Номер 52, страница 190 - гдз по геометрии 9 класс учебник Солтан, Солтан

Геометрия, 9 класс Учебник, авторы: Солтан Г Н, Солтан Алла Евгеньевна, Жумадилова Аманбала Жумадиловна, издательство Кокшетау, Алматы, 2019

Авторы: Солтан Г. Н., Солтан А. Е., Жумадилова А. Ж.

Тип: Учебник

Издательство: Кокшетау

Год издания: 2019 - 2026

ISBN: 978-601-317-432-7

Рекомендовано Министерством образования и науки Республики Казахстан

Задания для итоговой самопроверки - номер 52, страница 190.

App\Models\Task {#1030 // resources/views/models/task/default.blade.php
  #connection: "mysql"
  #table: "tasks"
  #primaryKey: "id"
  #keyType: "int"
  +incrementing: true
  #with: []
  #withCount: []
  +preventsLazyLoading: false
  #perPage: 15
  +exists: true
  +wasRecentlyCreated: false
  #escapeWhenCastingToString: false
  #attributes: array:24 [
    "id" => 1136790
    "created_at" => "2026-04-10 13:58:26"
    "updated_at" => null
    "field_page_start" => "190"
    "field_page_end" => null
    "field_url" => "/9-klass/geometrija/kokshetau-soltan-uchebnik/vopr-550"
    "field_display_title" => "52"
    "field_outside_task" => null
    "field_task_type" => Illuminate\Database\Eloquent\Collection {#1037
      #items: array:1 [
        0 => App\Models\Term {#1036
          #connection: "mysql"
          #table: "terms"
          #primaryKey: "id"
          #keyType: "int"
          +incrementing: false
          #with: []
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          +preventsLazyLoading: false
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          #escapeWhenCastingToString: false
          #attributes: array:6 [
            "id" => 26
            "created_at" => "2026-04-10 13:58:26"
            "updated_at" => null
            "name" => "номер"
            "field_cases" => array:6 [
              "field_accusative_case" => "номер"
              "field_creative_case" => "номером"
              "field_dative_case" => "номеру"
              "field_genitive_case" => "номера"
              "field_nominative_case" => "номер"
              "field_prepositional_case" => "номере"
            ]
            "field_short_name" => ""
          ]
          #original: array:6 [
            "id" => 26
            "created_at" => "2026-04-10 13:58:26"
            "updated_at" => null
            "name" => "номер"
            "field_cases" => array:6 [
              "field_accusative_case" => "номер"
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              "field_nominative_case" => "номер"
              "field_prepositional_case" => "номере"
            ]
            "field_short_name" => ""
          ]
          #changes: []
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          #visible: []
          #fillable: []
          #guarded: array:1 [
            0 => "*"
          ]
        }
      ]
      #escapeWhenCastingToString: false
    }
    "field_metatags_title" => null
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    "field_h1" => null
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    "field_description_bottom" => null
    "field_match" => null
    "breadcrumbs" => []
    "edition_groups" => Illuminate\Database\Eloquent\Collection {#1035
      #items: []
      #escapeWhenCastingToString: false
    }
    "top_parent_branch" => Illuminate\Database\Eloquent\Collection {#1046
      #items: array:1 [
        0 => App\Models\Branch {#1045
          #connection: "mysql"
          #table: "branches"
          #primaryKey: "id"
          #keyType: "int"
          +incrementing: true
          #with: []
          #withCount: []
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          #attributes: array:24 [
            "id" => 1136233
            "created_at" => "2026-04-10 13:58:26"
            "updated_at" => null
            "field_display_title" => "Задания для итоговой самопроверки"
            "field_branch_order" => null
            "field_url" => null
            "field_branch_type" => Illuminate\Database\Eloquent\Collection {#1047
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              #escapeWhenCastingToString: false
            }
            "field_page_start" => "182"
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            "field_branch_cover" => []
            "field_branch_covers" => []
            "book" => Illuminate\Database\Eloquent\Collection {#1078
              #items: array:1 [
                0 => App\Models\Book {#1048
                  #connection: "mysql"
                  #table: "books"
                  #primaryKey: "id"
                  #keyType: "int"
                  +incrementing: false
                  #with: []
                  #withCount: []
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                  #perPage: 15
                  +exists: true
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                  #escapeWhenCastingToString: false
                  #attributes: array:50 [
                    "id" => 4312
                    "created_at" => "2026-04-10 13:58:26"
                    "updated_at" => null
                    "field_tree_default_status" => "tasks"
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                    "field_subject" => Illuminate\Database\Eloquent\Collection {#1050
                      #items: array:1 [
                        0 => App\Models\Term {#1049
                          #connection: "mysql"
                          #table: "terms"
                          #primaryKey: "id"
                          #keyType: "int"
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                            "id" => 6477
                            "created_at" => "2026-04-10 13:58:26"
                            "updated_at" => null
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                              "field_prepositional_case" => "геометрии"
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                            "field_subject_type" => "technical_subject"
                            "field_translit" => "geometrija"
                          ]
                          #original: array:10 [
                            "id" => 6477
                            "created_at" => "2026-04-10 13:58:26"
                            "updated_at" => null
                            "name" => "Геометрия"
                            "field_abbreviated_name" => null
                            "field_cases" => array:6 [
                              "field_accusative_case" => "геометрию"
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                              "field_genitive_case" => "геометрии"
                              "field_nominative_case" => "геометрия"
                              "field_prepositional_case" => "геометрии"
                            ]
                            "field_foreign_lang_name" => null
                            "field_short_name" => null
                            "field_subject_type" => "technical_subject"
                            "field_translit" => "geometrija"
                          ]
                          #changes: []
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                          #dateFormat: null
                          #appends: []
                          #dispatchesEvents: []
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                          #touches: []
                          +timestamps: true
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                          #hidden: []
                          #visible: []
                          #fillable: []
                          #guarded: array:1 [
                            0 => "*"
                          ]
