\draw [color=ffhhhh , dotted, line width = 0.4](3.055,7.165)--(5.043,7.062)--(3.878,7.567)--(3.055,7.165)--cycle;\n          \\\\fill [color = ffhhhh, opacity = 0.2](3.055,7.165)--(5.043,7.062)--(3.878,7.567)--(3.055,7.165)--cycle;\n          \\\\draw [color=hhofhh , dotted, line width = 0.4](1.822,1.235)--(6.791,0.978)--(3.878,2.24)--(1.822,1.235)--cycle;\n          \\\\fill [color = hhhhff, opacity = 0.2](1.822,1.235)--(6.791,0.978)--(3.878,2.24)--(1.822,1.235)--cycle;\n          \\\\node at (3.395, 11.857) [align=left,below right ,black,,font= \\\\sf \\\\fontsize {0.403cm} {0.503cm} \\\\selectfont] {\\\\textbf{S}};\n          \\\\end{tikzpicture} \\n\\t \\\\end{minipage}`\n        }\n\n        texteCorr = numAlpha(0) + `The base area of the pyramid is: $\\\\dfrac{${texNombre(c, 1)}\\\\times${texNombre(c2, 1)}}{2}$ cm${texteExposant(2)} $= ${texNombre(c.mul(c2).div(2), 3)}$ cm${texteExposant(2)}.<br>`\n        texteCorr += numAlpha(1) + ` The volume of the pyramid is: $\\\\dfrac{A_\\\\text{base} \\\\times \\\\text{height}}{3}$ cm${texteExposant(3)} $= \\\\dfrac{${texNombre(c.mul(c2).div(2), 3)}\\\\times ${texNombre(h1, 1)}}{3}$ cm${texteExposant(3)} $\\\\approx ${texNombre(c.mul(c2).mul(h1).div(6), 3)}$ cm${texteExposant(3)}.<br>`\n        texteCorr += numAlpha(2) + ` The section is a reduction of the coefficient base $${texFractionReduite(h2.mul(10), h1.mul(10))}$.<br>`\n        texteCorr += `In a reduction of coefficient k, the areas are multiplied by k${texteExposant(2)}.<br>`\n        texteCorr += `So its area is $\\\\left(${texFractionReduite(h2.mul(10), h1.mul(10))}\\\\right)^2 \\\\times ${texNombre(c.mul(c2).div(2), 3)}$ cm${texteExposant(2)} $=${texFractionReduite(h2.mul(h2).mul(100).mul(c).mul(c2), h1.mul(h1).mul(200))}$ cm${texteExposant(2)} $\\\\approx ${texNombre(h2.div(h1).pow(2).mul(c).mul(c2).div(2), 2)}$ cm${texteExposant(2)}.<br>`\n        texteCorr += numAlpha(3) + ` In a reduction of coefficient k, the volumes are multiplied by k${texteExposant(3)}.<br>`\n        texteCorr += `So the volume of the pyramid SO'A'B' is: $\\\\left(${texFractionReduite(h2.mul(10), h1.mul(10))}\\\\right)^3 \\\\times \\\\dfrac{${texNombre(c.mul(c2).mul(h1).div(2), 3)}}{3}$ cm${texteExposant(3)} $\\\\approx ${texNombre(h2.pow(3).mul(c).mul(c2).div(h1.pow(2)).div(6), 2)}$ cm${texteExposant(3)} '.<br>`\n        texteCorr += numAlpha(4) + ' The volume of the trunk of the pyramid is: '\n        texteCorr += `$V_\\\\text{SABCD} - V_\\\\text{SA'B'C'D'}$<br>Or: <br>$${texNombre(c.mul(c2).mul(h1).div(6), 3)}$ cm${texteExposant(3)}$ - ${texNombre(h2.pow(3).mul(c).mul(c2).div(h1.pow(2)).div(6), 3)}$ cm${texteExposant(3)}$ \\\\approx ${texNombre(c.mul(c2).mul(h1).div(6).sub(h2.pow(3).mul(c).mul(c2).div(h1.pow(2)).div(6)), 2)}$ cm${texteExposant(3)}.<br>`\n        texteCorr += `Which represents $${texFractionReduite(h1.pow(3).sub(h2.pow(3)).mul(1000), h1.pow(3).mul(1000))}$ of the volume of SABCD.