Здравствуйте, помогите пожалуйста решить 3 задачки.. Created by Tomoki. geometriya-ru

3) ΔEBF подобен ΔCBA (2 -ой признак подобия) :∠B -общий и две стороны пропорциональны .Следовательно ∠A =∠BFE =40° как соответствующие углы.4) Треугольники ABC и A₁B₁C₁ подобны по третьему признаку подобия .Коэффициент подобия К=5/2. Отношения площадей подобных треугольников равен квадрату коэффициента подобия.S/S₁ = 25/4.{S/S₁ = 25/4 ;. S+S₁ =58. S = (58*/(25+4))*25 =50 (см²).S = (58*/(25+4))*4 =8 (см²).5) S(AOB)/S(ABE) =AO/AE =2/3.S(AOB)=(2/3) *S(ABE) =2/3) *S(ABC)/2 =S(ABC)/3.S(AOB)=12/3 =4 (см²).ответ : 4 см².Спасибо вам огромное !Новые вопросы в ГеометрияПж , 2 рисунок. По теореме Пифагора Помогите пожалуйста срочноСРОЧНО!! 50 БАЛЛОВУ трикутнику ABC проведено медіани BE і CF; M – точка перетину медіан. Доведіть, що площі трикутника BMC і чотирикутника AEMF рівні….СРОЧНО!!У трикутнику ABC проведено медіани BE і CF; M – точка перетину медіан. Доведіть, що площі трикутника BMC і чотирикутника AEMF рівні.Даю 40 балів за вирішення задачі з розв'язком.(Чорним написано дано, синім те що потрібно знайти)Подскажите пж дам 20 балов!!Задача! У прямокутному трикутнику катети дорівнюють 24 і 7 , знайдіть косинус гострого кута який лежить навпроти меншого катета? Розпишіть будь ласка!…!!У рівнобічну трапецію вписано коло. Знайдіть радіус кола, якщо основи трапеції відносяться як 2:3, а бічна сторона дорівнює 10 см. Віповідь: 2√6 смУ трикутнику ABC вписано ромб LMCK (на рис.). Знайдіть довжину сторони AC трикутника, якщо довжина його сторони BC дорівнює 7, а сторона ромба дорівню…є 5.У прямокутнику ABCD AB=15, ∠CAB>45 градусів. Яке з наведених чисел МОЖЕ бути довжиною сторони BC цього прямокутника? А 12. Б 16. В 5. Г 10. Д 15.ПредыдущийСледующийВместе мы знаем большеЭтот сайт использует cookies Политика Cookies . 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Здравствуйте, помогите пожалуйста решить 3 задачки...

Tomoki @Tomoki

July 2022 1 17 Report

Здравствуйте, помогите пожалуйста решить 3 задачки.

Answers & Comments

oganesbagoyan

Verified answer

3) ΔEBF подобен ΔCBA (2 -ой признак подобия) :∠B -общий и две стороны
пропорциональны .Следовательно ∠A =∠BFE =40° как соответствующие углы.
4) Треугольники ABC и A₁B₁C₁ подобны по третьему признаку подобия .
Коэффициент подобия К=5/2. Отношения площадей подобных треугольников равен квадрату коэффициента подобия.
S/S₁ = 25/4.
{S/S₁ = 25/4 ;. S+S₁ =58.
S = (58*/(25+4))*25 =50 (см²).
S = (58*/(25+4))*4 =8 (см²).
5) S(AOB)/S(ABE) =AO/AE =2/3.
S(AOB)=(2/3) *S(ABE) =2/3) *S(ABC)/2 =S(ABC)/3.

S(AOB)=12/3 =4 (см²).
ответ : 4 см².