                        }
                      ]
                      #escapeWhenCastingToString: false
                    }
                    "field_class" => Illuminate\Database\Eloquent\Collection {#1051
                      #items: array:1 [
                        0 => App\Models\Term {#1052
                          #connection: "mysql"
                          #table: "terms"
                          #primaryKey: "id"
                          #keyType: "int"
                          +incrementing: false
                          #with: []
                          #withCount: []
                          +preventsLazyLoading: false
                          #perPage: 15
                          +exists: true
                          +wasRecentlyCreated: false
                          #escapeWhenCastingToString: false
                          #attributes: array:6 [
                            "id" => 5458
                            "created_at" => "2026-04-10 13:58:26"
                            "updated_at" => null
                            "name" => "9"
                            "field_cases" => array:6 [
                              "field_accusative_case" => "девятый"
                              "field_creative_case" => "девятым"
                              "field_dative_case" => "девятому"
                              "field_genitive_case" => "девятого"
                              "field_nominative_case" => "девятый"
                              "field_prepositional_case" => "девятом"
                            ]
                            "field_translit" => "devjatyj"
                          ]
                          #original: array:6 [
                            "id" => 5458
                            "created_at" => "2026-04-10 13:58:26"
                            "updated_at" => null
                            "name" => "9"
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                              "field_genitive_case" => "девятого"
                              "field_nominative_case" => "девятый"
                              "field_prepositional_case" => "девятом"
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                          ]
                          #changes: []
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                      ]
                      #escapeWhenCastingToString: false
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                    "field_publisher" => Illuminate\Database\Eloquent\Collection {#1053
                      #items: array:1 [
                        0 => App\Models\Term {#1054
                          #connection: "mysql"
                          #table: "terms"
                          #primaryKey: "id"
                          #keyType: "int"
                          +incrementing: false
                          #with: []
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                          #attributes: array:6 [
                            "id" => 7002
                            "created_at" => "2026-04-10 13:58:26"
                            "updated_at" => null
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                            "field_cases" => array:6 [
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                              "field_nominative_case" => null
                              "field_prepositional_case" => null
                            ]
                            "field_translit" => "Kokshetau"
                          ]
                          #original: array:6 [
                            "id" => 7002
                            "created_at" => "2026-04-10 13:58:26"
                            "updated_at" => null
                            "name" => "Кокшетау"
                            "field_cases" => array:6 [
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                            "field_translit" => "Kokshetau"
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                      ]
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                    "field_author" => Illuminate\Database\Eloquent\Collection {#1055
                      #items: array:3 [
                        0 => App\Models\Term {#1056
                          #connection: "mysql"
                          #table: "terms"
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                        1 => App\Models\Term {#1057
                          #connection: "mysql"
                          #table: "terms"
                          #primaryKey: "id"
                          #keyType: "int"
                          +incrementing: false
                          #with: []
                          #withCount: []
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                            "created_at" => "2026-04-10 13:58:26"
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                            "field_name" => "Алла"
                            "field_patronymic" => "Евгеньевна"
                            "field_surname_rp" => null
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                            "id" => 7004
                            "created_at" => "2026-04-10 13:58:26"
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                        2 => App\Models\Term {#1058
                          #connection: "mysql"
                          #table: "terms"
                          #primaryKey: "id"
                          #keyType: "int"
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                          #attributes: array:12 [
                            "id" => 7005
                            "created_at" => "2026-04-10 13:58:26"
                            "updated_at" => null
                            "name" => "Жумадилова"
                            "field_bio" => null
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                            "field_foreign_lang_patronymic" => null
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                            "field_name" => "Аманбала"
                            "field_patronymic" => "Жумадиловна"
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                            "updated_at" => null
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                    "field_author_foreign" => Illuminate\Database\Eloquent\Collection {#1059
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                        0 => App\Models\Term {#1061
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                          #attributes: array:10 [
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                            "created_at" => "2026-04-10 13:58:26"
                            "updated_at" => null
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                    "field_country" => Illuminate\Database\Eloquent\Collection {#1062
                      #items: array:1 [
                        0 => App\Models\Term {#1063
                          #connection: "mysql"
                          #table: "terms"
                          #primaryKey: "id"
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                            "updated_at" => null
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                            "field_cases" => array:6 [
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            "task" => array:2 [
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                          +exists: true
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                          #attributes: array:6 [
                            "id" => 6714
