ответ:1056+1584√3 (см²)
Объяснение: 1)Пусть параллелограмм АВСД-нижнее основание призмы,А₁В₁С₁Д₁-верхнее основание; ∠А=30°, тогда ∠Д=180°-30°=150°. 2)Боковая поверхность призмы S= P·h, P= 2·(АД+СД)= 2( 16+24√3)=32+48√3. 3)Вычислим большую диагональ основания АС по теореме косинусов из ΔАДС: АС²= АД²+СД²- 2·АС·СД·CosД= 16²+(24√3)² - 2·16·24√3·Cos150°= 256+1728 - 2·16·24√3· (-Cos30°)=256+1728 + 2·16·24√3· √3/2 =256+1728 +1152=3136, ⇒АС = √3136= 56. 4)Рассмотрим прямоугольный треугольник АА₁С, по условию большая диагональ призмы А₁С=65 см.⇒h²= AA₁²= А₁С²- AC²65²-56²= 1089, h=√1089=33 (cм) 5) Боковая поверхность призмы S= P·h =(32+48√3) P= 2·(АД+СД)= 2( 16+24√3)=(32+48√3)· 33 =1056+1584√3 (см²)
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ответ:1056+1584√3 (см²)
Объяснение: 1)Пусть параллелограмм АВСД-нижнее основание призмы,А₁В₁С₁Д₁-верхнее основание; ∠А=30°, тогда ∠Д=180°-30°=150°. 2)Боковая поверхность призмы S= P·h, P= 2·(АД+СД)= 2( 16+24√3)=32+48√3. 3)Вычислим большую диагональ основания АС по теореме косинусов из ΔАДС: АС²= АД²+СД²- 2·АС·СД·CosД= 16²+(24√3)² - 2·16·24√3·Cos150°= 256+1728 - 2·16·24√3· (-Cos30°)=256+1728 + 2·16·24√3· √3/2 =256+1728 +1152=3136, ⇒АС = √3136= 56. 4)Рассмотрим прямоугольный треугольник АА₁С, по условию большая диагональ призмы А₁С=65 см.⇒h²= AA₁²= А₁С²- AC²65²-56²= 1089, h=√1089=33 (cм) 5) Боковая поверхность призмы S= P·h =(32+48√3) P= 2·(АД+СД)= 2( 16+24√3)=(32+48√3)· 33 =1056+1584√3 (см²)
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