`\n\n        this.MG32codeBase64 = codeBase64\n        this.mg32init = (mtg32App, idDoc) => {\n          mtg32App.giveFormula2(idDoc, 'c', c.toString())\n          mtg32App.giveFormula2(idDoc, 'h1', h1.toString())\n          mtg32App.giveFormula2(idDoc, 'h2', texNombre(h2, 0))\n          mtg32App.giveFormula2(idDoc, \"c'\", c2.toString())\n          mtg32App.calculate(idDoc, true)\n          return mtg32App.display(idDoc)\n        }\n        break\n      case 4: // Un tronc de cône étant donné (seau), calcul de la hauteur du cône dont il est issu, de son volume, puis du volume du seau. Lecture graphique du volume d'eau à mi hauteur et calcul de ce volume\n\n        r = new Decimal(randint(15, 20)).div(10)\n        r2 = new Decimal(randint(11, r * 10 - 3)).div(10)\n        h3 = new Decimal(randint(10, 15)).div(5)\n        h2 = r2.mul(h3).div(r.sub(r2))\n        h1 = h2.plus(h3)\n        while (!h2.div(h1).mul(10).eq(h2.div(h1).mul(10).round()) || !h3.div(2).plus(h2).div(h1).mul(10).eq(h3.div(2).plus(h2).div(h1).mul(10).round())) { // on impose des coefficients de réduction décimaux dans cet exercice.\n          r = new Decimal(randint(15, 20)).div(10)\n          r2 = new Decimal(randint(11, r * 10 - 3)).div(10)\n          h3 = new Decimal(randint(10, 15)).div(5)\n          h2 = r2.mul(h3).div(r.sub(r2))\n          h1 = h2.plus(h3)\n        }\n        codeBase64 = 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       if (!context.isHtml) { texte = '\\\\begin{minipage}{0.7 \\\\linewidth} \\n\\t' } else { texte = '' }\n        texte += 'A bucket is shaped like a truncated cone.<br>'\n        texte += `Its interior height is $${texNombre(h3, 1)}${sp()}$dm, its small base has an interior diameter of $${texNombre(r2 * 2)}${sp()}$dm, and its large base (the opening) has an interior diameter of $${texNombre(r * 2)}${sp()}$dm.<br>`\n        if (!context.isHtml) { texte += 'The figure is not in full size.<br>' }\n        texte += numAlpha(0) + ' Calculate the height of the cone obtained by extending the edges of the bucket.<br>'\n        texte += numAlpha(1) + ' Deduce the volume of this cone.<br>'\n        texte += numAlpha(2) + ' Deduce the total volume of the bucket.<br>'\n        if (context.isHtml) { texte += 'In these next two questions, we consider filling the bucket halfway with water.<br>' } else { texte += numAlpha(3) + ' Fill the bucket halfway with water. Calculate the corresponding volume of water.' }\n        if (context.isHtml) {\n          texte += numAlpha(3) + ' By graphic reading, after having correctly configured the figure, read the corresponding volume of water.<br>'\n          texte += numAlpha(4) + ' Find this result by calculation.