                            "created_at" => "2026-04-10 13:58:26"
                            "updated_at" => null
                            "name" => "Gdz"
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                            "updated_at" => null
                            "name" => "Gdz"
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                            "field_translit" => "gdz"
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                            0 => "*"
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                      ]
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            "text" => "<p><strong>Дано:</strong></p><p>треугольник $ABC$;</p><p>$AC = 2$ дм;</p><p>$\angle A = 30^\circ$;</p><p>$\angle C = 45^\circ$.</p><p>Перевод в СИ:</p><p>$AC = 2$ дм $ = 0.2$ м.</p><p><strong>Найти:</strong></p><p>высоту $BH$, проведенную к стороне $AC$.</p><p><strong>Решение:</strong></p><p>Пусть $BH$ — высота треугольника $ABC$, проведенная из вершины $B$ к стороне $AC$. Точка $H$ лежит на отрезке $AC$, так как углы $A$ и $C$ острые (меньше $90^\circ$).</p><p>Высота $BH$ перпендикулярна стороне $AC$, поэтому треугольники $ABH$ и $CBH$ являются прямоугольными.</p><p>Рассмотрим прямоугольный треугольник $ABH$:</p><p>$\angle A = 30^\circ$.</p><p>Тангенс угла $A$ определяется как отношение противолежащего катета $BH$ к прилежащему катету $AH$:</p><p>$\tan A = \frac{BH}{AH}$</p><p>Отсюда, $AH = \frac{BH}{\tan A} = \frac{BH}{\tan 30^\circ}$.</p><p>Известно, что $\tan 30^\circ = \frac{1}{\sqrt{3}}$.</p><p>Следовательно, $AH = \frac{BH}{1/\sqrt{3}} = BH\sqrt{3}$.</p><p>Рассмотрим прямоугольный треугольник $CBH$:</p><p>$\angle C = 45^\circ$.</p><p>Тангенс угла $C$ определяется как отношение противолежащего катета $BH$ к прилежащему катету $HC$:</p><p>$\tan C = \frac{BH}{HC}$</p><p>Отсюда, $HC = \frac{BH}{\tan C} = \frac{BH}{\tan 45^\circ}$.</p><p>Известно, что $\tan 45^\circ = 1$.</p><p>Следовательно, $HC = \frac{BH}{1} = BH$.</p><p>Длина стороны $AC$ равна сумме длин отрезков $AH$ и $HC$:</p><p>$AC = AH + HC$</p><p>Подставим найденные выражения для $AH$ и $HC$:</p><p>$AC = BH\sqrt{3} + BH$</p><p>$AC = BH(\sqrt{3} + 1)$</p><p>Выразим высоту $BH$:</p><p>$BH = \frac{AC}{\sqrt{3} + 1}$</p><p>Подставим значение $AC = 2$ дм:</p><p>$BH = \frac{2}{\sqrt{3} + 1}$</p><p>Чтобы избавиться от иррациональности в знаменателе, умножим числитель и знаменатель на сопряженное выражение $(\sqrt{3} - 1)$:</p><p>$BH = \frac{2}{(\sqrt{3} + 1)} \cdot \frac{(\sqrt{3} - 1)}{(\sqrt{3} - 1)}$</p><p>Используем формулу разности квадратов $(a+b)(a-b) = a^2 - b^2$:</p><p>$BH = \frac{2(\sqrt{3} - 1)}{(\sqrt{3})^2 - (1)^2}$</p><p>$BH = \frac{2(\sqrt{3} - 1)}{3 - 1}$</p><p>$BH = \frac{2(\sqrt{3} - 1)}{2}$</p><p>$BH = \sqrt{3} - 1$</p><p>Единица измерения — дециметры (дм).</p><p><strong>Ответ:</strong></p><p>Высота треугольника, проведенная к стороне $AC$, равна $(\sqrt{3} - 1)$ дм.</p>"
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            "edition" => Illuminate\Database\Eloquent\Collection {#1110}
            "task" => array:2 [
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              "type" => "task"
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            "text" => "<p><strong>Дано:</strong></p><p>треугольник $ABC$;</p><p>$AC = 2$ дм;</p><p>$\angle A = 30^\circ$;</p><p>$\angle C = 45^\circ$.</p><p>Перевод в СИ:</p><p>$AC = 2$ дм $ = 0.2$ м.</p><p><strong>Найти:</strong></p><p>высоту $BH$, проведенную к стороне $AC$.</p><p><strong>Решение:</strong></p><p>Пусть $BH$ — высота треугольника $ABC$, проведенная из вершины $B$ к стороне $AC$. Точка $H$ лежит на отрезке $AC$, так как углы $A$ и $C$ острые (меньше $90^\circ$).</p><p>Высота $BH$ перпендикулярна стороне $AC$, поэтому треугольники $ABH$ и $CBH$ являются прямоугольными.</p><p>Рассмотрим прямоугольный треугольник $ABH$:</p><p>$\angle A = 30^\circ$.</p><p>Тангенс угла $A$ определяется как отношение противолежащего катета $BH$ к прилежащему катету $AH$:</p><p>$\tan A = \frac{BH}{AH}$</p><p>Отсюда, $AH = \frac{BH}{\tan A} = \frac{BH}{\tan 30^\circ}$.</p><p>Известно, что $\tan 30^\circ = \frac{1}{\sqrt{3}}$.</p><p>Следовательно, $AH = \frac{BH}{1/\sqrt{3}} = BH\sqrt{3}$.</p><p>Рассмотрим прямоугольный треугольник $CBH$:</p><p>$\angle C = 45^\circ$.</p><p>Тангенс угла $C$ определяется как отношение противолежащего катета $BH$ к прилежащему катету $HC$:</p><p>$\tan C = \frac{BH}{HC}$</p><p>Отсюда, $HC = \frac{BH}{\tan C} = \frac{BH}{\tan 45^\circ}$.</p><p>Известно, что $\tan 45^\circ = 1$.</p><p>Следовательно, $HC = \frac{BH}{1} = BH$.</p><p>Длина стороны $AC$ равна сумме длин отрезков $AH$ и $HC$:</p><p>$AC = AH + HC$</p><p>Подставим найденные выражения для $AH$ и $HC$:</p><p>$AC = BH\sqrt{3} + BH$</p><p>$AC = BH(\sqrt{3} + 1)$</p><p>Выразим высоту $BH$:</p><p>$BH = \frac{AC}{\sqrt{3} + 1}$</p><p>Подставим значение $AC = 2$ дм:</p><p>$BH = \frac{2}{\sqrt{3} + 1}$</p><p>Чтобы избавиться от иррациональности в знаменателе, умножим числитель и знаменатель на сопряженное выражение $(\sqrt{3} - 1)$:</p><p>$BH = \frac{2}{(\sqrt{3} + 1)} \cdot \frac{(\sqrt{3} - 1)}{(\sqrt{3} - 1)}$</p><p>Используем формулу разности квадратов $(a+b)(a-b) = a^2 - b^2$:</p><p>$BH = \frac{2(\sqrt{3} - 1)}{(\sqrt{3})^2 - (1)^2}$</p><p>$BH = \frac{2(\sqrt{3} - 1)}{3 - 1}$</p><p>$BH = \frac{2(\sqrt{3} - 1)}{2}$</p><p>$BH = \sqrt{3} - 1$</p><p>Единица измерения — дециметры (дм).</p><p><strong>Ответ:</strong></p><p>Высота треугольника, проведенная к стороне $AC$, равна $(\sqrt{3} - 1)$ дм.</p>"
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                      #escapeWhenCastingToString: false
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                    "parent_branches" => Illuminate\Database\Eloquent\Collection {#1195
                      #items: array:1 [
                        0 => App\Models\Branch {#1045}
                      ]
                      #escapeWhenCastingToString: false
                    }
                    "content" => Illuminate\Database\Eloquent\Collection {#1198
                      #items: array:3 [
                        0 => App\Models\Element {#1207
                          #connection: "mysql"
                          #table: "elements"
                          #primaryKey: "id"
                          #keyType: "int"
                          +incrementing: true
                          #with: []
                          #withCount: []
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                          #perPage: 15
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                          #attributes: array:7 [
                            "id" => 1357566
                            "created_at" => "2026-04-10 13:58:26"
                            "updated_at" => null
                            "edition" => Illuminate\Database\Eloquent\Collection {#1208 …2}
                            "task" => array:2 [ …2]
                            "text" => "<strong>50. Длина гипотенузы прямоугольного треугольника равна 12 см.</strong><p>Найдите длины его катетов, если известно, что длина большего катета равна среднему арифметическому длин меньшего катета и гипотенузы.</p><p>(9,6 см, 7,2 см)</p>"
                            "img" => array:1 [ …1]
                          ]
                          #original: array:7 [
                            "id" => 1357566
                            "created_at" => "2026-04-10 13:58:26"
                            "updated_at" => null
                            "edition" => Illuminate\Database\Eloquent\Collection {#1208 …2}
                            "task" => array:2 [ …2]
