thenoname1105
25.07.2022 06:52

Р -
2. Самостійна робота з подальшою взаємоперевіркою
Варіант 1
1) З точки кола проведено дві хорди, які дорівнюють його радіусу. Знай-
діть кут між ними.
2) Діаметр кола з центром у точці Одорівнює 8 см.
D
Знайдіть периметр трикутника АОС, якщо хорда
C с
AC дорівнює 5 см.
3) На рисунку ВС і DF — діаметри кола з центром
у точці О. Чи правильно, що:
а) АBОD — рівносторонній;
В
б) АBОD рівнобедрений;
в) АBОD = ДСОF? Відповідь обгрунтуйте.
Варiант 2
1) З точки кола проведено діаметр і хорду, яка дорівнює його радіусу.
Знайдіть кут між ними.
2) Діаметр кола з центром у точці 0 дорівнює 10 см.
Знайдіть периметр трикутника Мок, якщо хор- D
да мк дорівнює 8 см.
3) На рисунку ВС і BD — рівні хорди кола з цен-
0
тром у точці О. Чи правильно, що:
а) АBОD - рівнобедрений;
б) АBОD — рівносторонній;
в) АBОD = ДВОС? Відповідь обґрунтуйте.
В​


Р -2. Самостійна робота з подальшою взаємоперевіркоюВаріант 11) З точки кола проведено дві хорди, як

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Ответ:
liza1449
26.11.2022 16:40
Я думаю, задание надо читать так: В основании пирамиды лежит прямоугольник со сторонОЙ 6 см.Основанием высоты пирамиды является центр описанной окружности с радиусом 5 см.Найдите объем пирамиды, если ее высота равна 9 см. Тогда решение следующее:
Vпир.=1/3Sосн.*h (одна третья  площади основания пирамиды на высоту пирамиды).
Чтобы найти площадь основания, надо найти вторую сторону прямоугольника. По т. Пифагора АВ²=АС²-ВС² АС=d=2c=10см.
АВ²=100-36=64⇒АВ=√64=8см.
S осн.=АВ*ВС=6*8=48см²
Vпир.=1/3*Sосн*h=1/3*48*9=144cм³
Восновании пирамиды дежит прямоугольник со сторонами 6 см.основанием высоты пирамиды-центр описаной
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Ответ:
ooficerova
26.05.2023 14:21

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