Найдите длину средней линии трапеции abcd с большим основанием ad.1)a(-1; 2), b(1; -2), c(2; 0), d(1; 6) 2)a(4; 0), b(-1; 0), c(-1; 3), d(4; 6) 3)a(-2; -2), b(3; 1), c(2.5; 2.5), d(-3; 1) 4)a(-3; 1), b(-2; -2), c(3; 1), d(7; 7)

Пусть m - это средняя линия трапеции. 1) ad^2 = (1 - (-1))^2 + (6 - 2)^2 = 2^2 + 4^2 = 4 + 16 = 20 ad = 2*sqrt(5) bc^2 = (2 - 1)^2 + (0 - (-2))^2 = 1^2 + 2^2 = 1 + 4 = 5 bc = sqrt(5) m = (ad + bc)/2 = (2sqrt(5) + sqrt(5))/2 = 3sqrt(5)/2 2) ad^2 = (4 - 4)^2 + (6 - 0)^2 = 0^2 + 6^2 = 36 ad = 6 bc^2 = (-1 - (-1))^2 + (3 - 0)^2 = 0^2 + 3^2 = 9 bc = 3 m = (ad + bc)/2 = (6 + 3)/2 = 9/2 3) ad^2 = (-))^2 + (1 - (-2))^2 = (-1)^2 + 3^2 = 1 + 9 = 10 ad = sqrt(10) bc^2 = (2,5 - 3)^2 + (2,5 - 1)^2 = 0,5^2 + 1,5^2 = 0,25 + 2,25 = 2,5 = 10/4 bc = sqrt(10) / 2 m = (ad + bc)/2 = (sqrt(10) + sqrt(10)/2) /2 = 3sqrt(10)/4 4) ad^2 = (7 - (-3))^2 + (7 - 1)^2 = 10^2 + 6^2 = 136 ad = 2sqrt(34) bc^2 = ())^2 + (1 - (-2))^2 = 5^2 + 3^2 = 25 + 9 = 34 bc = sqrt(34) m = (ad + bc)/2 = 3sqrt(34)/2 sqrt - это корень.

Вкубе 12 ребер.   12дм: 12=1дм