import Exercice from '../Exercice.js'
import Decimal from 'decimal.js'
import { context } from '../../modules/context.js'
import { listeQuestionsToContenu, randint, texteExposant, pgcd, fractionSimplifiee, texFractionReduite, texNombre, sp, katexPopup2, numAlpha } from '../../modules/outils.js'
export const titre = 'Connaître les effets des agrandissements/réductions sur les aires et les volumes'
/**
* Problèmes calculs d'aire et de volumes utilisant l'effet d'une réduction sur les aires et les volumes
* @author Jean-Claude Lhote
* 3G22
*/
export const uuid = '960f9'
export const ref = '3G22'
export default function AgrandissementReduction () {
'use strict'
Exercice.call(this) // Héritage de la classe Exercice()
this.titre = titre
this.consigne = ''
this.nbQuestions = 1
this.nbQuestionsModifiable = false
context.isHtml ? this.spacingCorr = 3.5 : this.spacingCorr = 1.5
context.isHtml ? this.spacing = 3 : this.spacing = 2
this.nbCols = 1
this.nbColsCorr = 1
this.quatrieme = false
this.sup = 1 //
this.sup2 = 1
this.pasDeVersionLatex = false
this.nouvelleVersion = function (numeroExercice) {
this.listeQuestions = []
this.listeCorrections = []
let texte, texteCorr, r, r2, h1, h2, h3, c, c2, kprime
const pi = Decimal.acos(-1)
// if (context.isHtml) {
this.typeExercice = 'MG32'
this.dimensionsDivMg32 = [600, 700]
let codeBase64
let choix
if (parseInt(this.sup) === 1) {
choix = randint(1, 3)
} else if (parseInt(this.sup) === 2) {
choix = randint(4, 5)
} else {
choix = randint(1, 5)
}
switch (choix) {
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
c = new Decimal(randint(30, 60)).div(10)
h1 = new Decimal(randint(12, 20)).div(2)
h2 = new Decimal(randint(3, h1.floor().sub(1).toNumber()))
if (this.sup2 < 3) {
if (parseInt(this.sup2) === 1) {
// on veut un coefficient de réduction décimal à 1 chiffre après la virgule
while (!h2.div(h1).mul(10).eq(h2.div(h1).mul(10).round())) {
c = new Decimal(randint(30, 60)).div(10)
h1 = new Decimal(randint(12, 20)).div(2)
h2 = new Decimal(randint(3, h1.floor().sub(1).toNumber()))
}
} else {
// coefficient qui peut être décimal avec plus d'un chiffre ou rationnel non décimal.
while (h2.div(h1).mul(10).eq(h2.div(h1).mul(10).round())) {
c = new Decimal(randint(30, 60)).div(10)
h1 = new Decimal(randint(12, 20)).div(2)
h2 = new Decimal(randint(3, h1.floor().sub(1)))
}
}
}
codeBase64 = 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if (!context.isHtml) {
texte = '\\begin{minipage}{0.65 \\linewidth} \n\t'
} else {
texte = ''
}
texte += `SABCD est une pyramide à base carrée de hauteur $SO${sp()}=${sp()}${texNombre(h1, 1)}${sp()}$cm et de côté de base $${texNombre(c, 1)}${sp()}$cm.<br>`
texte += ` Le point O' est situé sur la hauteur [SO] à ${texNombre(h2, 0)}${sp()}cm de S.`
texte += '<br>Un plan parallèle à la face ABCD passant par O\' coupe la pyramide en formant la section A\'B\'C\'D\'.<br>'
if (!context.isHtml) {
texte += 'La figure n\'est pas en vraie grandeur.<br>'
}
texte += numAlpha(0) + ' Calculer l\'' + katexPopup2(numeroExercice, 1, 'aire de base de la pyramide', 'Formule : Aire d\'un carré de côté c', `$Aire=\\text{c}$${texteExposant(2)}`) + '.<br>'
texte += numAlpha(1) + ' Calculer le ' + katexPopup2(numeroExercice + 5, 1, 'volume de la pyramide', 'Formule : volume d\'une pyramide d\'aire de base $B$ et de hauteur h', '$Volume= \\dfrac{B \\times \\text{h}}{3}$') + ' SABCD.<br>'
texte += numAlpha(2) + ' En déduire l\'aire de la ' + katexPopup2(numeroExercice + 10, 1, 'section', 'Définition : section plane d\'un solide', `La section d'un solide par un plan est une figure plane.<br>Dans le cas d'une section d'une pyramide par un plan parallèle à sa base, cette section est un polygone qui est une réduction de la base.<br>Dans une réduction de coefficient k, les aires sont multipliées par k${texteExposant(2)} et les volumes sont multipliés par k${texteExposant(3)}.`) + ` A'B'C'D' sachant que SO'${sp()}=${sp()}${texNombre(h2, 0)}${sp()}cm.<br>`
texte += numAlpha(3) + ' Calculer le volume de la pyramide SA\'B\'C\'D\'.<br>'
texte += numAlpha(4) + ' Calculer le volume du tronc de la pyramide (partie de la pyramide située entre la base et la section).'