<br>'\n        }\n        if (context.isHtml) { texte += '<br>You can move the cone with S and modify the values with their respective cursors.' } else {\n          texte += `\\n\\t \\\\end{minipage} \\n\\t \\\\begin{minipage}{0.3 \\\\linewidth} \\n\\t \\\\begin{tikzpicture}[scale=0.8]\n\\\\definecolor{hhhhhh}{rgb}{0,0,0}\n\\\\definecolor{phphph}{rgb}{0.5,0.5,0.5}\n\\\\definecolor{ffffff}{rgb}{1,1,1}\n\\\\definecolor{hhhhff}{rgb}{0,0,1}\n\\\\definecolor{hhofof}{rgb}{0,0.5,0.5}\n\\\\clip (7.96,0) rectangle (0,12.29);\n\\\\fill[color=black] (3.836,0.278) circle (0.053);\n\\\\node at (3.331, 0.65) [align=left,below right ,black,,font= \\\\sf \\\\fontsize {0.398cm} {0.498cm} \\\\selectfont] 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[color=black , dotted, line width = 0.4](3.836,9.678)--(5.514,9.678);\n\\\\draw [color=black , line width = 0.4](3.836,0.278)--(5.336,8.678);\n\\\\draw [color=black , line width = 0.4](3.836,0.278)--(2.336,8.678);\n\\\\node at (3.411, 8.97) [align=left,below right ,black,,font= \\\\sf \\\\fontsize {0.398cm} {0.498cm} \\\\selectfont] {\\\\textbf{B}};\n\\\\node at (5.362, 9.023) [align=left,below right ,black,,font= \\\\sf \\\\fontsize {0.398cm} {0.498cm} \\\\selectfont] {\\\\textbf{A}};\n\\\\fill[color=black] (3.836,0.278) circle (0.053);\n\\\\end{tikzpicture} \\n\\t \\\\end{minipage}`\n        }\n        texteCorr = numAlpha(0) + 'Triangles SBA and SHL are similar because the lines (BA) and (HL) are parallel in the plane of triangle SHL.<br>'\n        texteCorr += ' The tangent of the angle $\\\\widehat{HSL}$ is equal in the triangle SHL to $\\\\dfrac{\\\\text{HL}}{\\\\text{SH}}$ and in the triangle SBA to $ \\\\dfrac{\\\\text{BA}}{\\\\text{SB}}$.<br>'\n        texteCorr += ' hence $\\\\dfrac{\\\\text{HL}}{\\\\text{SH}}=\\\\dfrac{\\\\text{BA}}{\\\\text{SH}-\\\\text{BH}}$ '\n        texteCorr += ' and the equality of the cross products gives us: $\\\\text{HL}\\\\left(\\\\text{SH}-\\\\text{BH} \\\\right) =\\\\text{BA}\\\\times \\\\text{SH}$.<br>'\n        texteCorr += ` Or with numerical data: $${texNombre(r, 1)}\\\\left(\\\\text{SH}-${texNombre(h3, 1)}\\\\right)=${texNombre(r2, 1)}\\\\times \\\\text{SH}$.<br>`\n        texteCorr += `We deduce that SH$\\\\left(${texNombre(r, 1)}-${texNombre(r2, 1)}\\\\right)=${texNombre(r, 1)}\\\\times${texNombre(h3, 1)}$.<br>`\n        texteCorr += `Hence SH $=\\\\dfrac{${texNombre(r.mul(h3), 2)}}{${texNombre(r.sub(r2), 1)}} = ${texNombre(h1, 1)}$ dm (SB = $${texNombre(h2, 1)}$ dm).<br>`\n        texteCorr += numAlpha(1) + ` The volume of the cone is $\\\\dfrac{A_\\\\text{base}}{3}\\\\times \\\\text{height}$ dm${texteExposant(3)} $= \\\\dfrac{${texNombre(r.pow(2), 2)}\\\\pi}{3} \\\\times ${texNombre(h1, 1)} $ dm${texteExposant(3)} $= \\\\dfrac{${texNombre(r.mul(r).mul(h1), 3)}}{3}\\\\pi$ dm${texteExposant(3)} $\\\\approx ${texNombre(r.mul(r).mul(pi).mul(h1).div(3), 3)}$ dm${texteExposant(3)}.<br>`\n        texteCorr += numAlpha(2) + ' The bucket is a truncated cone. To calculate its volume, we will calculate the volume of the reduced cone of height SB and subtract it from the volume of the cone of height SH.