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 = 5BC = sqrt(5)m = (AD + BC)/2 = (2sqrt(5) + sqrt(5))/2 = 3sqrt(5)/22) AD^2 = (4 - 4)^2 + (6 - 0)^2 = 0^2 + 6^2 = 36AD = 6BC^2 = (-1 - (-1))^2 + (3 - 0)^2 = 0^2 + 3^2 = 9BC = 3m = (AD + BC)/2 = (6 + 3)/2 = 9/23) AD^2 = (-3-(-2))^2 + (1 - (-2))^2 = (-1)^2 + 3^2 = 1 + 9 = 10AD = sqrt(10)BC^2 = (2,5 - 3)^2 + (2,5 - 1)^2 = 0,5^2 + 1,5^2 = 0,25 + 2,25 = 2,5 = 10/4BC = sqrt(10) / 2m = (AD + BC)/2 = (sqrt(10) + sqrt(10)/2) /2 = 3sqrt(10)/44) AD^2 = (7 - (-3))^2 + (7 - 1)^2 = 10^2 + 6^2 = 136AD = 2sqrt(34)BC^2 = (3-(-2))^2 + (1 - (-2))^2 = 5^2 + 3^2 = 25 + 9 = 34BC = sqrt(34)m = (AD + BC)/2 = 3sqrt(34)/2sqrt - это корень.