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С! площадь основания конуса равна 12 см2, а площадь боковой поверхности 13 см2. найдите площадь осевого сечения конуса.

Геометрия

Ответы

Roman343247
Основание конуса - круг. Площадь круга   
S = \pi R^{2} = 12 cm^{2} \\ R^{2} = \frac{12}{ \pi } \\ R = \sqrt{ \frac{12}{ \pi } }
Площадь боковой поверхности 

S = \pi Rl = 13 cm^{2} \\ l = \frac{13}{ \pi R}   - образующая конуса.
l = \frac{13}{ \pi R}= \frac{13}{ \pi }* \sqrt{ \frac{ \pi }{12} } = 13\sqrt{ \frac{\pi }{ 12\pi ^{2} } } = \\ = 13 \sqrt{ \frac{1}{12 \pi } }

Осевое сечение - равнобедренный треугольник, В основании - диаметр. Высоту h можно найти из прямоугольного треугольника: h - вертикальный катет, R - горизонтальный катет, образующая l - гипотенуза.
h = \sqrt{ l^{2} - R^{2} } = \sqrt{ [13 \sqrt{ \frac{1}{12 \pi } } ] ^{2} - \sqrt{ \frac{12}{ \pi } }^{2} } = \sqrt{ \frac{169}{12 \pi} - \frac{12}{ \pi } }= \\ = \sqrt{ \frac{169-144}{12 \pi } } = \sqrt{ \frac{25}{12 \pi } }

Площадь сечения
S = \frac{1}{2} * (2R) * h = Rh = \sqrt{ \frac{12}{ \pi } } * \sqrt{\frac{25}{12 \pi } } = \\ = \sqrt{\frac{25}{\pi^2 } } = \frac{5}{ \pi }

Площадь осевого сечения конуса S = \frac{5}{ \pi }
sorokinae

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igor8809337
S=30*4=120
Р=(30+4)*2=68
пусть уменьшенная длина будет 30-у
уменьшенная ширина 4-х
новая площадь должна равняться 120/2
новый периметр 68-22=46
полупериметр 46/2=23
составим систему с 2-мя неизвестными:

(30-у)(4-х)=120/2
(30-у)+(4-х)=46/2

(30-у)(4-х)=60
30-у+4-х=23

(30-у)(4-х)=60
х+у=11

(30-у)(4-х)=60      (1)
х=11-у                  (2)

подставляем наш х в (1)
получаем
(30-у)(4-х(11-у))=60
(30-у)(у-7)=60
30у-210-у²+7у-60=0
-у²+37у-270=0
Д=37²-4(-1)(-270)=1369-1080=289=17²
у1=-27   нам не подходит т.к. сторона не может быть отрицательной
у2=10

подставляем в (2)
х=11-у=11-10=1
 
ширину надо уменьшить на 10 см, длину на 1 см

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С! площадь основания конуса равна 12 см2, а площадь боковой поверхности 13 см2. найдите площадь осевого сечения конуса.
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