<br>'\n        texteCorr += ` The SB height cone is a reduction of the SH height cone. The coefficient of this reduction is: $\\\\dfrac{${texNombre(r2, 1)}}{${texNombre(r, 1)}}`\n        if (!Number.isInteger(r) || pgcd(r2.mul(10), r * 10) > 1) { texteCorr += `=${texFractionReduite(r2.mul(10), r.mul(10))}$.<br>` } else { texteCorr += '.$<br>' }\n        texteCorr += `In a reduction of coefficient k, the volumes are multiplied by k${texteExposant(3)}.<br>`\n        texteCorr += `So the volume of the cone of height SB is: $\\\\left(${texFractionReduite(r2.mul(10), r.mul(10))}\\\\right)^3 \\\\times \\\\dfrac{${texNombre(r.mul(r).mul(h1), 3)}}{3}\\\\pi$ dm${texteExposant(3)} $\\\\approx ${texNombre(pi.mul(h2.div(h1).pow(3)).mul(r.pow(2).mul(h1)).div(3), 3)}$ dm${texteExposant(3)} '.<br>`\n        texteCorr += 'The volume of the truncated cone is: '\n        texteCorr += `$V_\\\\text{Cone} - V_\\\\text{ReducedCone}$<br>Or: <br>$\\\\dfrac{${texNombre(r.mul(r).mul(h1), 3)}}{3}\\\\pi$ dm${texteExposant(3)}$ - \\\\left(${texFractionReduite(r2.mul(10), r.mul(10))}\\\\right )^3 \\\\times \\\\dfrac{${texNombre(r.mul(r).mul(h1), 3)}}{3}\\\\pi$ dm${texteExposant(3)}`\n        texteCorr += `$ = \\\\left(1-\\\\left(${texFractionReduite(r2.mul(10), r.mul(10))}\\\\right)^3\\\\right)\\\\times \\\\dfrac{${texNombre(r.mul(r).mul(h1), 3)}}{3}\\\\pi$ dm${texteExposant(3)} `\n        texteCorr += `$ = \\\\left(1-\\\\dfrac{${fractionSimplifiee(r2.mul(10), r.mul(10))[0] ** 3}}{${fractionSimplifiee(r2.mul(10), r.mul(10))[1] ** 3}}\\\\right)\\\\times \\\\dfrac{${texNombre(r.mul(r).mul(h1), 3)}}{3}\\\\pi$ dm${texteExposant(3)} `\n        texteCorr += `$ = \\\\dfrac{${fractionSimplifiee(r2.mul(10), r.mul(10))[1] ** 3 - fractionSimplifiee(r2.mul(10), r.mul(10))[0] ** 3}}{${fractionSimplifiee(r2.mul(10), r.mul(10))[1] ** 3}}\\\\times \\\\dfrac{${texNombre(r.mul(r).mul(h1), 3)}}{3}\\\\pi$ dm${texteExposant(3)} `\n        kprime = new Decimal(r2).div(r).pow(3)\n        texteCorr += `$ \\\\approx ${texNombre(r.pow(2).mul(h1).mul(pi).div(3).mul(kprime.sub(1).mul(-1)), 3)}$ dm${texteExposant(3)}<br>`\n        c = h3.div(2)\n        kprime = c.plus(h2).div(h2)\n        if (context.isHtml) {\n          texteCorr += numAlpha(3) + `We must set HL to ${texNombre(r * 10)} cm; BA to ${texNombre(r2.mul(10), 0)} cm; BH to ${texNombre(h3.mul(10), 0)} cm and the water height to ${texNombre(h1.sub(h2).mul(5), 1)} cm.<br>`\n          texteCorr += `Reading $ y = V(x)$ gives us a volume of water of approximately $${texNombre(kprime.pow(3).sub(1).mul(r2.pow(2)).mul(h2).mul(pi).div(3), 1)}$ dm${texteExposant(3)} or approximately $${texNombre(kprime.pow(3).sub(1).mul(r2.pow(2)).mul(h2).mul(pi).div(3), 0)}$ liters of water.