Question

Номер 17 снизу, помогите пожалуйста!!!

Answer 1

Пусть CD = y; SΔABC = 0,5·AC·BC = 0,5·10·(6 + y) = 30 + 5y;

SΔDCA = 0,5·AC·DC = 0,5·10y = 5y; SΔDBE = 0,5·DB·BE·sinB =

[sinB = AC/AB = 10/5x] = 0,5·4x·6·sinB = 12x·10/5x = 12·2 = 24 ⇒

SΔADE = SΔABC - (SΔDBE + SΔDCA) = 30 + 5y - (5y + 24) = 30 - 24 = 6