obliczanie bokow trapezu dana a? b=6 h=3 pole trapezu=24

prosze o pomoc

a=10

P=0,5(a+b)/h 24=0,5(a+6)5 a+6=16 a=16-6=10

P=(a+b)*h/2 P=24 (a+b)*h/2=24 |*2 (a+6)*3=48 3a+18=48 3a=48-18 3a=30 |:3 a=10

p = a+b * h /2 24 = a + 6 * 3 /2 48 = a + 6 * 3 / : 3 16 = a + 6 a = 10