<br>`\n          texteCorr += numAlpha(4)\n        } else { texteCorr += numAlpha(3) }\n        texteCorr += ' We will determine the volume of the cone of height SE, then we will subtract the volume of the cone of height SB to obtain the volume of water.<br>'\n        texteCorr += ` The SE height cone is a reduction of the SH height cone. The coefficient of this reduction is: $${texFractionReduite(c.add(h2).mul(100), h1.mul(100))}$.<br>`\n        texteCorr += `So the volume $V$ of the cone of height SE is: $\\\\left(${texFractionReduite(h1.add(h2).mul(50), h1.mul(100))}\\\\right)^3 \\\\times \\\\dfrac{${texNombre(r.mul(r).mul(h1), 3)}}{3}\\\\pi$ dm${texteExposant(3)} $\\\\approx ${texNombre(r.pow(2).mul(pi).mul(h2.plus(c).div(h1).pow(3)).mul(h1).div(3), 3)}$ dm${texteExposant(3)}. <br>`\n        texteCorr += 'Let $V\\'$ denote the volume of the cone of height SB calculated in the question' + numAlpha(2) + '<br> The volume of water is therefore: '\n\n        texteCorr += `$V-V' \\\\approx ${texNombre(pi.div(3).mul(r.pow(2)).mul(h1).mul(h2.plus(c).div(h1).pow(3)), 3)}$ dm${texteExposant(3)}$ - ${texNombre(pi.div(3).mul(h2.div(h1).pow(3).mul(r.pow(2))).mul(h1), 3)}$ dm${texteExposant(3)} $\\\\approx ${texNombre(kprime.pow(3).sub(1).mul(r2.pow(2)).mul(h2).mul(pi).div(3), 1)}$ dm${texteExposant(3)}.<br>`\n        this.MG32codeBase64 = codeBase64\n        this.mg32init = (mtg32App, idDoc) => {\n          mtg32App.calculate(idDoc, true)\n          return mtg32App.display(idDoc)\n        }\n        break\n      case 5: // Un problème avec un cône Vanille Chocolat.\n        r = new Decimal(randint(20, 28)).div(10)\n        h1 = new Decimal(randint(20, 28)).div(2)\n        h2 = new Decimal(randint(1, 3))\n\n        if (this.sup2 < 3) {\n          if (this.sup2 === 1) { // coefficient de réduction décimal\n            while (!h2.div(h1).mul(10).eq(h2.div(h1).mul(10).round())) {\n              r = new Decimal(randint(20, 28)).div(10)\n              h1 = new Decimal(randint(20, 28)).div(2)\n              h2 = new Decimal(randint(1, 3))\n            }\n          } else { // coefficient de réduction rationnel\n            while (h2.div(h1).mul(10).eq(h2.div(h1).mul(10).round())) {\n              r = new Decimal(randint(20, 28)).div(10)\n              h1 = new Decimal(randint(20, 28)).div(2)\n              h2 = new Decimal(randint(1, 3))\n            }\n          }\n        }\n        r2 = r.mul(h2).div(h1)\n        codeBase64 = 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       if (!context.isHtml) { texte = '\\\\begin{minipage}{0.7 \\\\linewidth} \\n\\t' } else { texte = '' }\n        texte += `An ice cream cone from a famous brand has radius $${texNombre(r, 1)}${sp()}$cm and height $SO${sp()} = ${sp()}${texNombre(h1, 1)}${sp()}$cm.<br>`\n        texte += `The manufacturer has sunk chocolate to the bottom of this cone to a height SW' of ${texNombre(h2, 0)}${sp()}cm.<br>`\n        texte += numAlpha(0) + ' Calculate the total volume of the cone.