if (context.isHtml) {
texte += '<br>Le point O peut être déplacé et on peut changer l\'angle de vue φ '
} else {
texte += `\n\t \\end{minipage} \n\t \\begin{minipage}{0.35 \\linewidth} \n\t \\begin{tikzpicture}[scale=0.8] \n\t
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\\node at (1.914, 4.255) [align=left,below right ,black,,font= \\sf \\fontsize {0.469cm} {0.586cm} \\selectfont] {D};
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\\end{tikzpicture} \n\t \\end{minipage}`
}
texteCorr = numAlpha(0) + ` L'aire de base de la pyramide est : $${texNombre(c, 1)}^2$ cm${texteExposant(2)} $= ${texNombre(c * c)}$ cm${texteExposant(2)}.<br>`
texteCorr += numAlpha(1) + ` Le volume de la pyramide est : $\\dfrac{A_\\text{base} \\times \\text{hauteur}}{3}$ cm${texteExposant(3)} $= \\dfrac{${texNombre(c * c)}\\times ${texNombre(h1, 1)}}{3}$ cm${texteExposant(3)} $\\approx ${texNombre(c.mul(c).mul(h1).div(3), 3)}$ cm${texteExposant(3)}.<br>`
texteCorr += numAlpha(2) + ` La section est une réduction de la base de coefficient $\\dfrac{${texNombre(h2, 0)}}{${texNombre(h1, 1)}}`
if (!Number.isInteger(h1) || pgcd(h2, h1) > 1) {
texteCorr += `=${texFractionReduite(h2.mul(10), h1.mul(10))}$.<br>`
} else {
texteCorr += '.$<br>'
}
texteCorr += `Dans une réduction de coefficient k, les aires sont multipliés par k${texteExposant(2)}.<br>`
texteCorr += `Donc son aire est $\\left(${texFractionReduite(h2.mul(10), h1.mul(10))}\\right)^2 \\times ${texNombre(c * c)}$ cm${texteExposant(2)} $=${texFractionReduite(h2.mul(c).mul(10).pow(2), h1.mul(10).pow(2))}$ cm${texteExposant(2)} $\\approx ${texNombre(h2.mul(c).div(h1).pow(2), 2)}$ cm${texteExposant(2)}.<br>`
texteCorr += numAlpha(3) + ` Dans une réduction de coefficient k, les volumes sont multipliés par k ${texteExposant(3)}.<br>`
texteCorr += `Donc le volume de la pyramide SA'B'C'D' est : $\\left(${texFractionReduite(h2.mul(10), h1.mul(10))}\\right)^3 \\times \\dfrac{${texNombre(c * c * h1)}}{3}$ cm${texteExposant(3)} $\\approx ${texNombre(h2.pow(3).mul(c.pow(2)).div(h1.pow(2)).div(3), 3)}$ cm${texteExposant(3)}.<br>`
texteCorr += numAlpha(4) + ' Le volume du tronc de la pyramide est : '
texteCorr += `$V_\\text{SABCD} - V_\\text{SA'B'C'D'}$<br>Soit : <br>$${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>`
texteCorr += `Ce qui représente $${texFractionReduite(h1.pow(3).sub(h2.pow(3)).mul(1000), h1.pow(3).mul(1000))}$ du volume de SABCD.`
this.MG32codeBase64 = codeBase64
this.mg32init = (mtg32App, idDoc) => {
mtg32App.giveFormula2(idDoc, 'c', c.toString())
mtg32App.giveFormula2(idDoc, 'h1', h1.toString())
mtg32App.giveFormula2(idDoc, 'h2', h2.toString())
// le 2e argument à true sert à recalculer les rand() éventuels contenus dans les formules,
// à priori inutile ici (car on initialise une figure), mais ça mange pas de pain
mtg32App.calculate(idDoc, true)
return mtg32App.display(idDoc)