                            "text" => "<strong>50. Длина гипотенузы прямоугольного треугольника равна 12 см.</strong><p>Найдите длины его катетов, если известно, что длина большего катета равна среднему арифметическому длин меньшего катета и гипотенузы.</p><p>(9,6 см, 7,2 см)</p>"
                            "img" => array:1 [ …1]
                          ]
                          #changes: []
                          #casts: []
                          #classCastCache: []
                          #attributeCastCache: []
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                          #relations: []
                          #touches: []
                          +timestamps: true
                          +usesUniqueIds: false
                          #hidden: []
                          #visible: []
                          #fillable: []
                          #guarded: array:1 [
                            0 => "*"
                          ]
                        }
                        1 => App\Models\Element {#1209
                          #connection: "mysql"
                          #table: "elements"
                          #primaryKey: "id"
                          #keyType: "int"
                          +incrementing: true
                          #with: []
                          #withCount: []
                          +preventsLazyLoading: false
                          #perPage: 15
                          +exists: true
                          +wasRecentlyCreated: false
                          #escapeWhenCastingToString: false
                          #attributes: array:6 [
                            "id" => 1358166
                            "created_at" => "2026-04-10 13:58:26"
                            "updated_at" => null
                            "edition" => Illuminate\Database\Eloquent\Collection {#1210 …2}
                            "task" => array:2 [ …2]
                            "img" => array:1 [ …1]
                          ]
                          #original: array:6 [
                            "id" => 1358166
                            "created_at" => "2026-04-10 13:58:26"
                            "updated_at" => null
                            "edition" => Illuminate\Database\Eloquent\Collection {#1210 …2}
                            "task" => array:2 [ …2]
                            "img" => array:1 [ …1]
                          ]
                          #changes: []
                          #casts: []
                          #classCastCache: []
                          #attributeCastCache: []
                          #dateFormat: null
                          #appends: []
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                          #relations: []
                          #touches: []
                          +timestamps: true
                          +usesUniqueIds: false
                          #hidden: []
                          #visible: []
                          #fillable: []
                          #guarded: array:1 [
                            0 => "*"
                          ]
                        }
                        2 => App\Models\Element {#1211
                          #connection: "mysql"
                          #table: "elements"
                          #primaryKey: "id"
                          #keyType: "int"
                          +incrementing: true
                          #with: []
                          #withCount: []
                          +preventsLazyLoading: false
                          #perPage: 15
                          +exists: true
                          +wasRecentlyCreated: false
                          #escapeWhenCastingToString: false
                          #attributes: array:6 [
                            "id" => 1561764
                            "created_at" => "2026-04-10 13:58:26"
                            "updated_at" => null
                            "edition" => Illuminate\Database\Eloquent\Collection {#1212 …2}
                            "task" => array:2 [ …2]
                            "text" => "<p><strong>Дано:</strong></p><p>Длина гипотенузы прямоугольного треугольника: $c = 12$ см</p><p>Длина большего катета ($b$) равна среднему арифметическому длин меньшего катета ($a$) и гипотенузы ($c$): $b = \frac{a + c}{2}$</p><p><strong>Перевод в СИ:</strong></p><p>$c = 12$ см $= 0.12$ м</p><p>Длины катетов $a$ и $b$ будут получены в метрах при пересчете.</p><p><strong>Найти:</strong></p><p>Длины катетов $a$ и $b$.</p><p><strong>Решение:</strong></p><p>Обозначим длины катетов как $a$ (меньший катет) и $b$ (больший катет). Длина гипотенузы $c = 12$ см.</p><p>Согласно условию задачи, длина большего катета $b$ равна среднему арифметическому длин меньшего катета $a$ и гипотенузы $c$:</p><p>$b = \frac{a + c}{2}$</p><p>Подставим известное значение гипотенузы $c = 12$ см:</p><p>$b = \frac{a + 12}{2} \quad (1)$</p><p>Для прямоугольного треугольника справедлива теорема Пифагора:</p><p>$a^2 + b^2 = c^2$</p><p>Подставим значение $c = 12$ см:</p><p>$a^2 + b^2 = 12^2$</p><p>$a^2 + b^2 = 144 \quad (2)$</p><p>Теперь подставим выражение для $b$ из уравнения $(1)$ в уравнение $(2)$:</p><p>$a^2 + \left(\frac{a + 12}{2}\right)^2 = 144$</p><p>$a^2 + \frac{(a + 12)^2}{4} = 144$</p><p>Умножим обе части уравнения на 4, чтобы избавиться от знаменателя:</p><p>$4a^2 + (a + 12)^2 = 144 \times 4$</p><p>Раскроем скобки $(a + 12)^2 = a^2 + 2 \times a \times 12 + 12^2 = a^2 + 24a + 144$:</p><p>$4a^2 + a^2 + 24a + 144 = 576$</p><p>Приведем подобные члены и перенесем все слагаемые в левую часть уравнения:</p><p>$5a^2 + 24a + 144 - 576 = 0$</p><p>$5a^2 + 24a - 432 = 0$</p><p>Это квадратное уравнение вида $Ax^2 + Bx + C = 0$, где $A = 5$, $B = 24$, $C = -432$. Найдем дискриминант $D = B^2 - 4AC$:</p><p>$D = 24^2 - 4 \times 5 \times (-432)$</p><p>$D = 576 - 20 \times (-432)$</p><p>$D = 576 + 8640$</p><p>$D = 9216$</p><p>Найдем квадратный корень из дискриминанта:</p><p>$\sqrt{D} = \sqrt{9216} = 96$</p><p>Теперь найдем значения $a$ по формуле $a = \frac{-B \pm \sqrt{D}}{2A}$:</p><p>$a_1 = \frac{-24 + 96}{2 \times 5} = \frac{72}{10} = 7.2$</p><p>$a_2 = \frac{-24 - 96}{2 \times 5} = \frac{-120}{10} = -12$</p><p>Длина отрезка не может быть отрицательной, поэтому принимаем $a = 7.2$ см.</p><p>Теперь найдем длину большего катета $b$, используя уравнение $(1)$:</p><p>$b = \frac{a + 12}{2} = \frac{7.2 + 12}{2} = \frac{19.2}{2} = 9.6$</p><p>Таким образом, длины катетов равны 7.2 см и 9.6 см. Проверим, что $a$ является меньшим катетом, а $b$ большим: $7.2 &lt; 9.6$, что соответствует условию задачи.</p><p><strong>Ответ:</strong></p><p>Длины катетов составляют 9.6 см и 7.2 см.</p>"
                          ]
                          #original: array:6 [
                            "id" => 1561764
                            "created_at" => "2026-04-10 13:58:26"
                            "updated_at" => null
                            "edition" => Illuminate\Database\Eloquent\Collection {#1212 …2}
                            "task" => array:2 [ …2]
