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4) In trapezul  ABCD,m(<A)=m(<D)=90,ABllCD,AB>CD,iar bazele CD si AB sunt proportionale cu numerele 4 si 6,Stiind ca AC(perpendicular)BC,iar AD=12√2cm, Calculati:
a) lungimile bazelor [AB] si [CD]
b) aria trapezului ABCD.
c) lungimile diagonalelor trapezului,[AC] si [BD].

Răspuns :

AndraTrifanid
(a)AB,CD)~(4,6)
AB[tex] \frac{AB}{4} [/tex] = [tex] \frac{CD}{6} [/tex]=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 = [tex] \frac{60+24}{2} [/tex] 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    


Miky93id
ABCD-trapez
mas<A=mas<D=90
AC_|_BC     =>mas<ACB=90

AB;DC    dp cu 4;6

AB/4=DC/6=k
AB=4k
DC=6k

ΔABC, mas<C=90
fie CM _|_AB
CM-h
AM=DC=4k

AC²=AM*AB
AC²= 4k* 6k=24k²    =>AC= 2√6 k

ΔADC, mas<D=90

AD²=AC²- DC²

(12√2)²= (2√6 k)² + (4k)²

288= 24k² +16k²

288= 40k²

k²= 288/40=>   k= √288/10= 12√2/√10=  12√20/10= 6√20/5=  12√5/5

DC= 4 * 12√5/5=  48√5/5

AB= 6 * 12√5/5= 72√5/5

AC = 2√6 * 12√5/5=  24√30/5

A ABCD = (DC+AB)*AD/2 =  (48√5/5 + 72√5/5 ) * 12√2/2=

= 120√5/5 * 6√2=  24√5 * 6√2=  144√10

ΔADB, mas<A=90 

DB²=AD² +AB²

DB² = (12√2)² + (72√5/5)²

DB² = 288 + 25920/25

DB²= (7200+25 920)/25

DB² = 33120/25

DB = √33120/25=  12√230/5