}
break
case 2: // calcul de l'aire de base, du volume d'un cône. puis, calcul de la section, du volume du cône réduit et du volume du tronc
r = new Decimal(randint(12, 35)).div(10)
h1 = new Decimal(randint(12, 20)).div(2)
h2 = new Decimal(3, Math.floor(h1) - 1)
if (this.sup2 < 3) {
if (parseInt(this.sup2) === 1) { // coefficient de réduction décimal
while (!h2.div(h1).mul(10).eq(h2.div(h1).mul(10).round())) {
r = new Decimal(randint(12, 35)).div(10)
h1 = new Decimal(randint(12, 20)).div(2)
h2 = new Decimal(3, Math.floor(h1) - 1)
}
} else { // coefficient de réduction rationnel
while (h2.div(h1).mul(10).eq(h2.div(h1).mul(10).round())) {
r = new Decimal(randint(12, 35)).div(10)
h1 = new Decimal(randint(12, 20)).div(2)
h2 = new Decimal(3, Math.floor(h1) - 1)
}
}
}
codeBase64 = 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if (!context.isHtml) { texte = '\\begin{minipage}{0.7 \\linewidth} \n\t' } else { texte = '' }
texte += `Un cône a pour rayon $${texNombre(r, 1)}$${sp()}cm et pour hauteur $SO${sp()}=${sp()}${texNombre(h1, 1)}${sp()}$cm.<br>Le point O' est situé sur la hauteur [SO] à ${texNombre(h2, 0)}${sp()}cm de S.<br>`
texte += 'Un plan parallèle à la base passant par O\' coupe le cône.<br>'
texte += 'On obtient ainsi une section ' + katexPopup2(numeroExercice, 1, 'semblable', 'Définition : Figures semblables', 'Se dit de deux figures/solides qui ont les mêmes formes, les mêmes angles mais pas nécessairement les mêmes mesures.<br>Les représentations d\'une même figure à deux échelles différentes sont des figures semblables.') + ' à la base et un cône réduit semblable au grand cône.<br>'
if (!context.isHtml) { texte += 'La figure n\'est pas en vraie grandeur.<br>' }
texte += numAlpha(0) + ' Calculer l\'' + katexPopup2(numeroExercice + 1, 1, 'aire de base du cône', 'Formule : Aire du disque de rayon R', '$Aire=\\pi \\times \\text{R}^{2}$') + '.<br>'
texte += numAlpha(1) + ' Calculer le ' + katexPopup2(numeroExercice + 6, 1, 'volume du cône', 'Formule : Volume d\'un cône de rayon R et de hauteur h', '$Volume= \\dfrac{\\pi \\times \\text{R}^{2} \\times \\text{h}}{3}$') + '.<br>'
texte += numAlpha(2) + ' En déduire l\'aire de la ' + katexPopup2(numeroExercice + 11, 1, 'section', 'Définition : Section plane d\'un solide', `La section d'un solide par un plan est une figure plane.<br>Dans le cas d'une section d'un cône par un plan parallèle à sa base, cette section est un disque qui est une réduction de la base.<br>Dans une réduction de coefficient k, les aires sont multipliées par k${texteExposant(2)} et les volumes sont multipliés par k${texteExposant(3)}.`) + ` sachant que SO'${sp()}=${sp()}${texNombre(h2, 0)}${sp()}cm.<br>`
texte += numAlpha(3) + ' Calculer le volume du cône de hauteur SO\'.<br>'
texte += numAlpha(4) + ' Calculer le volume du tronc de cône (partie du cône située entre la base et la section).'
