Предмет: Геометрия, автор: Dимасuk

В окружность вписан четырехугольник ABCD. Найдите его площадь, если известно,
AB·CD = 10, BC·AD = 15, а угол между диагоналями прямой.


Аноним: 25/2
Dимасuk: нет)
Аноним: как это?
Dимасuk: Вот так вот :)
Аноним: по теореме Птолемея d1*d2=25
Аноним: площадь S=1/2*d1*d2*sin90
Аноним: что там не так?

Ответы

Автор ответа: siestarjoki
2
Произведение диагоналей вписанного четырёхугольника равно сумме произведений противоположных сторон (теорема Птолмея).
p1p2= 
AB·CD+BC·AD = 10+15 =25

Площадь 
четырёхугольника равна полупроизведению диагоналей на синус угла между ними.
S= 0,5*p1p2*sin(90) = 0,5*25 =12,5


Dимасuk: Спасибо)
Аноним: теорему было бы написать)
Автор ответа: Аноним
2
Используя теорему Птолемея AB\cdot CD+BC\cdot AD=AC\cdot BD, получим 10+15=AC\cdot BD  откуда   AC\cdot BD=25

Площадь четырёхугольника вычисляется по формуле S=0.5\cdot d_1\cdot d_2\sin \alpha , где d1 и d2 - диагонали, sin a - синус угла между диагоналями

S=0.5\cdot AC\cdot BD\cdot \sin90а=0.5\cdot25=12.5

Ответ: 12,5.
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