                            "text" => "<p><strong>Дано:</strong></p><p>Длина гипотенузы прямоугольного треугольника: $c = 12$ см</p><p>Длина большего катета ($b$) равна среднему арифметическому длин меньшего катета ($a$) и гипотенузы ($c$): $b = \frac{a + c}{2}$</p><p><strong>Перевод в СИ:</strong></p><p>$c = 12$ см $= 0.12$ м</p><p>Длины катетов $a$ и $b$ будут получены в метрах при пересчете.</p><p><strong>Найти:</strong></p><p>Длины катетов $a$ и $b$.</p><p><strong>Решение:</strong></p><p>Обозначим длины катетов как $a$ (меньший катет) и $b$ (больший катет). Длина гипотенузы $c = 12$ см.</p><p>Согласно условию задачи, длина большего катета $b$ равна среднему арифметическому длин меньшего катета $a$ и гипотенузы $c$:</p><p>$b = \frac{a + c}{2}$</p><p>Подставим известное значение гипотенузы $c = 12$ см:</p><p>$b = \frac{a + 12}{2} \quad (1)$</p><p>Для прямоугольного треугольника справедлива теорема Пифагора:</p><p>$a^2 + b^2 = c^2$</p><p>Подставим значение $c = 12$ см:</p><p>$a^2 + b^2 = 12^2$</p><p>$a^2 + b^2 = 144 \quad (2)$</p><p>Теперь подставим выражение для $b$ из уравнения $(1)$ в уравнение $(2)$:</p><p>$a^2 + \left(\frac{a + 12}{2}\right)^2 = 144$</p><p>$a^2 + \frac{(a + 12)^2}{4} = 144$</p><p>Умножим обе части уравнения на 4, чтобы избавиться от знаменателя:</p><p>$4a^2 + (a + 12)^2 = 144 \times 4$</p><p>Раскроем скобки $(a + 12)^2 = a^2 + 2 \times a \times 12 + 12^2 = a^2 + 24a + 144$:</p><p>$4a^2 + a^2 + 24a + 144 = 576$</p><p>Приведем подобные члены и перенесем все слагаемые в левую часть уравнения:</p><p>$5a^2 + 24a + 144 - 576 = 0$</p><p>$5a^2 + 24a - 432 = 0$</p><p>Это квадратное уравнение вида $Ax^2 + Bx + C = 0$, где $A = 5$, $B = 24$, $C = -432$. Найдем дискриминант $D = B^2 - 4AC$:</p><p>$D = 24^2 - 4 \times 5 \times (-432)$</p><p>$D = 576 - 20 \times (-432)$</p><p>$D = 576 + 8640$</p><p>$D = 9216$</p><p>Найдем квадратный корень из дискриминанта:</p><p>$\sqrt{D} = \sqrt{9216} = 96$</p><p>Теперь найдем значения $a$ по формуле $a = \frac{-B \pm \sqrt{D}}{2A}$:</p><p>$a_1 = \frac{-24 + 96}{2 \times 5} = \frac{72}{10} = 7.2$</p><p>$a_2 = \frac{-24 - 96}{2 \times 5} = \frac{-120}{10} = -12$</p><p>Длина отрезка не может быть отрицательной, поэтому принимаем $a = 7.2$ см.</p><p>Теперь найдем длину большего катета $b$, используя уравнение $(1)$:</p><p>$b = \frac{a + 12}{2} = \frac{7.2 + 12}{2} = \frac{19.2}{2} = 9.6$</p><p>Таким образом, длины катетов равны 7.2 см и 9.6 см. Проверим, что $a$ является меньшим катетом, а $b$ большим: $7.2 &lt; 9.6$, что соответствует условию задачи.</p><p><strong>Ответ:</strong></p><p>Длины катетов составляют 9.6 см и 7.2 см.</p>"
                          ]
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                    "next" => array:2 [
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                    "previous" => array:2 [
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                      "type" => "task"
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                    "book" => Illuminate\Database\Eloquent\Collection {#1204
                      #items: array:1 [
                        0 => App\Models\Book {#1048}
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                      #escapeWhenCastingToString: false
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                    "page" => array:2 [
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                    "updated_at" => null
                    "field_page_start" => "190"
                    "field_page_end" => null
                    "field_url" => "/9-klass/geometrija/kokshetau-soltan-uchebnik/vopr-548"
                    "field_display_title" => "50"
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                    "field_task_type" => Illuminate\Database\Eloquent\Collection {#1192}
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                    "parent_branches" => Illuminate\Database\Eloquent\Collection {#1195}
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                    "book" => Illuminate\Database\Eloquent\Collection {#1204}
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                    0 => "*"
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                1 => App\Models\Task {#1197
                  #connection: "mysql"
                  #table: "tasks"
                  #primaryKey: "id"
                  #keyType: "int"
                  +incrementing: true
                  #with: []
                  #withCount: []
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                    "created_at" => "2026-04-10 13:58:26"
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                    "field_page_end" => null
                    "field_url" => "/9-klass/geometrija/kokshetau-soltan-uchebnik/vopr-549"
                    "field_display_title" => "51"
                    "field_outside_task" => null
                    "field_task_type" => Illuminate\Database\Eloquent\Collection {#1196
                      #items: array:1 [
                        0 => App\Models\Term {#1036}
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                      #escapeWhenCastingToString: false
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                    "edition_groups" => Illuminate\Database\Eloquent\Collection {#1203
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                    "top_parent_branch" => Illuminate\Database\Eloquent\Collection {#1205
                      #items: array:1 [
                        0 => App\Models\Branch {#1045}
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                      #escapeWhenCastingToString: false
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                    "parent_branches" => Illuminate\Database\Eloquent\Collection {#1201
                      #items: array:1 [
                        0 => App\Models\Branch {#1045}
                      ]
                      #escapeWhenCastingToString: false
                    }
                    "content" => Illuminate\Database\Eloquent\Collection {#1200
                      #items: array:3 [
                        0 => App\Models\Element {#1221
                          #connection: "mysql"
                          #table: "elements"
                          #primaryKey: "id"
                          #keyType: "int"
                          +incrementing: true
                          #with: []
                          #withCount: []
                          +preventsLazyLoading: false
                          #perPage: 15
                          +exists: true
                          +wasRecentlyCreated: false
                          #escapeWhenCastingToString: false
                          #attributes: array:7 [
                            "id" => 1357567
                            "created_at" => "2026-04-10 13:58:26"
                            "updated_at" => null
                            "edition" => Illuminate\Database\Eloquent\Collection {#1222 …2}
                            "task" => array:2 [ …2]
                            "text" => "<p><strong>51. a)</strong> Найдите наибольшую площадь четырехугольника, вписанного в окружность, радиус которой 9 см.</p><p><strong>б)</strong> На расстоянии 8 м от центра $O$ окружности радиуса 6 м взята точка $M$. Найдите наибольшую площадь треугольника $OMX$, где $X$ – произвольная точка окружности.</p><p>(a) 162 $см^2$; б) 24 $м^2$)</p>"
                            "img" => array:1 [ …1]
                          ]
                          #original: array:7 [
                            "id" => 1357567
                            "created_at" => "2026-04-10 13:58:26"
                            "updated_at" => null
                            "edition" => Illuminate\Database\Eloquent\Collection {#1222 …2}
                            "task" => array:2 [ …2]
                            "text" => "<p><strong>51. a)</strong> Найдите наибольшую площадь четырехугольника, вписанного в окружность, радиус которой 9 см.</p><p><strong>б)</strong> На расстоянии 8 м от центра $O$ окружности радиуса 6 м взята точка $M$. Найдите наибольшую площадь треугольника $OMX$, где $X$ – произвольная точка окружности.</p><p>(a) 162 $см^2$; б) 24 $м^2$)</p>"
                            "img" => array:1 [ …1]
                          ]
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                          #hidden: []