if (context.isHtml) { texte += '<br>Le point O peut être déplacé et on peut changer l\'angle de vue φ ' } else {
texte += `\n\t \\end{minipage} \n\t \\begin{minipage}{0.3 \\linewidth} \n\t \\begin{tikzpicture}[scale=0.7] \n\t
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\\node at (3.922, 22.559) [align=left,below right ,black,,font= \\sf \\fontsize {0.469cm} {0.586cm} \\selectfont] {S};
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\\end{tikzpicture} \n\t \\end{minipage}`
}
texteCorr = numAlpha(0) + ` L'aire de base du cône est : $\\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>`
texteCorr += numAlpha(1) + ` Le volume du cône est $\\dfrac{A_\\text{base}}{3}\\times \\text{hauteur}$ 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>`
texteCorr += numAlpha(2) + ` La section est une réduction de la base de coefficient $\\dfrac{${texNombre(h2, 1)}}{${texNombre(h1, 1)}}`
if (!Number.isInteger(h1) || pgcd(h2, h1) > 1) { texteCorr += `=${texFractionReduite(h2.mul(10), h1.mul(10))}$.<br>` } else { texteCorr += '.$<br>' }
texteCorr += `Dans une réduction de coefficient k, les aires sont multipliés par k${texteExposant(2)}.<br>`
texteCorr += `Donc son aire est $\\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>`
texteCorr += numAlpha(3) + ` Dans une réduction de coefficient k, les volumes sont multipliés par k${texteExposant(3)}.<br>`
texteCorr += `Donc le volume du cône de hauteur SO' est : $\\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>`
texteCorr += numAlpha(4) + ' Le volume du tronc de cône est : '
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)} `
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)} `
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)} `
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)} `
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>`
this.MG32codeBase64 = codeBase64
this.mg32init = (mtg32App, idDoc) => {
mtg32App.giveFormula2(idDoc, 'r', r.toString())
mtg32App.giveFormula2(idDoc, 'h1', h1.toString())
mtg32App.giveFormula2(idDoc, 'h2', h2.toString())
mtg32App.calculate(idDoc, true)
return mtg32App.display(idDoc)
}
break
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
c = new Decimal(randint(30, 60)).div(10)
c2 = new Decimal(randint(30, 60)).div(10)
h1 = new Decimal(randint(12, 20) / 2)
h2 = new Decimal(randint(3, h1.floor().sub(1)))
if (this.sup2 < 3) {
if (parseInt(this.sup2) === 1) { // coefficient de réduction décimal
while (!h2.div(h1).mul(10).eq(h2.div(h1).mul(10).round())) {
c = new Decimal(randint(30, 60)).div(10)
c2 = new Decimal(randint(30, 60)).div(10)
h1 = new Decimal(randint(12, 20) / 2)
h2 = new Decimal(randint(3, h1.floor().sub(1)))
}
} else { // coefficient de réduction rationnel
while (h2.div(h1).mul(10).eq(h2.div(h1).mul(10).round())) {
c = new Decimal(randint(30, 60)).div(10)
c2 = new Decimal(randint(30, 60)).div(10)
h1 = new Decimal(randint(12, 20) / 2)
h2 = new Decimal(randint(3, h1.floor().sub(1)))
}
}
}
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if (!context.isHtml) { texte = '\\begin{minipage}{0.7 \\linewidth} \n\t' } else { texte = '' }
texte += `SOAB est une pyramide à base triangulaire de hauteur $SO${sp()}=${sp()}${texNombre(h1, 1)}${sp()}$cm.<br> Sa base est un triangle OAB rectangle en O tel que $OA${sp()}=${sp()}${texNombre(c, 1)}${sp()}$ cm et $OB${sp()}=${sp()}${texNombre(c2, 1)}${sp()}$ cm.<br>`
texte += ` Le point O' est situé sur la hauteur [SO] à ${texNombre(h2, 0)}${sp()}cm de S.`
texte += '<br>Un plan parallèle à la face OAB passant par O\' coupe la pyramide en formant la section O\'A\'B\'.<br>'
if (!context.isHtml) { texte += 'La figure n\'est pas en vraie grandeur.<br>' }
texte += numAlpha(0) + ' Calculer l\'' + katexPopup2(numeroExercice, 1, 'aire de base de la pyramide', 'Formule : Aire d\'un triangle rectangle', '$Aire=\\dfrac{\\text{c}\\times\\text{c\'}}{2}$') + '.<br>'
texte += numAlpha(1) + ' Calculer le ' + katexPopup2(numeroExercice + 5, 1, 'volume de la pyramide', 'Formule : volume d\'une pyramide d\'aire de base $B$ et de hauteur h', '$Volume= \\dfrac{B \\times \\text{h}}{3}$') + ' SOAB.<br>'
texte += numAlpha(2) + ' En déduire l\'aire de la ' + katexPopup2(numeroExercice + 10, 1, 'section', 'Définition : section plane d\'un solide', `La section d'un solide par un plan est une figure plane.<br>Dans le cas d'une section d'une pyramide par un plan parallèle à sa base, cette section est un polygone qui est une réduction de la base.<br>Dans une réduction de coefficient k, les aires sont multipliées par k${texteExposant(2)} et les volumes sont multipliés par k${texteExposant(3)}.`) + ` O'A'B' sachant que SO'${sp()}=${sp()}${texNombre(h2, 0)}${sp()}cm.<br>`
texte += numAlpha(3) + ' Calculer le volume de la pyramide SO\'A\'B\'.<br>'
texte += numAlpha(4) + ' Calculer le volume du tronc de la pyramide (partie de la pyramide située entre la base et la section).'
