Question

In trapezul ABCD,m(A)=90,AB//CD,AB>CD,iar bazele AB si CD sunt proportionale cu numerele 4 si 6.Stiind ca AC e perpendicular pe BC,iar AD=12 radical din 2,calculati:
a)lungimile bazelor AB si CD
b)aria trapezului ABCD
c)lungimile diagonalelor trapezului AC si BD

Answer 1

(a)AB,CD)~(4,6)
AB = =K
AB supra 4=k => AB= 4k
CD supra 6 =K -> CD=6k
In Δ ACB-m(B)= 90grade => (prin teorema inaltimii 1) CE²=AE XEB
(12√2)² =4k² x 2k² 
288=8k²
K²=36 => k= √36=6cm
AE=4k => AE=24 cm
EB=6k => EB=36 cm
AB=AE+EB=24 +36= 60 cm 
DC=AE= 24 cm 
b)A= AB =DC supra 2 x CE =  x 12√2 = 42 x 12√2 =504√2 cm²

c)I n ΔACE- m(E)= 90 grade => (ptin PT)  CE²+AE²=AC ²
(12√2)² + 24² = AC²
288+576=AC²
864=AC² =>AC=12√6
In Δ DAB- m(A)= 90 grade => AD² +AB²=BD² 
12√2² +  60²=BD ²
288+3600=BD²
3888=BD² => BD=36√3 cm