                          #visible: []
                          #fillable: []
                          #guarded: array:1 [
                            0 => "*"
                          ]
                        }
                        1 => App\Models\Element {#1223
                          #connection: "mysql"
                          #table: "elements"
                          #primaryKey: "id"
                          #keyType: "int"
                          +incrementing: true
                          #with: []
                          #withCount: []
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                          #perPage: 15
                          +exists: true
                          +wasRecentlyCreated: false
                          #escapeWhenCastingToString: false
                          #attributes: array:6 [
                            "id" => 1358167
                            "created_at" => "2026-04-10 13:58:26"
                            "updated_at" => null
                            "edition" => Illuminate\Database\Eloquent\Collection {#1224 …2}
                            "task" => array:2 [ …2]
                            "img" => array:2 [ …2]
                          ]
                          #original: array:6 [
                            "id" => 1358167
                            "created_at" => "2026-04-10 13:58:26"
                            "updated_at" => null
                            "edition" => Illuminate\Database\Eloquent\Collection {#1224 …2}
                            "task" => array:2 [ …2]
                            "img" => array:2 [ …2]
                          ]
                          #changes: []
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                          #visible: []
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                          #guarded: array:1 [
                            0 => "*"
                          ]
                        }
                        2 => App\Models\Element {#1225
                          #connection: "mysql"
                          #table: "elements"
                          #primaryKey: "id"
                          #keyType: "int"
                          +incrementing: true
                          #with: []
                          #withCount: []
                          +preventsLazyLoading: false
                          #perPage: 15
                          +exists: true
                          +wasRecentlyCreated: false
                          #escapeWhenCastingToString: false
                          #attributes: array:6 [
                            "id" => 1561765
                            "created_at" => "2026-04-10 13:58:26"
                            "updated_at" => null
                            "edition" => Illuminate\Database\Eloquent\Collection {#1226 …2}
                            "task" => array:2 [ …2]
                            "text" => "<p><strong>a)</strong></p><p><strong>Дано:</strong></p><p>Радиус окружности $R = 9$ см</p><p><strong>Перевод в СИ:</strong></p><p>$R = 9 \text{ см} = 0.09 \text{ м}$</p><p><strong>Найти:</strong></p><p>Наибольшую площадь четырехугольника $S_{max}$</p><p><strong>Решение:</strong></p><p>Среди всех четырехугольников, вписанных в окружность, наибольшую площадь имеет квадрат. Диагональ такого квадрата равна диаметру окружности.</p><p>Пусть сторона квадрата равна $a$. Тогда его диагональ $d$ связана со стороной формулой $d = a\sqrt{2}$.</p><p>Диаметр окружности $D$ равен двум радиусам, то есть $D = 2R$.</p><p>Поскольку диагональ квадрата является диаметром окружности, имеем $a\sqrt{2} = 2R$.</p><p>Выразим сторону квадрата $a$ через радиус $R$: $a = \frac{2R}{\sqrt{2}} = R\sqrt{2}$.</p><p>Площадь квадрата $S$ вычисляется как $S = a^2$. Подставим выражение для $a$:</p><p>$S_{max} = (R\sqrt{2})^2 = R^2 \cdot (\sqrt{2})^2 = 2R^2$.</p><p>Подставим заданное значение радиуса $R = 9$ см:</p><p>$S_{max} = 2 \cdot (9 \text{ см})^2 = 2 \cdot 81 \text{ см}^2 = 162 \text{ см}^2$.</p><p><strong>Ответ:</strong> 162 см$^2$</p><p><strong>б)</strong></p><p><strong>Дано:</strong></p><p>Расстояние от центра $O$ окружности до точки $M$: $OM = 8$ м</p><p>Радиус окружности: $R = 6$ м</p><p>Точка $X$ – произвольная точка окружности.</p><p><strong>Перевод в СИ:</strong></p><p>Все величины даны в системе СИ.</p><p><strong>Найти:</strong></p><p>Наибольшую площадь треугольника $OMX$, $S_{OMX,max}$</p><p><strong>Решение:</strong></p><p>Площадь треугольника $OMX$ определяется по формуле $S_{OMX} = \frac{1}{2} \cdot \text{основание} \cdot \text{высота}$.</p><p>В качестве основания треугольника $OMX$ возьмем отрезок $OM$. Его длина фиксирована и равна $OM = 8$ м.</p><p>Тогда площадь треугольника $S_{OMX} = \frac{1}{2} \cdot OM \cdot h_X$, где $h_X$ – высота треугольника, опущенная из вершины $X$ на прямую, содержащую основание $OM$.</p><p>Для того чтобы площадь треугольника была наибольшей, необходимо максимизировать высоту $h_X$.</p><p>Высота $h_X$ – это перпендикулярное расстояние от точки $X$ (лежащей на окружности) до прямой $OM$.</p><p>Максимальное расстояние от точки на окружности до прямой, проходящей через центр этой окружности $O$ (прямая $OM$), достигается, когда радиус $OX$ перпендикулярен прямой $OM$.</p><p>В этом случае высота $h_X$ будет равна радиусу окружности $R$.</p><p>Таким образом, максимальная высота $h_{X,max} = R = 6$ м.</p><p>Подставим значения $OM$ и $R$ в формулу площади:</p><p>$S_{OMX,max} = \frac{1}{2} \cdot OM \cdot R$</p><p>$S_{OMX,max} = \frac{1}{2} \cdot 8 \text{ м} \cdot 6 \text{ м}$</p><p>$S_{OMX,max} = \frac{1}{2} \cdot 48 \text{ м}^2$</p><p>$S_{OMX,max} = 24 \text{ м}^2$</p><p><strong>Ответ:</strong> 24 м$^2$</p>"
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                            "edition" => Illuminate\Database\Eloquent\Collection {#1226 …2}
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                            "text" => "<p><strong>a)</strong></p><p><strong>Дано:</strong></p><p>Радиус окружности $R = 9$ см</p><p><strong>Перевод в СИ:</strong></p><p>$R = 9 \text{ см} = 0.09 \text{ м}$</p><p><strong>Найти:</strong></p><p>Наибольшую площадь четырехугольника $S_{max}$</p><p><strong>Решение:</strong></p><p>Среди всех четырехугольников, вписанных в окружность, наибольшую площадь имеет квадрат. Диагональ такого квадрата равна диаметру окружности.</p><p>Пусть сторона квадрата равна $a$. Тогда его диагональ $d$ связана со стороной формулой $d = a\sqrt{2}$.</p><p>Диаметр окружности $D$ равен двум радиусам, то есть $D = 2R$.</p><p>Поскольку диагональ квадрата является диаметром окружности, имеем $a\sqrt{2} = 2R$.</p><p>Выразим сторону квадрата $a$ через радиус $R$: $a = \frac{2R}{\sqrt{2}} = R\sqrt{2}$.</p><p>Площадь квадрата $S$ вычисляется как $S = a^2$. Подставим выражение для $a$:</p><p>$S_{max} = (R\sqrt{2})^2 = R^2 \cdot (\sqrt{2})^2 = 2R^2$.</p><p>Подставим заданное значение радиуса $R = 9$ см:</p><p>$S_{max} = 2 \cdot (9 \text{ см})^2 = 2 \cdot 81 \text{ см}^2 = 162 \text{ см}^2$.</p><p><strong>Ответ:</strong> 162 см$^2$</p><p><strong>б)</strong></p><p><strong>Дано:</strong></p><p>Расстояние от центра $O$ окружности до точки $M$: $OM = 8$ м</p><p>Радиус окружности: $R = 6$ м</p><p>Точка $X$ – произвольная точка окружности.</p><p><strong>Перевод в СИ:</strong></p><p>Все величины даны в системе СИ.</p><p><strong>Найти:</strong></p><p>Наибольшую площадь треугольника $OMX$, $S_{OMX,max}$</p><p><strong>Решение:</strong></p><p>Площадь треугольника $OMX$ определяется по формуле $S_{OMX} = \frac{1}{2} \cdot \text{основание} \cdot \text{высота}$.</p><p>В качестве основания треугольника $OMX$ возьмем отрезок $OM$. Его длина фиксирована и равна $OM = 8$ м.</p><p>Тогда площадь треугольника $S_{OMX} = \frac{1}{2} \cdot OM \cdot h_X$, где $h_X$ – высота треугольника, опущенная из вершины $X$ на прямую, содержащую основание $OM$.</p><p>Для того чтобы площадь треугольника была наибольшей, необходимо максимизировать высоту $h_X$.</p><p>Высота $h_X$ – это перпендикулярное расстояние от точки $X$ (лежащей на окружности) до прямой $OM$.</p><p>Максимальное расстояние от точки на окружности до прямой, проходящей через центр этой окружности $O$ (прямая $OM$), достигается, когда радиус $OX$ перпендикулярен прямой $OM$.</p><p>В этом случае высота $h_X$ будет равна радиусу окружности $R$.</p><p>Таким образом, максимальная высота $h_{X,max} = R = 6$ м.</p><p>Подставим значения $OM$ и $R$ в формулу площади:</p><p>$S_{OMX,max} = \frac{1}{2} \cdot OM \cdot R$</p><p>$S_{OMX,max} = \frac{1}{2} \cdot 8 \text{ м} \cdot 6 \text{ м}$</p><p>$S_{OMX,max} = \frac{1}{2} \cdot 48 \text{ м}^2$</p><p>$S_{OMX,max} = 24 \text{ м}^2$</p><p><strong>Ответ:</strong> 24 м$^2$</p>"