if (context.isHtml) { texte += '<br>Le point O peut être déplacé et on peut changer l\'angle de vue φ ' } else {
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}
texteCorr = numAlpha(0) + ` L'aire de base de la pyramide est : $\\dfrac{${texNombre(c, 1)}\\times${texNombre(c2, 1)}}{2}$ cm${texteExposant(2)} $= ${texNombre(c.mul(c2).div(2), 3)}$ cm${texteExposant(2)}.<br>`
texteCorr += numAlpha(1) + ` Le volume de la pyramide est : $\\dfrac{A_\\text{base} \\times \\text{hauteur}}{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>`
texteCorr += numAlpha(2) + ` La section est une réduction de la base de coefficient $${texFractionReduite(h2.mul(10), h1.mul(10))}$.<br>`
texteCorr += `Dans une réduction de coefficient k, les aires sont multipliés par k${texteExposant(2)}.<br>`
texteCorr += `Donc son aire est $\\left(${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${texteExposant(2)}.<br>`
texteCorr += numAlpha(3) + ` Dans une réduction de coefficient k, les volumes sont multipliés par k${texteExposant(3)}.<br>`
texteCorr += `Donc le volume de la pyramide SO'A'B' est : $\\left(${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${texteExposant(3)} '.<br>`
texteCorr += numAlpha(4) + ' Le volume du tronc de la pyramide est : '
texteCorr += `$V_\\text{SABCD} - V_\\text{SA'B'C'D'}$<br>Soit : <br>$${texNombre(c.mul(c2).mul(h1).div(6), 3)}$ cm${texteExposant(3)}$ - ${texNombre(h2.pow(3).mul(c).mul(c2).div(h1.pow(2)).div(6), 3)}$ cm${texteExposant(3)}$ \\approx ${texNombre(c.mul(c2).mul(h1).div(6).sub(h2.pow(3).mul(c).mul(c2).div(h1.pow(2)).div(6)), 2)}$ cm${texteExposant(3)}.<br>`
texteCorr += `Ce qui représente $${texFractionReduite(h1.pow(3).sub(h2.pow(3)).mul(1000), h1.pow(3).mul(1000))}$ du volume de SABCD.`
this.MG32codeBase64 = codeBase64
this.mg32init = (mtg32App, idDoc) => {
mtg32App.giveFormula2(idDoc, 'c', c.toString())
mtg32App.giveFormula2(idDoc, 'h1', h1.toString())
mtg32App.giveFormula2(idDoc, 'h2', texNombre(h2, 0))
mtg32App.giveFormula2(idDoc, "c'", c2.toString())
mtg32App.calculate(idDoc, true)
return mtg32App.display(idDoc)
}
break
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
r = new Decimal(randint(15, 20)).div(10)
r2 = new Decimal(randint(11, r * 10 - 3)).div(10)
h3 = new Decimal(randint(10, 15)).div(5)
h2 = r2.mul(h3).div(r.sub(r2))
h1 = h2.plus(h3)
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.