<br>'\n        texte += numAlpha(1) + ' Deduce the volume of chocolate present in the bottom of the cone.<br>'\n        texte += numAlpha(2) + ' Deduce from the first two questions the volume of ice needed to fill the cone.<br>'\n        texte += numAlpha(3) + ' If the ice cream had been put in the cone before the chocolate, how high would the ice cream reach in the cone?<br>'\n        texte += numAlpha(4) + ' What would be the thickness of chocolate above the ice cream?<br>'\n        if (context.isHtml) { texte += 'Point O can be moved and the viewing angle can be changed &#x3C6;' } else {\n          texte += `\\n\\t \\\\end{minipage} \\n\\t \\\\begin{minipage}{0.3 \\\\linewidth} \\n\\t \\\\begin{tikzpicture}[scale=0.8]\n          \\\\definecolor{hhhhhh}{rgb}{0,0,0}\n          \\\\definecolor{phphph}{rgb}{0.5,0.5,0.5}\n          \\\\definecolor{ofofof}{rgb}{0.5,0.5,0.5}\n          \\\\definecolor{ffffhh}{rgb}{1,1,0}\n          \\\\definecolor{enenen}{rgb}{0.9,0.9,0.9}\n          \\\\definecolor{dpdpdp}{rgb}{0.85,0.85,0.85}\n          \\\\definecolor{bdbdbd}{rgb}{0.74,0.74,0.74}\n          \\\\definecolor{alalal}{rgb}{0.64,0.64,0.64}\n          \\\\definecolor{ffhhhh}{rgb}{1,0,0}\n          \\\\clip (40.89,0) rectangle (0,22.58);\n          \\\\fill[color=black] (4.641,21.063) circle (0.063);\n          \\\\node at (4.234, 20.625) [align=left,inner sep = 0pt, outer sep = 0pt,below right,black,font= \\\\sf \\\\fontsize {0.438cm} {0.547cm} \\\\selectfont] {$\\\\text{O}$};\n          \\\\draw 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0.4](4.641,13.987)--(4.641,13.603)--(5.041,13.603)--(5.041,13.987)--(4.641,13.987)--cycle;\n          \\\\draw [color=black , dotted, line width = 0.4](4.641,21.447)--(4.641,21.063);\n          \\\\draw [color=black , dotted, line width = 0.4](4.641,21.063)--(5.041,21.063);\n          \\\\draw [color=black , dotted, line width = 0.4](5.041,21.063)--(5.041,21.447);\n          \\\\draw [color=black , dotted, line width = 0.4](5.041,21.447)--(4.641,21.447);\n          \\\\draw [color=black , dotted, line width = 0.4](4.641,21.447)--(4.641,21.063)--(5.041,21.063)--(5.041,21.447)--(4.641,21.447)--cycle;\n          \\\\end{tikzpicture} \\n\\t \\\\end{minipage}`\n        }\n        texteCorr = numAlpha(0) + `The volume of the cone is $\\\\dfrac{A_\\\\text{base}}{3}\\\\times \\\\text{height}$ cm${texteExposant(3)} $= \\\\dfrac{${texNombre(r.pow(2), 2)}\\\\pi}{3} \\\\times ${texNombre(h1, 1)} $ cm${texteExposant(3)} $= \\\\dfrac{${texNombre(r.mul(r).mul(h1), 3)}}{3}\\\\pi$ cm${texteExposant(3)} $\\\\approx ${texNombre(r.mul(r).mul(h1).mul(pi).div(3), 3)}$ cm${texteExposant(3)}.<br>`\n        texteCorr += numAlpha(1) + ` The chocolate cone is a reduction of the full cone. The reduction coefficient is $\\\\dfrac{${texNombre(h2, 1)}}{${texNombre(h1, 1)}}`\n        if (!Number.isInteger(h1) || pgcd(h2, h1) > 1) { texteCorr += `=${texFractionReduite(h2.mul(10), h1.mul(10))}$.