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                            "edition" => Illuminate\Database\Eloquent\Collection {#1243 …2}
                            "task" => array:2 [ …2]
                            "text" => "<p><strong>53.</strong> Катеты прямоугольного треугольника равны 9 см и 40 см. Найдите расстояние от центра вписанной в треугольник окружности до центра описанной около него окружности. $(2,5\sqrt{41} \text{ см})$</p>"
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                            "edition" => Illuminate\Database\Eloquent\Collection {#1243 …2}
                            "task" => array:2 [ …2]
                            "text" => "<p><strong>53.</strong> Катеты прямоугольного треугольника равны 9 см и 40 см. Найдите расстояние от центра вписанной в треугольник окружности до центра описанной около него окружности. $(2,5\sqrt{41} \text{ см})$</p>"
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                        1 => App\Models\Element {#1244
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                        2 => App\Models\Element {#1246
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                            "text" => "<p><strong>Дано</strong></p><p>Прямоугольный треугольник со следующими длинами катетов:</p><p>$a = 9 \, \text{см}$<br>$b = 40 \, \text{см}$</p><p><strong>Найти</strong></p><p>Расстояние $d$ от центра вписанной окружности до центра описанной окружности.</p><p><strong>Решение</strong></p><p><strong>1.</strong> Найдем длину гипотенузы $c$ прямоугольного треугольника, используя теорему Пифагора: $c^2 = a^2 + b^2$ $c^2 = 9^2 + 40^2$ $c^2 = 81 + 1600$ $c^2 = 1681$ $c = \sqrt{1681}$ $c = 41 \, \text{см}$</p><p><strong>2.</strong> Найдем радиус $r$ вписанной окружности для прямоугольного треугольника. Формула для радиуса вписанной окружности в прямоугольный треугольник: $r = \frac{a + b - c}{2}$ $r = \frac{9 + 40 - 41}{2}$ $r = \frac{49 - 41}{2}$ $r = \frac{8}{2}$ $r = 4 \, \text{см}$</p><p><strong>3.</strong> Найдем радиус $R$ описанной окружности для прямоугольного треугольника. Центр описанной окружности прямоугольного треугольника лежит на середине гипотенузы, а ее радиус равен половине длины гипотенузы: $R = \frac{c}{2}$ $R = \frac{41}{2}$ $R = 20.5 \, \text{см}$</p><p><strong>4.</strong> Определим координаты центров окружностей. Пусть вершины прямоугольного треугольника находятся в точках $A(0, a)$, $B(b, 0)$ и $C(0, 0)$ (с прямым углом в точке $C$). В нашем случае, $A(0, 9)$, $B(40, 0)$, $C(0, 0)$ (или $A(0, 40)$, $B(9, 0)$, $C(0, 0)$ - это не повлияет на результат, так как оси симметричны). Центр вписанной окружности $O_i$ имеет координаты $(r, r)$, так как он равноудален от катетов и находится внутри угла: $O_i = (4, 4)$ Центр описанной окружности $O_o$ является серединой гипотенузы. Если вершины гипотенузы $A(0, a)$ и $B(b, 0)$, то координаты $O_o$ будут: $O_o = \left( \frac{0 + b}{2}, \frac{a + 0}{2} \right)$ $O_o = \left( \frac{40}{2}, \frac{9}{2} \right)$ $O_o = (20, 4.5)$</p><p><strong>5.</strong> Найдем расстояние $d$ между центрами $O_i(x_1, y_1)$ и $O_o(x_2, y_2)$ по формуле расстояния между двумя точками: $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$ $d = \sqrt{(20 - 4)^2 + (4.5 - 4)^2}$ $d = \sqrt{(16)^2 + (0.5)^2}$ $d = \sqrt{256 + 0.25}$ $d = \sqrt{256.25}$</p><p>Представим $256.25$ в виде дроби: $256.25 = \frac{25625}{100} = \frac{1025}{4}$. $d = \sqrt{\frac{1025}{4}}$ $d = \frac{\sqrt{1025}}{\sqrt{4}}$ $d = \frac{\sqrt{25 \cdot 41}}{2}$ $d = \frac{5\sqrt{41}}{2}$ $d = 2.5\sqrt{41} \, \text{см}$</p><p><strong>Ответ</strong></p><p>Расстояние от центра вписанной окружности до центра описанной окружности составляет $2.5\sqrt{41} \, \text{см}$.</p>"
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                            "edition" => Illuminate\Database\Eloquent\Collection {#1247 …2}
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                            "text" => "<p><strong>Дано</strong></p><p>Прямоугольный треугольник со следующими длинами катетов:</p><p>$a = 9 \, \text{см}$<br>$b = 40 \, \text{см}$</p><p><strong>Найти</strong></p><p>Расстояние $d$ от центра вписанной окружности до центра описанной окружности.</p><p><strong>Решение</strong></p><p><strong>1.</strong> Найдем длину гипотенузы $c$ прямоугольного треугольника, используя теорему Пифагора: $c^2 = a^2 + b^2$ $c^2 = 9^2 + 40^2$ $c^2 = 81 + 1600$ $c^2 = 1681$ $c = \sqrt{1681}$ $c = 41 \, \text{см}$</p><p><strong>2.</strong> Найдем радиус $r$ вписанной окружности для прямоугольного треугольника. Формула для радиуса вписанной окружности в прямоугольный треугольник: $r = \frac{a + b - c}{2}$ $r = \frac{9 + 40 - 41}{2}$ $r = \frac{49 - 41}{2}$ $r = \frac{8}{2}$ $r = 4 \, \text{см}$</p><p><strong>3.</strong> Найдем радиус $R$ описанной окружности для прямоугольного треугольника. Центр описанной окружности прямоугольного треугольника лежит на середине гипотенузы, а ее радиус равен половине длины гипотенузы: $R = \frac{c}{2}$ $R = \frac{41}{2}$ $R = 20.5 \, \text{см}$</p><p><strong>4.</strong> Определим координаты центров окружностей. Пусть вершины прямоугольного треугольника находятся в точках $A(0, a)$, $B(b, 0)$ и $C(0, 0)$ (с прямым углом в точке $C$). В нашем случае, $A(0, 9)$, $B(40, 0)$, $C(0, 0)$ (или $A(0, 40)$, $B(9, 0)$, $C(0, 0)$ - это не повлияет на результат, так как оси симметричны). Центр вписанной окружности $O_i$ имеет координаты $(r, r)$, так как он равноудален от катетов и находится внутри угла: $O_i = (4, 4)$ Центр описанной окружности $O_o$ является серединой гипотенузы. Если вершины гипотенузы $A(0, a)$ и $B(b, 0)$, то координаты $O_o$ будут: $O_o = \left( \frac{0 + b}{2}, \frac{a + 0}{2} \right)$ $O_o = \left( \frac{40}{2}, \frac{9}{2} \right)$ $O_o = (20, 4.5)$</p><p><strong>5.</strong> Найдем расстояние $d$ между центрами $O_i(x_1, y_1)$ и $O_o(x_2, y_2)$ по формуле расстояния между двумя точками: $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$ $d = \sqrt{(20 - 4)^2 + (4.5 - 4)^2}$ $d = \sqrt{(16)^2 + (0.5)^2}$ $d = \sqrt{256 + 0.25}$ $d = \sqrt{256.25}$</p><p>Представим $256.25$ в виде дроби: $256.25 = \frac{25625}{100} = \frac{1025}{4}$. $d = \sqrt{\frac{1025}{4}}$ $d = \frac{\sqrt{1025}}{\sqrt{4}}$ $d = \frac{\sqrt{25 \cdot 41}}{2}$ $d = \frac{5\sqrt{41}}{2}$ $d = 2.5\sqrt{41} \, \text{см}$</p><p><strong>Ответ</strong></p><p>Расстояние от центра вписанной окружности до центра описанной окружности составляет $2.5\sqrt{41} \, \text{см}$.</p>"
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№52 (с. 190)
Условие. №52 (с. 190)
скриншот условия
Геометрия, 9 класс Учебник, авторы: Солтан Г Н, Солтан Алла Евгеньевна, Жумадилова Аманбала Жумадиловна, издательство Кокшетау, Алматы, 2019, страница 190, номер 52, Условие