r = new Decimal(randint(15, 20)).div(10)
r2 = new Decimal(randint(11, r * 10 - 3)).div(10)
h3 = new Decimal(randint(10, 15)).div(5)
h2 = r2.mul(h3).div(r.sub(r2))
h1 = h2.plus(h3)
}
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if (!context.isHtml) { texte = '\\begin{minipage}{0.7 \\linewidth} \n\t' } else { texte = '' }
texte += 'Un seau a la forme d\'un tronc de cône.<br>'
texte += `Sa hauteur intérieure est de $${texNombre(h3, 1)}${sp()}$dm, sa petite base a un diamètre intérieur de $${texNombre(r2 * 2)}${sp()}$dm et sa grande base (l'ouverture) a un diamètre intérieur de $${texNombre(r * 2)}${sp()}$dm.<br>`
if (!context.isHtml) { texte += 'La figure n\'est pas en vraie grandeur.<br>' }
texte += numAlpha(0) + ' Calculer la hauteur du cône obtenu en prolongeant les bords du seau.<br>'
texte += numAlpha(1) + ' En déduire le volume de ce cône.<br>'
texte += numAlpha(2) + ' En déduire le volume total du seau.<br>'
if (context.isHtml) { texte += 'Dans ces deux prochaines questions, on considère qu\'on remplit le seau à mi-hauteur d\'eau.<br>' } else { texte += numAlpha(3) + ' On remplit le seau à mi-hauteur d\'eau. Calculer le volume d\'eau correspondant.' }
if (context.isHtml) {
texte += numAlpha(3) + ' Par lecture graphique, après avoir correctement paramétré la figure, lire le volume d\'eau correspondant.<br>'
texte += numAlpha(4) + ' Retrouver ce résultat par le calcul.<br>'
}
if (context.isHtml) { texte += '<br>On peut déplacer le cône avec S et modifier les valeurs avec leurs curseurs respectifs.' } else {
texte += `\n\t \\end{minipage} \n\t \\begin{minipage}{0.3 \\linewidth} \n\t \\begin{tikzpicture}[scale=0.8]
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\\node at (3.331, 0.65) [align=left,below right ,black,,font= \\sf \\fontsize {0.398cm} {0.498cm} \\selectfont] {\\textbf{S}};
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}
texteCorr = numAlpha(0) + ' Les triangles SBA et SHL sont semblables car les droites (BA) et (HL) sont parallèles dans le plan du triangle SHL.<br>'
texteCorr += ' La tangente de l\'angle $\\widehat{HSL}$ est égale dans le triangle SHL à $\\dfrac{\\text{HL}}{\\text{SH}}$ et dans le triangle SBA à $\\dfrac{\\text{BA}}{\\text{SB}}$.<br>'
texteCorr += ' d\'où $\\dfrac{\\text{HL}}{\\text{SH}}=\\dfrac{\\text{BA}}{\\text{SH}-\\text{BH}}$ '
texteCorr += ' et l\'égalité des produits en croix nous donne : $\\text{HL}\\left(\\text{SH}-\\text{BH} \\right) =\\text{BA}\\times \\text{SH}$.<br>'
texteCorr += ` Soit avec les données numériques : $${texNombre(r, 1)}\\left(\\text{SH}-${texNombre(h3, 1)}\\right)=${texNombre(r2, 1)}\\times \\text{SH}$.<br>`
texteCorr += `On en déduit que SH$\\left(${texNombre(r, 1)}-${texNombre(r2, 1)}\\right)=${texNombre(r, 1)}\\times${texNombre(h3, 1)}$.<br>`
texteCorr += `D'où SH $=\\dfrac{${texNombre(r.mul(h3), 2)}}{${texNombre(r.sub(r2), 1)}}=${texNombre(h1, 1)}$ dm ( SB = $${texNombre(h2, 1)}$ dm).<br>`
texteCorr += numAlpha(1) + ` Le volume du cône est $\\dfrac{A_\\text{base}}{3}\\times \\text{hauteur}$ 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>`
texteCorr += numAlpha(2) + ' Le seau est un tronc de cône. Pour calculer son volume, on va calculer le volume du cône réduit de hauteur SB et le soustraire du volume du cône de hauteur SH.<br>'
texteCorr += ` Le cône de hauteur SB est une réduction du cône de hauteur SH. Le coefficient de cette réduction est : $\\dfrac{${texNombre(r2, 1)}}{${texNombre(r, 1)}}`