<br>` } else { texteCorr += '.$<br>' }\n        texteCorr += ` In a reduction of coefficient k, the volumes are multiplied by k${texteExposant(3)}.<br>`\n        texteCorr += `So the volume of the cone of height SO' is: $\\\\left(${texFractionReduite(h2.mul(10), h1.mul(10))}\\\\right)^3 \\\\times \\\\dfrac{${texNombre(r.mul(r).mul(h1), 3)}}{3}\\\\pi$ cm${texteExposant(3)} $ \\\\approx ${texNombre(pi.mul(h2.pow(3)).mul(r.pow(2)).div(h1.pow(2)).div(3), 3)}$ cm${texteExposant(3)}.<br>`\n        texteCorr += numAlpha(2) + ' The ice volume is the difference between the two previous volumes:<br>'\n        texteCorr += `$${texNombre(r.pow(2).mul(h1).mul(pi).div(3), 3)}$ cm${texteExposant(3)}$ - ${texNombre(pi.mul(h2.pow(3)).mul(r.div(h1).pow(2)).div(3), 3)}$ cm${texteExposant(3)} $ \\\\approx ${texNombre(pi.div(3).mul(r.pow(2)).mul(h1.sub(h2.pow(3).div(h1.pow(2)))), 2)}$ cm${texteExposant(3)}.<br>`\n        texteCorr += numAlpha(3) + ' If we pour the ice cream at the bottom of the cone, we obtain a further reduction of the complete cone.<br>'\n        texteCorr += `Let k' be the coefficient of this reduction, we have: k'${texteExposant(3)} $= 1- \\\\text{k}^3$`\n        texteCorr += ', hence k\\' '\n        texteCorr += '$= \\\\sqrt[3]{1-{\\\\text{k}^3}}$.<br>'\n        kprime = new Decimal(1).sub(h2.div(h1).pow(3)).cbrt()\n        texteCorr += `So k' = $\\\\sqrt[3]{1 - \\\\left(${texFractionReduite(h2.mul(10), h1.mul(10))}\\\\right)^3} \\\\approx ${texNombre(kprime, 5)}$.<br>`\n        texteCorr += `We deduce that the ice height is approximately: $${texNombre(kprime, 5)} \\\\times ${texNombre(h1, 1)}$ cm $\\\\approx ${texNombre(kprime.mul(h1), 4)}$ cm.<br>`\n        texteCorr += numAlpha(4) + ` The thickness of the chocolate is then: $${texNombre(h1, 1)}\\\\text{ cm}-${texNombre(kprime.mul(h1), 4)} \\\\text{ cm}\\\\approx ${texNombre(h1.sub(kprime.mul(h1)).mul(10), 2)}$ mm!`\n        this.MG32codeBase64 = codeBase64\n        this.mg32init = (mtg32App, idDoc) => {\n          mtg32App.giveFormula2(idDoc, 'r', r.toString())\n          mtg32App.giveFormula2(idDoc, 'h1', h1.toString())\n          mtg32App.giveFormula2(idDoc, 'h2', h2.toString())\n          mtg32App.calculate(idDoc, true)\n          return mtg32App.display(idDoc)\n        }\n        break\n    }\n\n    if (context.isHtml) {\n      const divMtg = document.createElement('div')\n      const svg = document.createElement('svg')\n      svg.id = `svgMtg${this.numeroExercice}`\n      divMtg.id = `MG32div${this.numeroExercice}`\n      // divMtg.style.display = 'none'\n      divMtg.appendChild(svg)\n      // divMtg.setAttribute('MG32codeBase64', codeBase64)\n      texteCorr += divMtg.outerHTML\n\n      // eslint-disable-next-line @typescript-eslint/no-this-alias\n      const that = this\n\n    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