52. В треугольнике $ABC$ $AC = 2$ дм, $\angle A = 30^\circ$, $\angle C = 45^\circ$. Найдите высоту треугольника, проведенную к стороне $AC$.

$((\sqrt{3} - 1)$ дм)

Решение. №52 (с. 190)
Геометрия, 9 класс Учебник, авторы: Солтан Г Н, Солтан Алла Евгеньевна, Жумадилова Аманбала Жумадиловна, издательство Кокшетау, Алматы, 2019, страница 190, номер 52, Решение
Решение 2. №52 (с. 190)

Дано:

треугольник $ABC$;

$AC = 2$ дм;

$\angle A = 30^\circ$;

$\angle C = 45^\circ$.

Перевод в СИ:

$AC = 2$ дм $ = 0.2$ м.

Найти:

высоту $BH$, проведенную к стороне $AC$.

Решение:

Пусть $BH$ — высота треугольника $ABC$, проведенная из вершины $B$ к стороне $AC$. Точка $H$ лежит на отрезке $AC$, так как углы $A$ и $C$ острые (меньше $90^\circ$).

Высота $BH$ перпендикулярна стороне $AC$, поэтому треугольники $ABH$ и $CBH$ являются прямоугольными.

Рассмотрим прямоугольный треугольник $ABH$:

$\angle A = 30^\circ$.

Тангенс угла $A$ определяется как отношение противолежащего катета $BH$ к прилежащему катету $AH$:

$\tan A = \frac{BH}{AH}$

Отсюда, $AH = \frac{BH}{\tan A} = \frac{BH}{\tan 30^\circ}$.

Известно, что $\tan 30^\circ = \frac{1}{\sqrt{3}}$.

Следовательно, $AH = \frac{BH}{1/\sqrt{3}} = BH\sqrt{3}$.

Рассмотрим прямоугольный треугольник $CBH$:

$\angle C = 45^\circ$.

Тангенс угла $C$ определяется как отношение противолежащего катета $BH$ к прилежащему катету $HC$:

$\tan C = \frac{BH}{HC}$

Отсюда, $HC = \frac{BH}{\tan C} = \frac{BH}{\tan 45^\circ}$.

Известно, что $\tan 45^\circ = 1$.

Следовательно, $HC = \frac{BH}{1} = BH$.

Длина стороны $AC$ равна сумме длин отрезков $AH$ и $HC$:

$AC = AH + HC$

Подставим найденные выражения для $AH$ и $HC$:

$AC = BH\sqrt{3} + BH$

$AC = BH(\sqrt{3} + 1)$

Выразим высоту $BH$:

$BH = \frac{AC}{\sqrt{3} + 1}$

Подставим значение $AC = 2$ дм:

$BH = \frac{2}{\sqrt{3} + 1}$

Чтобы избавиться от иррациональности в знаменателе, умножим числитель и знаменатель на сопряженное выражение $(\sqrt{3} - 1)$:

$BH = \frac{2}{(\sqrt{3} + 1)} \cdot \frac{(\sqrt{3} - 1)}{(\sqrt{3} - 1)}$

Используем формулу разности квадратов $(a+b)(a-b) = a^2 - b^2$:

$BH = \frac{2(\sqrt{3} - 1)}{(\sqrt{3})^2 - (1)^2}$

$BH = \frac{2(\sqrt{3} - 1)}{3 - 1}$

$BH = \frac{2(\sqrt{3} - 1)}{2}$

$BH = \sqrt{3} - 1$

Единица измерения — дециметры (дм).

Ответ:

Высота треугольника, проведенная к стороне $AC$, равна $(\sqrt{3} - 1)$ дм.

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