if (!Number.isInteger(r) || pgcd(r2.mul(10), r * 10) > 1) { texteCorr += `=${texFractionReduite(r2.mul(10), r.mul(10))}$.<br>` } else { texteCorr += '.$<br>' }
texteCorr += `Dans une réduction de coefficient k, les volumes sont multipliés par k${texteExposant(3)}.<br>`
texteCorr += `Donc le volume du cône de hauteur SB est : $\\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>`
texteCorr += 'Le volume du tronc de cône est : '
texteCorr += `$V_\\text{Cône} - V_\\text{CôneRéduit}$<br>Soit : <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)} `
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)} `
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)} `
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)} `
kprime = new Decimal(r2).div(r).pow(3)
texteCorr += `$ \\approx ${texNombre(r.pow(2).mul(h1).mul(pi).div(3).mul(kprime.sub(1).mul(-1)), 3)}$ dm${texteExposant(3)}<br>`
c = h3.div(2)
kprime = c.plus(h2).div(h2)
if (context.isHtml) {
texteCorr += numAlpha(3) + ` Il faut fixer HL à ${texNombre(r * 10)} cm ; BA à ${texNombre(r2.mul(10), 0)} cm ; BH à ${texNombre(h3.mul(10), 0)} cm et la hauteur d'eau à ${texNombre(h1.sub(h2).mul(5), 1)} cm.<br>`
texteCorr += `La lecture de $ y = V(x)$ nous donne un volume d'eau d'environ $${texNombre(kprime.pow(3).sub(1).mul(r2.pow(2)).mul(h2).mul(pi).div(3), 1)}$ dm${texteExposant(3)} soit environ $${texNombre(kprime.pow(3).sub(1).mul(r2.pow(2)).mul(h2).mul(pi).div(3), 0)}$ litres d'eau.<br>`
texteCorr += numAlpha(4)
} else { texteCorr += numAlpha(3) }
texteCorr += ' Nous allons déterminer le volume du cône de hauteur SE, puis nous soustrairons le volume du cône de hauteur SB pour obtenir le volume d\'eau.<br>'
texteCorr += ` Le cône de hauteur SE est une réduction du cône de hauteur SH. Le coefficient de cette réduction est : $${texFractionReduite(c.add(h2).mul(100), h1.mul(100))}$.<br>`
texteCorr += `Donc le volume $V$ du cône de hauteur SE est : $\\left(${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>`
texteCorr += 'Notons $V\'$ le volume du cône de hauteur SB calculé à la question ' + numAlpha(2) + '<br> Le volume d\'eau est donc : '
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>`
this.MG32codeBase64 = codeBase64
this.mg32init = (mtg32App, idDoc) => {
mtg32App.calculate(idDoc, true)
return mtg32App.display(idDoc)
}
break
case 5: // Un problème avec un cône Vanille Chocolat.
r = new Decimal(randint(20, 28)).div(10)
h1 = new Decimal(randint(20, 28)).div(2)
h2 = new Decimal(randint(1, 3))
if (this.sup2 < 3) {
if (parseInt(this.sup2) === 1) { // coefficient de réduction décimal
while (!h2.div(h1).mul(10).eq(h2.div(h1).mul(10).round())) {
r = new Decimal(randint(20, 28)).div(10)
h1 = new Decimal(randint(20, 28)).div(2)
h2 = new Decimal(randint(1, 3))
}
} else { // coefficient de réduction rationnel
while (h2.div(h1).mul(10).eq(h2.div(h1).mul(10).round())) {
r = new Decimal(randint(20, 28)).div(10)
h1 = new Decimal(randint(20, 28)).div(2)
h2 = new Decimal(randint(1, 3))
}
}
}
r2 = r.mul(h2).div(h1)
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if (!context.isHtml) { texte = '\\begin{minipage}{0.7 \\linewidth} \n\t' } else { texte = '' }
texte += `Un cône de glace d'une marque célèbre a pour rayon $${texNombre(r, 1)}${sp()}$cm et pour hauteur $SO${sp()}=${sp()}${texNombre(h1, 1)}${sp()}$cm.<br>`
