Dane: P = 55√8 cm² = 110√2 cm² c = 10 cm α = 30 a = ? [cm] b = ? [cm] a = b + 2x h/c = sinα /*c h = c*sinα h = (10 cm)*sin30 h = 0,5*(10 cm) h = 5 cm x/c = cosα /*c x = c*cosα x = (10 cm)*cos30 x = √3/2*(10 cm) x = 5√3 cm 0,5(a + b)*h = P /*2 (a + b)*h = 2P (a + b)*(5 cm) = (220√2 cm²) /:(5 cm) { a + b = 44√2 cm { a = b + 2x { a + b = 44√2 cm { a = b + 10√3 cm { a + b = 44√2 cm { a - b = 10√3 cm { 2a = 44√2 cm + 10√3 cm /:2 { b = a - 10√3 cm { a = 22√2 cm + 5√3 cm { b = a - 10√3 cm { a = 22√2 cm + 5√3 cm { b = 22√2 cm + 5√3 cm - 10√3 cm { a = 22√2 cm + 5√3 cm { b = 22√2 cm - 5√3 cm Odp: Podstawy trapezu mają długość: 22√2 cm + 5√3 cm i 22√2 cm - 5√3 cm.
a,b-dł. podstaw trapezu; b=a+2x; c-dł. ramienia trapezu; c=10cm; h-wys. trapezu; sin30°=h/10; 1/2=h/10; 2h=10/:2; h=5cm; 5²+x²=10²; x²=100-25=75; x=√75=5√3cm; P=55√8cm²; 55√8=1/2(a+b)×h; 55√8=1/2(a+a+2x)×h; 55√8=1/2(2a+2×5√3)×5; 55√8=1/2(2a+10√3)×5/×2; 110√8=(2a+10√3)×5; 110√8=10a+50√3; 10a=110√8-50√3; 10a=220√2-50√3; 10a=10(22√2-5√3)/:10; a=22√2-5√3; b=a+2x=22√2-5√3+2×5√3=22√2-5√3+10√3=22√2+5√3; Dł. podstaw trapezu: a=22√2-5√3 cm; b=22√2+5√3cm;
