Предмет: Геометрия, автор: safinvalentijn

ДАЮ 50 БАЛЛОВ ГЕОМЕТРИЯ 10 КЛАСС
Найдите площадь боковой поверхности правильной треугольной усечённой пирамиды, стороны оснований которой равны 10 см и 18 см, а боковое ребро — 5 см.

Simba2017: это три одинаковые равнобедренные трапеции
Simba2017: высота каждой 3

Ответы

Автор ответа: ReMiDa
3

Ответ:

Площадь боковой поверхности правильной треугольной усеченной пирамиды равна 126 см²

Объяснение:

Боковая поверхность правильной треугольной усеченной пирамиды состоит из трёх равнобедренных трапеций ( т.к. основания усеченной пирамиды параллельны друг другу, а все боковые рёбра равны между собой)

  • Площадь трапеции равна полусумме её оснований умноженное на высоту.

S = \dfrac{a + b}{2} \times h

а = KL= 10 см, b = AB = 18 см, найдём высоту h.

Опустим перпендикуляры из вершины верхнего основания на нижнее. КН⟂АВ, LP⟂AB. Следовательно HKLP - прямоугольник. HP=KL=10 см, как стороны прямоугольника.

△АКН - прямоугольный (∠Н=90°), △BLP - прямоугольный (∠P=90°).

△AKH=△ BLP по гипотенузе и катету. (AK=LB - как рёбра правильной усеченной пирамиды; KH=LP - как высоты равнобедренной трапеции).

Следовательно АН=PB=(AB-KL):2=(18-10):2=4см

По теореме Пифагора найдём катет АН.

KH = \sqrt{ {AK}^{2} - {AH}^{2} } = \sqrt{ {5}^{2} - {4}^{2} } = \sqrt{ 25 - 16 } = \sqrt{9 } = 3 см

Площадь трапеции AKLB равна:

S = \dfrac{KL + AB}{2} \times KH = \dfrac{10 + 18}{2} \times 3 = 42 см²

Тогда площадь боковой поверхности правильной треугольной усеченной пирамиды равна:

Sб=3×S(AKLB)=3×42= 126 см²

#SPJ1

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