texte += `Le fabricant a coulé au fond de ce cône du chocolat sur une hauteur SO' de ${texNombre(h2, 0)}${sp()}cm.<br>`
texte += numAlpha(0) + ' Calculer le volume total du cône.<br>'
texte += numAlpha(1) + ' En déduire le volume de chocolat présent dans le fond du cône.<br>'
texte += numAlpha(2) + ' Déduire des deux premières questions le volume de glace permettant de remplir le cône.<br>'
texte += numAlpha(3) + ' Si la glace avait été mise dans le cône avant le chocolat, quelle serait la hauteur atteinte par la glace dans le cône ?<br>'
texte += numAlpha(4) + ' Quelle serait alors l\'épaisseur de chocolat au dessus de la glace ?<br>'
if (context.isHtml) { texte += 'Le point O peut être déplacé et on peut changer l\'angle de vue φ ' } else {
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\\node at (3.953, 8.694) [align=left,below right ,black,,font= \\sf \\fontsize {0.469cm} {0.586cm} \\selectfont] {S};
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\\end{tikzpicture} \n\t \\end{minipage}`
}
texteCorr = numAlpha(0) + ` Le volume du cône est $\\dfrac{A_\\text{base}}{3}\\times \\text{hauteur}$ 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>`
texteCorr += numAlpha(1) + ` Le cône de chocolat est une réduction du cône complet. Le coefficient de réduction est $\\dfrac{${texNombre(h2, 1)}}{${texNombre(h1, 1)}}`
if (!Number.isInteger(h1) || pgcd(h2, h1) > 1) { texteCorr += `=${texFractionReduite(h2.mul(10), h1.mul(10))}$.<br>` } else { texteCorr += '.$<br>' }
texteCorr += ` Dans une réduction de coefficient k, les volumes sont multipliés par k${texteExposant(3)}.<br>`
texteCorr += `Donc le volume du cône de hauteur SO' est : $\\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>`
texteCorr += numAlpha(2) + ' Le volume de glace est la différence entre les deux volumes précédents :<br>'
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${texteExposant(3)}.<br>`
texteCorr += numAlpha(3) + ' Si on verse la glace au fond du cône, on obtient une nouvelle réduction du cône complet.<br>'
texteCorr += `Soit k' le coefficient de cette réduction, on a : k'${texteExposant(3)} $= 1- \\text{k}^3$`
texteCorr += ', d\'où k\' '
texteCorr += '$= \\sqrt[3]{1-{\\text{k}^3}}$.<br>'
kprime = new Decimal(1).sub(h2.div(h1).pow(3)).cbrt()
texteCorr += `Donc k' = $\\sqrt[3]{1 - \\left(${texFractionReduite(h2.mul(10), h1.mul(10))}\\right)^3} \\approx ${texNombre(kprime, 5)}$.<br>`
texteCorr += `On en déduit que la hauteur de glace est approximativement : $${texNombre(kprime, 5)} \\times ${texNombre(h1, 1)}$ cm $\\approx ${texNombre(kprime.mul(h1), 4)}$ cm.<br>`
texteCorr += numAlpha(4) + ` L'épaisseur de chocolat est alors de : $${texNombre(h1, 1)}\\text{ cm}-${texNombre(kprime.mul(h1), 4)} \\text{ cm}\\approx ${texNombre(h1.sub(kprime.mul(h1)).mul(10), 2)}$ mm !`
this.MG32codeBase64 = codeBase64
this.mg32init = (mtg32App, idDoc) => {
mtg32App.giveFormula2(idDoc, 'r', r.toString())
mtg32App.giveFormula2(idDoc, 'h1', h1.toString())
mtg32App.giveFormula2(idDoc, 'h2', h2.toString())
mtg32App.calculate(idDoc, true)
return mtg32App.display(idDoc)
}
break
}
this.listeQuestions.push(texte)
this.listeCorrections.push(texteCorr)
listeQuestionsToContenu(this)
}
this.besoinFormulaireNumerique = ['Type d\'exercices', 3, ' 1 : Calcul d\' aire et de volumes\n 2 : Problème complexe\n 3 : Mélange']
this.besoinFormulaire2Numerique = ['Coefficient de réduction (problèmes de type 1)', 3, ' 1 : Décimal\n 2 : Non décimal\n 3 : Mélange']
}