pole trapezu Akita : Oblicz pole trapezu ABCD o podstawach AB i CD, jeśli AB = 27, BC=20, CD = 6, AD = 13

M:

mat3: P= 198

Janek191: a = 27 b = 6 c = 13 d = 20 Mamy a = x + b + y = 27 x + 6 + y = 27 więc y = 21 − x h − wysokość trapezu Mamy ( 21 − x)² + h² = 20² x² + h² = 13² −−−−−−−−−−−−−−−−−−−−−−−− 441 − 42 x + x² + h² = 400 x² + h² = 169 −−−−−−−−−−−−−−− odejmujemy stronami 441 − 42 x = 231 42 x = 210 x = 5 5² + h² = 169 h² = 169 − 25 = 144 h =12 ==== P = 0,5*( 27 + 6 )*12 = 198 =====================