Вопрос школьника по предмету Геометрия
abcda1b1c1d1 прямоугольный параллелепипед b1c 20 b1a 13 ad-ab 11 найти aa1
Ответ учителя по предмету Геометрия
ad-ab =11
ab=x ad=x+11
ad=cb
aa1=cc1
aa1^2=b1a^2 — ab^2= 13^2-x^2
cc1^2=b1c^2 — cb^2 = b1c^2 — ad^2 = 20^2-(x+11)^2
169-x^2 = 400-(x^2+22x+121)
22x+121+169-400=0
22x=110
x=5
aa1^2 = 13^2-x^2=169 -25 = 144
aa1=12
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